calculus made ridiculusly easy

VCEcoverageArea of studyUnits 3 & 4 • Calculus

5In this chapter5A The derivative of tan kx

5B Second derivatives

5C Analysing the behaviour of functions using the second derivative

5D Derivatives of inverse circular functions

5E Antidifferentiation involving inverse circular functions

Differentialcalculus

Chap 05 SM 71 Thursday, October 12, 2000 10:22 AM

172 S p e c i a l i s t M a t h e m a t i c s

IntroductionIn this chapter we extend the functions which can be differentiated to includef(x) = tan kx and the inverse circular functions Sin−1x, Cos−1x and Tan−1x. It is assumedthat the standard results for the differential calculus are familiar to the student as are theproduct rule, quotient rule and chain rule. These results are listed in the table below.

The derivative of tan kxSince tan , its derivative can be found using the quotient rule:

If f(x) = tan kx,

=

where k is a constant, then using the quotient rule with u = sin kx and v = cos kx:

f ′(x) =

=

= (by factorising the numerator)

= (since cos2kx + sin2kx = 1)

= k sec2kx

If f(x) = tan kx,then f ′(x) = k sec2 kx.

f(x) f′(x)

axn anxn – 1

loge kx

ekx kekx

sin kx k cos kx

cos kx –k sin kx

u(x) × v(x) u′(x) × v(x) + u(x) × v′(x) (product rule)

u′(x) × v(x) - u(x) × (quotient rule)

g[h(x)] h′(x) × g′[h(x)] (chain rule)

1x---

u x( )v x( )----------- v′ x( )

v2------------

kxsin kxcos kx---------------=

ddx------ u

v---

v

dudx------ u

dvdx------–

v2-------------------------=

sin kxcos kx---------------

k cos kx cos kx( ) sin kx k sin kx–( )–cos2kx

-----------------------------------------------------------------------------------------

k cos2kx k sin2kx+cos2kx

------------------------------------------------

k cos2kx sin2kx+( )cos2kx

----------------------------------------------

k 1( )cos2kx----------------

Chap 05 SM Page 172 Thursday, October 12, 2000 10:22 AM

C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s 173

If y = tan [f(x)]

then = f ′(x) sec2 [f(x)].

Differentiate the following expressions with respect to x.

a y = tan 6x b y = 2 tan THINK WRITEa Differentiate by rule: y = tan kx,

then = ksec2kx where k = 6.

a y = tan 6x

= 6 sec26x

b Differentiate tan by rule where

k = .

b y = 2 tan

=

Multiply the derivative by 2. =

4x3

------

dydx------ dy

dx------

14x3

------

43---

4x3

------

dydx------ 2

43--- sec2 4x

3------

2dydx------ 8

3--- sec2 4x

3------

1WORKEDExample

If f(x) = tan (x2 + 5x), find f ′(x).THINK WRITE

Let u = x2 + 5x to apply the chain rule. f(x) = tan(x2 + 5x)Let u = x2 + 5x.

Find . = 2x + 5

Express f(x) in terms of u. So f(x) = y = tan u.

Find . = sec2u

Find using the chain rule. =

= (2x + 5) sec2uReplace u with the expression x2 + 5x. f ′(x) = (2x + 5) sec2(x2 + 5x)

1

2dudx------ du

dx------

3

4dydu------ dy

du------

5dydx------ dy

dx------ dy

du------ du

dx------×

6

2WORKEDExample

d ydx------

Find if y = loge [tan (−3x)].

THINK WRITELet u = tan (−3x) and consequently apply the chain rule.

y = loge [tan(−3x)]Let u = tan(−3x).

Find . = −3 sec2(−3x)

d ydx------

1

2dudx------ du

dx------

3WORKEDExample

Continued over page



THINK WRITE

Express y in terms of u. y = logeu

Find . =

Find using the chain rule. So = ×

= × −3 sec2 (–3x)

Replace u with the expression tan (−3x). =

Express in terms of sin (−3x) and

cos (−3x) only in order to simplify the rational expression.

= ÷

Express the division of rational expressions as a multiplication.

= ×

Cancel out the factor of cos (−3x). =

Use the double angle formula to simplify sin (−3x) cos (−3x).

=

State the answer in simplest form. and hence = – or −6 cosec(−6x).

3

4dydu------ dy

du------ 1

u---

5dydx------ dy

dx------ dy

du------ du

dx------

1u---

63sec2 3x–( )–

3x–( )tan-------------------------------

7dydx------ 3–

cos2 3x–( )------------------------- 3x–( )sin

3x–( )cos-----------------------

83–

cos2 3x–( )------------------------- 3x–( )cos

3x–( )sin-----------------------

93–

3x–( )sin 3x–( )cos---------------------------------------------

10 3–12---sin 6x–( )------------------------

11dydx------ 6

6x–( )sin----------------------

Find the equation of the tangent to the curve y = 3x + cos 2x + tan x where x = .

THINK WRITE

To find the equation of a tangent line to a curve at a point, the coordinate of the point is needed as is the gradient of the curve at that point. Find y when x = to establish the coordinate.

y = 3x + cos 2x + tan x

If x = , y = + cos + tan

= + 0 + 1

= 1 +

The coordinate is ( , 1 + )Find to establish the gradient

function.

= 3 − 2 sin 2x + sec2x

π4---

1

π4---

π4--- 3π

4------ π

2--- π

4---

3π4

------

3π4

------

π4--- 3π

4------

2dydx------ dy

dx------

4WORKEDExample



The derivative of tan kx

1 Differentiate each of the following with respect to x.a y = tan 4x b y = tan 5x c y = 3 tan 7xd y = 4 tan 2x e y = tan(−3x) f y = tan(−12x)g y = 3 tan(−4x) h y = −2 tan 3x i y = −5 tan(−2x)

j y = tan k y = tan l y = tan

m y = tan n y = 8 tan o y = 6 tan

p y = −3 tan q y = 5 tan r y = −4 tan

THINK WRITE

Evaluate when x = to find the

gradient of the tangent at x = .

If x = , = 3 − 2 sin + sec2

= 3 − 2 +

= 1 +

= 3so the equation of the tangent is:

Substitute y1 = , x1 = and

m = 3 into the equation for a straight line: y − y1 = m(x − x1) where m is the gradient and (x1, y1) is a point on the line.

y − (1 + ) = 3(x − )

Simplify the equation and make y the subject.

y = 3x − + 1 +

y = 3x + 1

3dydx------ π

4---

π4---

π4--- dy

dx------ π

2--- π

4---

1

cos2 π4---

---------------

1

(12---)

---------

4 13π4

------+ π4--- 3π

4------ π

4---

53π4

------ 3π4

------

remember1. If y = tan kx then = k sec2 kx.

2. If y = tan f(x) then = f ′(x) sec2 [f (x)].

dydx------

dydx------

remember

5AWORKEDExample

1

Mathcad

Differentiator

x5--- x

2--- 3x

5------

2x7

------ 3x4

------ 4x9

------

5x6

------ 8x5

------ 7x2

------



2 For each of the following find f ′(x) if f(x) equals:

3If y = etan 2x, then:

a can be found by using:

b is equal to:

4 For each of the following find if y equals:

5 For each of the following find f ′(x) if f (x) equals

6 If f: [0, ] → R, f (x) = tan x, find the coordinates of the points on the graph wherethe gradient is equal to:

7 If f: [− , ] → R, f(x) = 4 tan , find the coordinates of the points on the graphwhere the gradient is equal to:

8 Find the equation of the tangent to the curve with equation y = 3 tan 2x at x = .

9 If f(x) = x tan x find the equation of i the tangent and ii the normal at the point on thecurve where x = .

10 Show that there are no stationary points for the graph of y = tan x for all values of x.

11 Explain why the gradient of tan x is always positive.

12 If f:[− , ] → R, f (x) = tan x − x:

a find any stationary points and state their natureb if the domain is changed to R, show that the gradient can never be negative.

a tan(x2 + 3x) b tan(x + 2) c tan(5x − 4)d tan(2x2 − 3x) e tan(3x + 2) f tan 8xg tan(7 − 4x) h tan(1 − 5x2)

A a direct rule B the product rule C the chain ruleD the quotient rule E graphical methods

A 2 sec22xetan 2x B 2 sec2xetan 2x C 4 sec2xetan 2x

D 2 tan2xesec 2x E ex sec22x

a loge(tan 6x) b etan 3x

c tan(e3x) d sin(tan 5x)e cos(tan 2x) f (tan 5x)g sin 4x − tan 3x h loge2x2 + 4 tani tan2x j tan42x

a x2 tan 3x b cos 2x tan c (5x3 − 6x) tan 4x

d e4xtan(−3x) e e4x2tan 8x f

g h

a 1 b c 4 d 10

a 4 b 2 c 0

WWORKEDORKEDEExamplexample

2

WORKEDExample

2

mmultiple choiceultiple choice

dydx------

dydx------

WORKEDExample

3

dydx------

12---

6x7

------

4x5

------

tan x2x2-----------

4x3

tan 2x--------------

sin 4xtan 4x--------------

π

43---

π π x2---

WORKEDExample

4

π6---

π4---

π2--- π

2---



Second derivativesSuppose that y = f (x).

Then the derivative, also known as the first derivative, is written as f ′(x) or .

By differentiating a second time, the second derivative, f ′′(x) or is obtained.

The process of obtaining f ′′(x) or from y = f (x) is also called double differen-tiation.

The second derivative is commonly called ‘f double dash’ or ‘d squared y, dx squared’.

dydx------

d2y

dx2--------

d2y

dx2--------

Find if y = + 2x4 − x−1.

THINK WRITE

Express as . y = + 2x4 − x−1

y = + 2x4 − x−1

Find the first derivative . = + 8x3 + x−2

Differentiate to obtain . = − + 24x2 − 2x−3

or − + 24x2 −

d2 ydx2--------- x

1 x x12--- x

x12---

2dydx------

dydx------ 1

2--- x

12---–

3dydx------ d2y

dx2-------- d2y

dx2-------- 1

4--- x

32---–

1

4x32---

-------- 2x3-----

5WORKEDExample

Find i f ′(x) and ii f ′′(x) if f (x) is equal to:

a ecos 2x + logex b .

THINK WRITE

a i Differentiate ecos 2x by the chain rule short cut and loge x by rule to obtain f ′(x).

a i f(x) = ecos 2x + logexf ′(x) = −2 sin 2x ecos 2x +

ii Express in index notation

so it can be differentiated.

ii f ′(x) = −2 sin 2x ecos 2x + x−1

Differentiate f ′(x) to obtain f ′′(x). Use the product rule to differentiate −2 sin 2x ecos 2x.

f ′′(x) = −4 cos 2x ecos 2x + 4 sin2 2x ecos 2x −x−2

or = 4ecos 2x (sin2 2x − cos 2x) −

sin x

x-----------

1x---

11x---

21x2-----

6WORKEDExample

Continued over page



THINK WRITE

b i Express in index notation. b i f(x) =

f(x) =

Express f(x) as a product. = sin x

Differentiate f(x) using the product rule to obtain f ′(x).

f ′(x) = − sin x + cos x

ii Differentiate f ′(x) using the product rule to obtain f ′′(x).

ii f ′′(x) = x sin x − x cos x

− x cos x − x sin x

Simplify by collecting like terms. = x sin x − x cos x − x sin x

Simplify f ′′(x) by taking out

the factor x .

= x (3 sin x − 4x cos x − 4x2 sin x)

or

1 xsin x

x-----------

sin x

x12---

-----------

2 x12---–

312--- x

32---–

x12---–

134---

52---– 1

2---

32---–

12---

32---– 1

2---–

2 34---

52---– 3

2---– 1

2---–

3

14---

52---–

14---

–52---

3 sin x 4x cos x– 4x2 sin x–

4x52---

--------------------------------------------------------------------

If y = and , find the value of k.

THINK WRITE

Find from y = . y =

=

Find . =

Substitute and into the equation

− 3 + 2y = 0.

so

Take out as a factor. (k2 − 6k + 8) = 0

Factorise the quadratic function of k.

(Note: cannot equal zero.)

(k − 4)(k − 2) = 0

State the solutions. Therefore k = 2 or k = 4.

ekx2

------ d2 ydx2--------- 3

d ydx------– 2 y+ 0=

1dydx------ e

kx2-----

ekx2-----

dydx------ ke

kx2-----

2----------

2d2ydx2-------- d2y

dx2-------- k2e

kx2-----

4------------

3dydx------ d2y

dx2--------

d2ydx2-------- dy

dx------

k2ekx2-----

4------------ 3ke

kx2-----

2-------------– 2e

kx2-----

+ 0=

414---e

kx2----- 1

4---e

kx2-----

5

14---e

kx2-----

14---e

kx2-----

6

7WORKEDExample



Second derivatives

1 Find if

2

If y = sin , then:

a is equal to:

b is equal to:

3 Find i f ′(x) and ii f ′′(x) if f(x) is equal to:

a y = 4x2 + 7x − 3 b y = 5x3 − x2 + 3x c y = 6x

d y = x4 + 2x3 − 3x + 1 e y = x6 + 2x4 − 3x f y =

g y = 4x3 + h y = + 5 i y = x2 + 5x−3

j y = k y = 2x−2 + x−1

A cos B C

D sin E cos

A x cos B C

D sin E

a loge2x b x2 − logex c sin 2x + 4 cos x

d 2 sin e 2e5x − 3ex + 1 f 4e−3x + 3x3

g tan 2x h −3 tan (−4x) + 1 i

j k e2 sin x l cos e5x

m n o loge(cos x)

remember1. is the second derivative of y with respect to x. It is found by differentiating

y twice.2. If y = f(x) then = f ′′(x).

d2y

dx2--------

d2y

dx2--------

remember

5BWORKEDExample

5

d2ydx2-------- Mathcad

Secondderivatives

x72---

x32---

+

2 x 1– 2x92---

6x53---

–2

x------- 3x 1–+


xdydx------

x xcos x

2 x--------------- –cos x

x------------------

1

2 x---------- 1

2 x----------

d2ydx2--------

xx sin x cos x–

x x------------------------------------------- x sin x cos x–

2x x-------------------------------------------

1

4x x-------------–

cos – x x sin x–

4x x--------------------------------------------------

WORKEDExample

6x2--- x

43---

+cos 2x

x---------------

x52---

sin 2x--------------

sin 2x cos 4x( )52---



4 For each of the following functions f(x), match the graph which could represent f ′′(x).

5 If y = cos x, show that .

6 If y = sin 3x, show that: + 9y = 0.

7 If y = x loge x, show that: .

8 If y = xex, show that: .

9 Find the value of k if y = e−kx and .

10 If y = ekx and , find the value of k.

11 The position of a particle travelling in a straight line is given by the equation:x(t) = t3 − t2 − t + 7, where x has units in cm and t is in seconds. Find:

a , that is, the velocity at any time t

b , that is, the acceleration at any time t

c when and where the particle momentarily stops; that is, when = 0d the minimum velocity.

a f (x) = x3 + 2x b f (x) = 4x2 c f (x) = x4 − x2

d f (x) = sin x e f (x) = logex f f (x) = e2x

g f (x) = A B C

D E F

G H I

4 x

x0f"(x)

y

x0f"(x)

y

x0

f"(x)

y

x0f"(x)

y

x0

f"(x)

y

x0

f"(x)

y

x0

f"(x)y

x0

f"(x)

y

x0

f"(x)

y

d2ydx2-------- y–=

d2ydx2--------

dydx------ x y+( ) d2y

dx2--------×=

d2ydx2-------- dy

dx------ y

x--+=

WORKEDExample

7

d2ydx2-------- 25y=

d2ydx2-------- 3

dydx------– 4y– 0=

dxdt------

d2xdt2--------

dxdt------



Analysing the behaviour of functions using the second derivative

The second derivative of a function can be used for testing the nature of stationarypoints, but is not a requirement of this course. It is provided as an alternative, orextension, to the first derivative test which is a requirement to Mathematical MethodsUnits 3 and 4.

First derivative functionWe have seen how the first derivative of a function, f ′(x) or , can tell us where a

function has a positive gradient, a negative gradient or a zero gradient (stationary point).For example, let us look at the functions f(x) = x2 and f(x) = x3.

Function f(x) = x2

Examine the graph at right. Since f (x) = x2,f ′(x) = 2x

and f ′(x) = 0 at x = 0

1. If f ′(x) = 0At x = 0, f (x) is a stationary point.

2. If f ′(x) < 0This occurs when 2x < 0. So if x < 0, then f (x) isa decreasing function, one with a negativegradient for all x < 0.

3. If f ′(x) > 0This occurs when 2x > 0. So if x > 0, then f (x) is an increasing function, one with apositive gradient for all x > 0.Consequently, at x = 0, a minimum stationary point occurs.

Function f(x) = x3

Examine the graph at right. Since f (x) = x3,f ′(x) = 3x2

and f ′(x) = 0 at x = 0

1. If f ′(x) = 0At x = 0, f (x) is a stationary point.

2. Here, f ′(x) > 0 for all values of x except zero.Consequently, at x = 0, a stationary point of inflectionoccurs.

Second derivative functionSimilarly the second derivative (f ′′(x) or ) can tell us where the gradient function

( f ′(x) or ) is increasing or decreasing or neither (that is, changing from increasing

to decreasing or vice-versa). Let us look at the situation when f ′′(x) is greater than, lessthan and equal to zero.

dydx------

x0

x > 0f (x) is increasing

x < 0f (x) is decreasing

x = 0 f (x) is neitherincreasing nor decreasing

f (x) = x2y

x0

x > 0f(x) isincreasing

x < 0f(x) is increasing

x = 0f(x) is neitherincreasing nor decreasing

f(x) = x3y

d2ydx2--------

dydx------



Function f ′′(x) > 0Examine the graph at right. When f ′′(x) > 0 then thegradient function f (x) is increasing. That is, as xincreases then f ′(x) increases.Again consider: f (x) = x2

f ′(x) = 2xf ′′(x) = 2

Function f ′′(x) < 0Examine the graph at right. When f ′′(x) < 0, then thegradient of f(x) is decreasing. That is, as x increasesthen f ′(x) decreases.

For example, iff(x) = −x2

f ′(x) = −2x (therefore there is a stationary point at x = 0)

f ′′(x) = −2that is, f ′′(x) < 0 for all values of x.

This means that the gradient function f ′(x) isalways decreasing.

Function f ′′(x) = 0The gradient of f (x) is changing from increasing todecreasing or vice-versa (providing the sign of f ′′(x)changes at the point where f ′′(x) = 0). Again consider:

f(x) = x3

f ′(x) = 3x2 (therefore there is a stationary pointat x = 0)

f ′′(x) = 6xso f ′′(x) = 0 at x = 0

and f ′′(x) < 0 if x < 0,f ′′(x) > 0 if x > 0.

Therefore the gradient is decreasing left of x = 0 and increasing right of x = 0.The point (0, 0) is called a point of inflection.As f ′(0) = 0, the point (0, 0) is also a stationary point, so it is called a stationary

point of inflection.

Points of inflectionPoints of inflection occur when the second derivative changes sign.

A tangent drawn at a point of inflection crosses the graph at that point (see figure 1).Sometimes points of inflection are also stationary points (see figure 2).

x0

f (x) = x2

f ' (3) = 6f ' (–3) = 6

f ' (2) = 4f ' (–2) = –4

f ' (–1) = –2 f ' (1) = 2

y

1–1–2–3 2 3

Gradient is always increasingfrom left to right

x0

f(x) = –x2

f ' (3) = –6f ' (–3) = 6

f ' (2) = –4

f ' (1) = –2f ' (–1) = 2

f ' (–2) = 4

f ' (0) = 0y

1–1–2–3 2 3

Gradient is always decreasingfrom left to right

x < 0f ' (x) is decreasing

x > 0f ' (x) is increasing

f (x) = x3

x

yf ' (2) = 12

f ' (–2) = 12

f ' (–1) = 3 f ' (1) = 3

0 1 2–2 –1

Point ofinflection[ f"(x) = 0]

Tang

ent

Tangent

f"(x) < 0

f"(x) 0<

Stationary points ofinflection

[f " (x) = 0 and f ' (x) = 0]

Tangent Tangent

f "(x) < 0

f " (x) 0<

Figure 1 Figure 2


C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s 183The second derivative can tell us the nature of any stationary points or where

points of inflection occur on a graph. In general:

1. If f ′(a) = 0 and f ′′(a) > 0, then a local minimum stationary point occurs at x = a.2. If f ′(a) = 0 and f ′′(a) < 0, then a local maximum stationary point occurs at x = a.3. If f ′(a) = 0 and f ′′(a) = 0, and f ′′(x) changes sign at x = a, then a stationary point

of inflection occurs at x = a.4. If f ′′(a) = 0 and f ′′(x) changes sign at x = a, then a point of inflection occurs at

x = a.

Notes:1. Types of stationary points (1, 2, 3 above) can also be determined using the first

derivative test either side of x = a.2. The second derivative test is usually more efficient than the first derivative test in

determining maximum or minimum stationary points, but not for stationary points ofinflection.

3. Displaying a curve on a graphics calculator will assist in determining the nature ofany stationary points.

a Find any stationary points, and their nature, if f (x) = x2(x − 1)(x + 1).b Sketch the graph of f(x), clearly indicating all stationary points and axes intercepts.THINK WRITEa Expand f (x) so it can be differentiated

easily.a f(x) = x2(x − 1)(x + 1)

f(x) = x2(x2 − 1)= x4 − x2

Find f ′(x). f ′(x) = 4x3 − 2xFind f ′′(x). f ′′(x) = 12x2 − 2Solve for x where f ′(x) = 0. For stationary points, f ′(x) = 0

2x(2x2 − 1) = 0x = 0 or x2 =

x = 0, – or

Find f (0) and f ′′(0) to determine one stationary point.

At x = 0, f (0) = 0, f ′(0) = 0 and f ′′(0) = −2.Therefore (0, 0) is a local maximum stationary point.

Find f (− ) and f ′′(− ) for the second stationary point.

At x = – , f (– ) = − = –

and f ′′(– ) = 6 − 2

= 4Therefore (– , − ) is a local minimum

stationary point.

Find f ( ) and f ′′( ) for the third stationary point.

At x = , f ( ) = − = −and f ′′( ) = 6 − 2

= 4Therefore ( , − ) is a local minimum

stationary point.

1

234

12---

1

2------- 1

2-------

5

61

2------- 1

2-------

1

2------- 1

2------- 1

4--- 1

2--- 1

4---

1

2-------

1

2------- 1

4---

71

2------- 1

2-------

1

2------- 1

2------- 1

4--- 1

2--- 1

4---

1

2-------

1

2------- 1

4---

8WORKEDExample

Continued over page



THINK WRITE

b Evaluate f (0) for the y-intercept. b f(0) = 0 so the y-intercept is 0.

Solve the factorised form of f (x) = 0 for the x-intercepts.

f(x) = x2(x − 1)(x + 1) = 0has solutions x = 0, 1, −1so the x-intercepts are −1, 0 and 1.

Sketch the graph of f (x).Check the graph using a graphics calculator.

1

2

3

x0

f (x) = x2(x – 1)(x + 1)y

1–2

1–4

( , – )(– , – )1–2

1–4

–1

1

1–1

4

If y = x4 − x3 + 2 find:

a any stationary points of inflection b any other points of inflection.

THINK WRITE

a Find . a y = x4 − x3 + 2

= x3 − 3x2

Solve for x where = 0 (that is, the

stationary points).

For stationary points, = 0.

x3 − 3x2 = 0x2(x − 3) = 0

x = 0 or x = 3

Find . = 3x2 − 6x

Evaluate where x = 0. At x = 0, = 3(0)2 − 6(0)

= 0

Find y where x = 0. and y = (0)4 − (0)3 + 2

= 2

Evaluate where x = 3. At x = 3, = 3(3)2 − 6(3)

= 27 − 18= 9

State any stationary points of inflection. Therefore the point (0, 2) is a stationary point of inflection.

Check using a graphics calculator.

14---

1dydx------ 1

4---

dydx------

2dydx------ dy

dx------

3d2ydx2-------- d2y

dx2--------

4d2ydx2-------- d2y

dx2--------

514---

6d2ydx2-------- d2y

dx2--------

7

8

9WORKEDExample



For worked example 9, the second derivative sign diagramverifies that at x = 0 and x = 2 there are indeed points ofinflection as f ′′(x) changes sign at these points.

This verification is not usually required but in rare casesit will show that what seems to be a point of inflection is infact not.

For example, if f (x) = x4

f ′(x) = 4x3

= 0 when x = 0.Hence a stationary point occurs at x = 0.

For f ′′(x) = 12x2 = 0, a solution occurs when x = 0.Thus it appears, there is a stationary point of inflection at x = 0.But the sign diagram of the second derivative (see figure

at right) shows that f ′′(x) does not change sign at x = 0.Therefore, there is not a stationary point of inflection at

x = 0.The first derivative test will verify that at x = 0 there is a local minimum stationary

point as shown in the figure below (left).

The sketch graph of f (x) = x4 is shown in the figure below (right).

THINK WRITE

b Solve = 0 b For points of inflection, = 0

3x2 − 6x = 03x(x − 2) = 0x = 0 or x = 2

Find y at x = 2 only, since a stationary point has already been determined at the point (0, 2).

At x = 2, y = (2)4 − 23 + 2

= 4 − 8 + 2= −2

Check the sign of either side of

x = 2.

If x = 1, = 3(1)2 − 6(1)

= −3

If x = 3, = 9 (from part a)

State the other point of inflection. Therefore the point (2, −2) is a point of inflection (not stationary.)

Check using a graphics calculator.

1d2ydx2-------- d2y

dx2--------

214---

3d2ydx2-------- d2y

dx2--------

d2ydx2--------

4

5

x0–1 1 2– –

sign ofd2y–––dx2

+ +

x0–1 1– –

f"(x)+ +

x0–1 1– –

f ' (x)+ +

x0

f (x) = x4y



For the function f(x) = 3 logex − x3 − 5, find:a all stationary points b any points of inflection c and sketch the graph of f (x).

THINK WRITE

a Find f ′(x) and set it equal to zero. a f ′(x) = − 3x2

For stationery points, f ′(x) = 0

− 3x2 = 0

Multiply the equation by x. or 3 − 3x3 = 0, x ≠ 0

Solve f ′(x) = 0. 3x3 = 3x3 = 1

Thus x = 1.

Evaluate f (1). f(1) = 3(0) − 13 − 5= −6

Find f ′′(x). f ′′(x) = −3x−2 − 6x

Evaluate f ′′(1). f ′′(1) = − − 6

= −9

State any stationary point and its type. Therefore (1, −6) is a local maximum.

b Set f ′′(x) = 0 and multiply the equation

by x2.b f ′′(x) = −3x−2 − 6x

For points of inflection, f ′′(x) = 0−3x−2 − 6x = 0or −3 − 6x3 = 0, x ≠ 0

Solve f ′′(x) = 0. 6x3 = −3

x3 = −

Thus x = − or approximately −0.79.

Since x = −0.79 is the only solution there are no points of inflection because the implied domain is x > 0.

Therefore there are no points of inflection as −0.79 is outside the implied domain, x > 0.

c Use a graphics calculator to determine intercepts.

Sketch the graph of f (x).

Check the graph using a graphics calculator.

13x---

3x---

2

3

4

5

6 31---

7

1

212---

12---

13---

3

1

x0

f (x) = 3 logex – x3 – 5

y

–61(1, –6)

2

3

10WORKEDExample



Different technology tools, includingthe graphics calculator, can be usedto verify or obtain the coordinates ofstationary points as well as draw therelevant graph.

The screen at right shows anexample from the prepared Mathcadfile ‘Stationary points’ found on theMaths Quest CD-ROM.

Find the minimum value of the gradient to the curve:f(x) = x3 + 4x2 − 7x.THINK WRITE

Find the equation of the gradient function f ′(x).

f ′(x) = 3x2 + 8x − 7

Find the rate of change of the gradient function, f ′′(x).

f ′′(x) = 6x + 8

Set f ′′(x) = 0. Let f ′′(x) = 6x + 8 = 0Solve f ′′(x) = 0. 6x = −8

x = −

Evaluate f ′(− ) which is the minimum value of the gradient as f ′(x) is a positive quadratic.

f ′(− ) = 3( ) + 8(− ) − 7

= − − 7

= −12

which is a minimum as f ′(x) is a positive quadratic equation.

State the minimum value of the gradient. Therefore the minimum gradient of f(x) is −12 .Verify using a graphics calculator.

1

2

34

43---

543--- 4

3--- 16

9------ 4

3---

163------ 32

3------

13---

613---

7

11WORKEDExample

Mathcad

Stationarypoints

remember1. If f ′(a) = 0 and f ′′(a) > 0, then a local minimum stationary point occurs at x = a.2. If f ′(a) = 0 and f ′′(a) < 0, then a local maximum stationary point occurs at x = a.3. If f ′(a) = 0 and f ′′(a) = 0, and f ′′(x) changes sign at x = a, then a stationary point

of inflection occurs at x = a.4. If f ′′(a) = 0 and f ′′(x) changes sign at x = a, then a point of inflection occurs at

x = a.

remember



Analysing the behaviour of functions using the second derivative

Use a graphics calculator to assist you to draw the graphs.

1 For each of the following functions find the stationary points and the nature of thestationary points using the second derivative test.

2 Sketch the graph of each function in question 1, clearly indicating all stationarypoints and axes intercepts.

3The function f(x) = x4 + 4x3 + 3:a has a point or points of inflection when x is equal to:

b has a stationary point of inflection:

4The minimum gradient of the curve with equation y = x3 − 6x2 − 8x is:

5If f ′(x) = 0 when x = 3 and x = −2, and f ′′(3) = 4 and f ′′(−2) = −5, then f(x) has:A stationary points of inflection when x = −2 and x = 3B a stationary point of inflection when x = 3 and a local maximum when x = −2C a local maximum when x = −2 and a local minimum when x = 3D a local minimum when x = −2 and a local maximum when x = 3E a stationary point of inflection when x = −2 and a local minimum when x = 3

6 Give i any stationary points of inflection and ii any other points of inflection for eachof the following functions.

7 Show that y = x logex does not have any points of inflection.

a f(x) = x2 − 4x b g(x) = 12 − x2

c y = x2(x + 2) d y = x(x − 1)(x + 2)e h(x) = (x − 3)(x + 3)(x + 1) f f(x) = x3 + 4x2 − 4x − 16g g(x) = x3 h f(x) = x3 + 3i h(x) = x3 − 3x j y = x4 + x3

A 0 and −4 B 0 only C −2 onlyD 0 and −2 E 0 and

A (−4, 3) B (0, 3) C (−2, −13)D (−2, 32) E (−4, 0)

A −20 B −32 C 2D 0 E −12

a y = x3 − 6x b f(x) = x2 − c y =

d g(x) = x3 + 2x2 + 1 e y = xex f y = 2x4 − x3

g f(x) = x2ex h g(x) = 8x2 − logex

5C

WORKEDExample

8a

Mathca

d

Stationarypoints

WORKEDExample

8b


12---



WORKEDExample

91x--- 1

x--- 1

x2-----–

Chap 05 SM 8 Thursday, October 12, 2000 10:22 AM

C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s 1898 Show that the graph of the function g(x) = ex2 has no points of inflection. Find its

stationary point.

9 Find i stationary points and ii points of inflection, and iii sketch the graphs with thefollowing rules:

Verify with a graphics calculator.

10 Show that the maximum value of the gradient to the curve f(x) = x3 + 3x2 + 2 is −3.

11 Find the maximum value of the gradient to the curve y = 10 + 3x2 − 2x3.

12 Sketch the graph of the function y = x4 − x2 − 12, showing all intercepts with the axes,the stationary points and any points of inflection. (Verify with a graphics calculator.)

13 The downward displacement of a meteor t seconds after hitting the surface of theocean is given by:

d = , where 0 ≤ t ≤ 20 and d is in metres.

a Find the depth of the meteor after 10 seconds.b Find the velocity, v, at any time t.c Find the maximum velocity of the meteor.d If the depth of the ocean where the meteor strikes is 600 metres and the

meteor disintegrates 20 seconds after hitting the ocean, does the meteor reach the ocean floor?

a f(x) = x3 − x b y = x3 + 8x c f(x) =

d g(x) =

WORKEDExample

10x

1x2-----–

x2 2x---–

WORKEDExample

11

40tt2

4---- t3

24------–+

WorkS

HEET 5.1



Derivatives of inverse circular functionsThe derivative of Sin−1 , a > 0

If y = Sin−1 , −a < x < a and − < y <

then = sin y, − < y <

x = a sin y

So = a cos y

or = , cos y ≠ 0 and − < y <

Since sin y =

cos y = (from diagram at right)

or a cos y =

so = , −a < x < a.

Therefore, if f (x) = Sin–1

then f ′(x) = , –a < x < a.

Note: Example 12(b) could also be done by the chain rule using the substitution u = 6x.

xa---

xa--- π

2--- π

2---

x0

yy = Sin–1 x–a

a–a

–2– π

–2π

xa--- π

2--- π

2---

dxdy------

dydx------ 1

a ycos--------------- π

2--- π

2---

xa---

x0

ax

y

a2 – x2

a2 x2–a

--------------------

a2 x2–dydx------ 1

a2 x2–

--------------------

xa---

1

a2 x2–---------------------

Find the derivative of: a Sin−1 b Sin−1 6x.THINK WRITEa Differentiate by rule where a = 4. a If y = Sin−1

then =

b Express 6x as . b If y = Sin−1 6x

then y = Sin−1

Differentiate by rule where a = . =

Take = out as a factor of the denominator to remove the fraction from the square root.

=

Simplify the derivative. =

=

x4---

x4---

dydx------ 1

16 x2–---------------------

1x

(16---)

---------x

(16---)

---------

216---

dydx------ 1

136------ x2–

-------------------

3136------

16---

1136------ 1 36x2–( )

----------------------------------

41

16--- 1 36x2–---------------------------

6

1 36x2–------------------------

12WORKEDExample



The derivative of Cos–1 , a > 0

If y = Cos−1 , −a ≤ x ≤ a and 0 ≤ y ≤

then = cos y, 0 ≤ y ≤

x = a cos y

Thus, = −a sin y

or = , sin y ≠ 0 and 0 < y < .

Since cos y =

sin y = (from diagram at right)

or a sin y =

So = – , −a < x < a.

Therefore, if f (x) = Cos–1

then f ′(x) = , –a < x < a

xa---

xa--- π

x0

y

y = Cos–1 x–a

a–a

π

–2πx

a--- π

dxdy------

dydx------ 1–

a sin y--------------- π

xa--- a

xy

a2 – x2

a2 x2–a

--------------------

a2 x2–dydx------ 1

a2 x2–--------------------

xa---

1–

a2 x2–---------------------

Find f ′(x) if f(x) is equal to: a Cos−1 b Cos−1 .THINK WRITE

a Differentiate by rule where a = . a f(x) = Cos−1

f ′(x) =

b Express as . b f(x) = Cos−1

= Cos−1

Differentiate by rule where a = . f ′(x) =

Take = out as a factor of the

denominator.

=

=

Simplify f ′(x). =

x

3------- 2x

5------

3x3---

1–3 x2–--------------

12x5

------ x

(52---)

--------- 2x5

------

x

(52---)

---------

252---

1–254------ x2–

-------------------

314---

12---

1–14--- 25 4x2–( )

--------------------------------

–112--- 25 x2–------------------------

42–

25 x2–---------------------

13WORKEDExample



The derivative of Tan−1

If y = Tan−1 , x ∈ R and − < y <

then = tan y, – < y <

x = a tan y

= a sec2y

= a(1 + tan2y)

= (from the diagram at right)

=

=

or =

Therefore, if f(x) = Tan–1

then f ′(x) = , x ∈ R

xa---

xa--- π

2--- π

2---

x0

y

y = Tan–1 x–a

–2π

–2π–

xa--- π

2--- π

2---

x0 a

xy

a2 + x2

dxdy------

a 1x2

a2-----+

a a2 x2+( )a2

-------------------------

a2 x2+a

-----------------

dydx------ a

a2 x2+-----------------

xa---

aa2 x2+-----------------

Find f ′(x) if f(x) is equal to:

a Tan−1 b Tan−1 .

THINK WRITE

a Differentiate by rule where a = 5. a f(x) = Tan−1

Thus, f ′(x) =

b Express as . b f(x) = Tan−1

= Tan−1

Differentiate by rule where a = . f ′(x) =

Take out as a factor of the denominator. =

Simplify the derivative by dividing by . =

x5--- 8x

3------

x5---

525 x2+-----------------

18x3

------ x

(38---)

--------- 8x3

------

x

(38---)

---------

238---

38---

964------ x2+----------------

3164------

38---

164------ 9 64x2+( )------------------------------

438--- 1

64------

249 64x2+--------------------

14WORKEDExample



Find if y is equal to: a Sin−1(4x + 7) b Cos−1(5 − 3x) c sin (Tan−1 ).THINK WRITE

a Let u = 4x + 7 so the chain rule can be applied. a y = Sin−1 (4x + 7)Let u = 4x + 7.

Find . = 4

Express y in terms of u. y = Sin−1 u

Find . =

Apply the chain rule. So = ×

=

Replace u with 4x + 7. =

b Let u = 5 − 3x so the chain rule can be applied. b y = Cos−1 (5 − 3x)Let u = 5 − 3x.

Find . = −3

Express y in terms of u. y = Cos−1 u

Find . =


=

Replace u with 5 − 3x. =

c Let u = Tan−1 so the chain rule can be applied. c y = sin (Tan−1 )Let u = Tan−1 .

Find . =

Express y in terms of u. y = sin u

Find . = cos u

d ydx------ x

5---

1

2dudx------ du

dx------

3

4dydu------ dy

du------ 1

1 u2–------------------

5dydx------ dy

du------ du

dx------

4

1 u2–------------------

64

1 4x 7+( )2–------------------------------------

1

2dudx------ du

dx------

3

4dydu------ dy

du------ 1–

1 u2–------------------

5dydx------ dy

du------ du

dx------

3

1 u2–------------------

63

1 5 3x–( )2–-----------------------------------

1x5--- x

5---

x5---

2dudx------ du

dx------ 5

25 x2+-----------------

3

4dydu------ dy

du------

15WORKEDExample

Continued over page



THINK WRITE


=

Replace u with Tan−1 .=

5dydx------ dy

du------ du

dx------

5 cos u25 x2+-----------------

6x5---

5 Tan–1x5---

cos

25 x2+------------------------------------

Find the equation of the normal to the curve with equation:

y = 2 Cos−1 at the point where x = .

THINK WRITE

Find y when x = . y = 2 Cos−1

When x = y = 2 Cos−1

= 2

=

Find . =

Substitute x = into to find the

gradient of the tangent at x = .

When x = , the gradient of the tangent is

=

=

= −2

Find the gradient of the normal

.

The gradient of the normal =

=

Substitute x1 = , y1 = and m =

into the equation of a straight line rule:(y − y1) = m(x − x1) where m is the gradient and (x1, y1) is a point on the line.

Therefore, the equation of the normal is:

y − = (x − )

=

Simplify the equation. y =

(or 6y = 3x + 2 − 3 )

x2--- 3

1 3x2---

332---

π6---

π3---

2dydx------ dy

dx------ 2–

4 x2–------------------

3 3dydx------

3

3dydx------ 2–

4 3–----------------

2–1

------

4

gradient of normal1–

gradient of tangent---------------------------------------------=

1–2–

------

12---

5 3π3--- 1

2---

π3--- 1

2--- 3

x2--- 3

2-------–

6 x2--- π

3--- 3

2-------–+π 3

16WORKEDExample



Derivatives of inverse circular functions

1 Find the derivative of each of the following expressions with respect to x:

2 Find if y is equal to:

3 Using the results of question 2, or otherwise, state the derivative of:

where b is a real, positive constant.

4

Consider the function with the rule f (x) = Sin−1 .

a The maximal domain of f (x) is:

a Sin−1 b Sin−1 c Sin−1

d Cos−1 e Cos−1 f Cos−1

g Tan−1 h Tan−1 i Tan−1

j Tan−1 k Sin−1 l Sin−1

m Cos−1 n Cos−1 o Tan−1

p Tan−1 q Sin−1 r Tan−1

a Sin−1 2x b Sin−1 5x c Sin−1 3x d Sin−1 8x

e Cos−1 4x f Cos−1 6x g Cos−1 7x h Cos−1 10x

i Tan−1 3x j Tan−1 9x k Tan−1 4x l Tan−1 5x

a Sin−1 bx b Cos−1 bx c Tan−1 bx

A [− , ] B [− , ] C [−1, 1]

D [−3, 3] E [− , ]

remember1. If f (x) = Sin−1 then f ′(x) = , −a < x < a

2. If f (x) = Cos−1 then f ′(x) = , −a < x < a

3. If f (x) = Tan−1 then f ′(x) = , x ∈ R

xa--- 1

a2 x2–--------------------

xa--- 1–

a2 x2–--------------------

xa--- a

a2 x2+-----------------

remember

5DMathcad

Inversetrig.

derivatives

WORKEDExample

12ax2--- x

5--- x

8---

WORKEDExample

13ax3--- x

4--- x

6---

WORKEDExample

14a

x2--- x

4--- x

7---

x3--- x

5------- x

0.2-------

x2.5------- x

7------- x

3-------

x0.8------- x

6------- x

10----------

dydx------

WORKEDExample

12b

mmultiple choiceultiple choicex3---

π2--- π

2--- 3π

2------ 3π

2------

13--- 1

3---



b The maximal domain of f ′(x) is:

5

Let f (x) = Cos−1 .

a When expressed in the form f (x) = Cos−1 , the value of a is:

b f ′(x) is equal to:

6The derivative of Tan−1 is equal to:

7 Find f ′(x) if f (x) is equal to:

8 Using the results of question 7, or otherwise, state the derivative of y with respect to x:

where a and b are real, positive constants.

9 Find if y is equal to:

A [− , ] B [−3, 3] C (−3, 3)

D [−1, 1] E (−1, 1)

A B 7 C D E 3

A B C

D E

A B C

D E

a Cos−1 b Cos−1 c Cos−1 d Cos−1

e Tan−1 f Tan−1 g Tan−1 h Tan−1

i Sin−1 j Sin−1 k Sin−1 l Sin−1

a y = Sin−1 b y = Cos−1 c y = Tan−1

a Sin−1(2x + 3) b Sin−1(3x − 5) c Cos−1(4x − 3)

d Cos−1(5x + 8) e Tan−1(3x + 2) f Tan−1(6x − 7)

g Sin−1 h Cos−1 i Tan−1

j Sin−1(4 − 3x) k Cos−1(7 − 2x) l Tan−1(8 − 5x)

m Sin−1 n Cos−1 o Tan−1

3π2

------ 3π2

------

mmultiple choiceultiple choice3x7

------

xa---

73--- 3

7--- 1

7---

7–

49 9x2–------------------------ 7–

9 49x2–------------------------ 3–

9 49x2–------------------------

1–

49 x2–--------------------- 3–

49 9x2–------------------------

mmultiple choiceultiple choice8x5

------40

25 64x2+------------------------ 64

64 25x2+----------------------- 40

64 25x2+-----------------------

525 x2+----------------- 8

64 x2+-----------------

WORKEDExample

13b3x4

------ 7x4

------ 9x5

------ 5x8

------

WORKEDExample

14b4x5

------ 3x8

------ 7x2

------ 9x5

------

2x3

------ 5x2

------ 6x7

------ 8x5

------

bxa

------ bxa

------ bxa

------

WORKEDExample

15

dydx------

x 3+2

------------ 2x 1+

3---------------

4x 3–5

---------------

3 4x–5

--------------- 6 3x–

7---------------

2 3x–4

---------------


C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s 19710 Find the derivative of each of the following expressions with respect to x.

11 a Find the gradient of the graph of y = Sin−1 at the origin.

b Hence, find the equation of the tangent to this curve at the origin.

12 Find the equation of the normal to the curve y = Cos−1 2x at the point where it crossesthe y-axis.

13 Find the equation of the tangent to the curve y = Sin−1 when x = 1.

14 a Find the coordinates of the point where the maximum gradient of f (x) = Tan−1

occurs (a is a constant).b Find the maximum gradient of f (x).

1. To find the gradient of Tan–1x at x = 1, press, select 8:nDeriv(, enter

tan–1(X),X,1) and press . Check thatMODE is set to radians.Alternatively, if the function is in Y1, then press

, select 8:nDeriv(, enter Y1,X,1) andpress . Remember: To enter Y1, press

, select Y–VARS, 1:Function..., and 1:Y1(similarly any Y variable).

2. To increase the accuracy, add an extra argumentfor h in nDeriv, for example h = 10–10. Press

and (using the above method) set up8:nDeriv(Y1,X,1,E–10) and press . Toenter E for E–10, press [EE].)(Note: The inbuilt method actually calculatesthe gradient of a chord distance h either side ofthe point required. The default value of h is0.001.)

a x3 + Cos−1 2x b 4x2 − Sin−1

c sin 4x + Tan−1 d loge 6x + x − Cos−1

e e7x + 3 + Tan−1 5x + 3 f

g (Tan−1 )4 h 3x Cos−1 2x

i Sin−1 x + Cos−1 x j sin (Cos−1 x)

k tan (Sin−1 ) l cos (Tan−1 )

x3---

x4---

92--- 2x

3------

Sin 1–x

x2---

x3--- x

4---

x4---

WORKEDExample

16x2---

xa---

Graphics CalculatorGraphics Calculator tip!tip! Finding numerical derivatives

MATHENTER

MATHENTER

VARS

MATHENTER

2nd



Antidifferentiation involving inverse circular functions

We now know that:

(Sin−1 ) = , −a < x < a

(Cos−1 ) = , −a < x < a

(Tan−1 ) = , x ∈ R

It therefore follows that:

dx = Sin–1 + c, a > 0

dx = Cos–1 + c, a > 0

dx = Tan–1 + c

When finding antiderivatives of inverse circular functions, the integrand should beexpressed in one of the standard forms above and then integrated.

ddx------ x

a--- 1

a2 x2–--------------------

ddx------ x

a--- 1–

a2 x2–--------------------

ddx------ x

a--- a

a2 x2+-----------------

1

a2 x2–---------------------∫ x

a---

1–

a2 x2–---------------------∫ x

a---

aa2 x2+-----------------∫ x

a---

Differentiate Tan−1 and hence find dx.

THINK WRITE

Write the equation. y = Tan−1

Differentiate Tan−1 by rule where a = 2. =

Express the result using integral notation. Therefore dx = Tan−1 + c.

x2--- 2

4 x2+---------------∫

1x2---

2x2--- dy

dx------ 2

4 x2+--------------

32

4 x2+--------------∫ x

2---

17WORKEDExample

Find the antiderivative for each of the following expressions:

a b c .

THINK WRITE

a The antiderivative is an inverse sine function

of the form Sin−1 where a = 5.a dx = Sin−1 + c

1

25 x2–--------------------- –3

49 x2–--------------------- 20

16 x2+------------------

xa---

1

25 x2–---------------------∫ x

5---

18WORKEDExample



THINK WRITE

b Take 3 out as a factor. b dx = 3 dx

The antiderivative of

is an inverse cos function of the form

Cos−1 where a = 7.

= 3 Cos−1 + c

(or −3 Sin−1 + c)

c Take 5 out as a factor. c dx = 5 dx


is an inverse tan function of the form

Tan−1 where a = 4.

= 5 Tan−1 + c

1–3

49 x2–---------------------∫ 1–

49 x2–---------------------∫

21–

49 x2–---------------------

xa---

x7---

x7---

120

16 x2+-----------------∫ 4

16 x2+-----------------∫

24

16 x2+---------------------

xa---

x4---

Find each of the following indefinite integrals:

a dx b dx c dx.

THINK WRITE

a Use the substitution u = 5x. a dx

Let u = 5x.

Find . = 5

Make dx the subject. dx =

Rewrite the integral in terms of u. dx =

= du

Antidifferentiate by rule. = Sin−1 u + c

Replace u with 5x. = Sin−1 5x + c

1

1 25x2–-------------------------∫ 1–

9 16x2–-------------------------∫ 8

9 5x2+------------------∫

11

1 25x2–------------------------∫

2dudx------ du

dx------

3du5

------

41

1 25x2–------------------------∫ 1

1 u2–------------------∫

du5

------

15---

1

1 u2–------------------∫

515---

615---

19WORKEDExample

Continued over page



THINK WRITE

b Use the substitution u = 4x. b dx

Let u = 4x.

Find . = 4

Make dx the subject. or dx =


= du

Antidifferentiate by rule. = Cos−1 + c

Replace u with 4x. = Cos−1 + c

c Use the substitution u = x. c dx

Let u = .

Find . =

Make dx the subject. dx =


= du

Express the integral in standard form. = du

Antidifferentiate by rule. =

Replace u with x and simplify the surd factor.

=

11–

9 16x2–------------------------∫

2dudx------ du

dx------

3du4

------

41–

9 16x2–------------------------∫ 1–

9 u2–------------------

du4

------∫14---

1–

9 u2–------------------∫

514---

u3---

614---

4x3

------

1 58

9 5x2+-----------------∫

5x

2dudx------ du

dx------ 5

3du

5-------

48

9 5x2+-----------------∫ 8

9 u2+-------------- du

5-------×∫

8

5------- 1

9 + u2---------------∫

58

3 5---------- 3

9 u2+--------------∫

68

3 5---------- Tan 1– u

3--- c+

7 58 515

---------- Tan 1– 5x3

---------- c+


C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s 201Antiderivatives involving all three inverse trigonometric functions can be obtained using the Mathcad file ‘Antidifferentiation involving trigonometric functions’. The screen at right shows the answer for part (c) of worked example 19.

Mathcad

Antiderivatives

Find the antiderivative of by first simplifying the rational expression.THINK WRITE

Divide x2 + 4 into x3 + 4x – 16 to make the rational expression antidifferentiable.

Simplify by long division:

x3 + 4x– 16

so = x –

Rewrite dx as two separate

integrals.

dx = dx

Express dx in standard form. = x dx – 8 dx

Antidifferentiate both integrals. = x2 – 8 Tan–1 + c

x3 4x 16–+

x2 4+------------------------------

1x

3 4x 16–+

x2 4+

-----------------------------

x

x2 4 x

3 4x 16–++

x3 4x 16–+

x2 4+

----------------------------- 16

x2 4+

--------------

2x

3 4x 16–+

x2 4+

-----------------------------∫ x3 4x 16–+

x2 4+

-----------------------------∫ x xd –16

x2 4+

--------------∫ ∫3

16

x2 4+

--------------∫ ∫ 2

x2 4+

--------------∫4

12---

x2---

20WORKEDExample

remember1. dx = Sin−1 + c, a > 0

2. dx = Cos−1 + c, a > 0

3. dx = Tan−1 + c

1a2 x2–----------------∫ x

a---

1–a2 x2–----------------∫ x

a---

aa2 x2+-----------------∫ x

a---

remember



Antidifferentiation involving inverse circular functions

1 a Differentiate Tan−1 and hence find dx.

b Differentiate Cos−1 and hence find dx.

c Differentiate Sin−1 and hence find dx.

2 Find the antiderivative for each of the following expressions:

3 a Differentiate Sin−1 2x and hence find dx.

b Differentiate Tan−1 4x and hence find dx.

4 Find each of the following indefinite integrals:

5

a The equivalent expression to is:

a b c

d e f

g h i

j

a dx b dx c dx

d dx e dx f dx

g h i dx

j dx k dx l dx

A B × C

D E

5E

Mathca

d

Inversetrig.derivatives

WORKEDExample

17

x6--- 6

36 x2+-----------------∫

x3---

1–

9 x2–------------------∫

x5---

1

25 x2–---------------------∫

Mathca

d

Antidiff’ninvolvinginversetrig.functions

WORKEDExample

18 14 x2–-------------- 1–

1 x2–------------------ 1

9 x2–------------------

1–

16 x2–--------------------- 2

4 x2+-------------- 5

25 x2+-----------------

4

25 x2–--------------------- 3–

9 x2–------------------ 6

9 x2+--------------

2016 x2+-----------------

1

1 4x2–---------------------∫

116x2 1+--------------------∫

WORKEDExample

19a 1

1 9x2–---------------------∫ 1

1 16x2–------------------------∫ 1–

1 4x2–---------------------∫

1–

1 25x2–------------------------∫ 3

1 9x2+-----------------∫ 5

1 25x2+--------------------∫

dx

1 25x2–------------------------∫ dx–

1 36x2–------------------------∫ 8

1 64x2+--------------------∫

6

1 49x2–------------------------∫ 10–

1 16x2–------------------------∫ 24

1 4x2+-----------------∫

mmultiple choiceultiple choice1–

1 4x2–---------------------

1–

(12---)2

x2–--------------------------- 1

2---

1–

(12---)2

x2–--------------------------- 2 1–×

(12---)2

x2–---------------------------

12--- 1–×

4 x2–------------------

2 1–×4 x2–

------------------



b The antiderivative of is:

6

The antiderivative of is:

7 Find f (x) if f ′(x) is equal to:

8 Find the antiderivative of each of the following expressions by first simplifying therational expression.

9 Differentiate Sin−1 x2 and hence find dx.

10 Differentiate Cos−1 and hence find dx.

A 2 Cos−1 2x + c B 2 Cos−1 + c C Cos−1 2x + c

D Cos−1 + c E Sin−1 + c

A Tan−1 + c B Tan−1 + c C Tan−1 + c

D Tan−1 + c E Tan−1 + c

a b c

d e f

g h i

j k l

m n o

p q r

s t u

v

a b c

d e f

g

1–

1 4x2–---------------------

x2--- 1

2---

12---

x2--- 1

2---

x2---

mmultiple choiceultiple choice2

49 25x2+-----------------------

145------

5x7

------ 25---

7x5

------ 25---

x5---

235------

5x7

------ 27---

x7---

WORKEDExample

19b 1

16 9x2–------------------------ 1

9 25x2–------------------------ 1–

9 4x2–---------------------

1–

4 25x2–------------------------ 1

4 9x2+----------------- 1

25 16x2+-----------------------

1

5 x2–------------------ 1–

6 x2–------------------ 1

1 7x2+-----------------

45 3x2+----------------- 1

7 4x2–--------------------- 1–

10 9x2–------------------------

1

1 x 3+( )2–-------------------------------- 1–

1 x 2–( )2–-------------------------------- 1

4 x 1–( )2–--------------------------------

1–

9 x 4+( )2–-------------------------------- 1

1 x 3–( )2+---------------------------- 1

16 x 5+( )2+-------------------------------

1

4 14---x2–

--------------------- 1–

8 12---x2–

---------------------32---

9 14---x2+

-----------------

6

5 15---x2+

-----------------

WORKEDExample

202 x2+1 x2+-------------- x2 7+

4 x2+-------------- x3 x 1+ +

x2 1+-----------------------

x3 x 5+ +x2 1+

----------------------- x3 9x 3+ +x2 9+

--------------------------- 3x3 12x 8–+x2 4+

---------------------------------

2x3 32x 1–+x2 16+

---------------------------------

2x

1 x4–------------------∫

x1–

x 1 x–( )------------------------∫



11 Differentiate Tan−1 and hence find dx.

12 A curve has a gradient given by = and its graph passes through the

origin. Find the equation of the curve.

13 A curve has a gradient given by = and its graph passes through

( , ). Find the equation of the curve.

x1

1 x+( ) x------------------------∫

dydx------ 2

1

1 x2–------------------+

dydx------ 3x2 1

1 4x2+-----------------+

12--- π

8---

History of mathematicsSOFIA KOVALEVSKAYA (15 .1 .1850 – 10 .2 .1891)

During her lifetime ...The Salvation Army is formed.Edison develops the electric light bulb.Queensland becomes a separate State.Abraham Lincoln is assassinated.

Sofia Kovalevskaya was born in Moscow. Her familywere members of the Russian nobility and theirfriends included the author Dostoyevsky. She wasinterested in mathematics as a child and by the age of11 her bedroom walls were covered with notes aboutcalculus. When she was 14 years old a neighbourpresented the family with a physics textbook he hadwritten. She read it with interest and tried tounderstand the formulas by deducing them from firstprinciples. The author was impressed and urged herfather to let her study mathematics.

Sofia wanted to continue her education atuniversity. Women were not allowed to attenduniversity in Russia at this time and young unmarriedwomen were not allowed to travel alone. Partly toresolve this problem, she married VladimirKovalevsky in 1868. Shortly afterwards she went to

Germany to attend Heidelberg University. Once thereshe found that women weren’t allowed to attendlectures or matriculate.

In 1870 she met Karl Weierstrass who was one ofthe most renowned mathematicians of the time. Heset a series of problems for her and found that hermathematical abilities were remarkable. She studiedwith him in Berlin for four years and received herPhD in 1874. She was then dismayed to find thatbecause of her sex the best job she could get wasteaching mathematics at primary school. She wenthome to Russia and in 1878 her daughter Fufa wasborn. In 1883 she returned to Berlin but later thatyear she received the terrible news that her husband,Vladimir, had committed suicide.

At last in 1884 she received an invitation to lectureat the University of Stockholm. It was only atemporary teaching post but her abilities became sorespected that after five years Kovalevskaya wasappointed Chair of Mathematics. Only two otherwomen, the physicist Laura Bassi and themathematician Maria Agnesi, had ever achieved sucha position at a European university.

On Christmas Eve 1888, she won the famous PrixBordin. The rules specified that the entries were to bejudged without the panel knowing who wrote thepapers. The judges selected Kovalevskaya’s paper asthe winner. They also increased the prize money from3000 francs to 5000 francs ‘on account of the quiteextraordinary service rendered to mathematicalphysics by this work’. Many people were surprisedwhen it was revealed that the winner was a woman. Alittle over two years later she died of influenzacomplicated by pneumonia.

Questions1. What did Kovalevskaya find when she arrived at

Heidelberg University?2. What was the best job she could get after

receiving her PhD?3. Why was her appointment to the Chair of

Mathematics at the University of Stockholm unusual for the times?



Rate of change of angleA hot-air balloon is released at ground level and rises at a steady vertical speed of 2.0 m/s. One hundred metres from the launch site a photographer controls a camera aimed directly at the balloon. As the balloon rises the camera is tilted and the axis of the camera makes an angle θ to the horizontal. At time t the balloon is a height y(t) above the ground and the camera makes an angle θ(t) to the horizontal. This is shown in the figure.

a Find the height of the balloon after 5 seconds and consequently find the angle θ at that time.

b Find the height of the balloon after 10 seconds and thus the angle θ at that time.c Find an expression for the height, y, in terms of the time, t, elapsed since

launching.d Give an expression for the angle θ in terms of y and hence in terms of t.e Use a graphics calculator or Graphmatica to sketch a graph of θ(t); t ≥ 0.f At what time will θ = 30°?

g Use calculus to find and consequently find the time and the height of the

balloon when = 0.020°/s.

The camera is moved closer to the launch site.h Use a series of values for the distance of the camera from the launch site and

plot θ(t) for the different values on the one axis.i Comment on the effect that reducing the distance of the camera from the launch

site has on the rate that the camera angle changes per second as the balloon rises.

θ

Camera

y(t)

= 2

t

2.0 m/s

100 m

dθdt------

dθdt------



Derivatives of the tangent and inverse circular functions

• (tan ax) = a sec2ax

• [tan f (x)] = f ′(x) sec2 [f (x)]

• (Sin−1 ) = , a > 0

• (Cos−1 ) = , a > 0

• (Tan−1 ) =

Using the second derivative to analyse the behaviour of functions

• If y = f(x), then the first derivative is denoted by or f ′(x).

• The second derivative is denoted by or f ′′(x).

• Stationary points and points of inflection can be established from the second derivative:

1. If f ′(a) = 0 and f ′′(a) > 0, then a local minimum stationary point occurs at x = a.

2. If f ′(a) = 0 and f ′′(a) < 0, then a local maximum stationary point occurs at x = a.

3. If f ′(a) = 0 and f ′′(a) = 0, and f ′′(x) changes sign at x = a, then a stationary point of inflection occurs at x = a.

4. If f ′′(a) = 0 and f ′′(x) changes sign at x = a, then a point of inflection occurs at x = a.

Antiderivatives involving inverse circular functions

• dx = Sin−1 + c, a > 0

• dx = Cos−1 + c, a > 0

• dx = Tan−1 + c

summaryd

dx------

ddx------

ddx------ x

a--- 1

a2 x2–--------------------

ddx------ x

a--- 1–

a2 x2–--------------------

ddx------ x

a--- a

a2 x2+-----------------

dydx------

d2ydx2--------

1

a2 x2–--------------------∫ x

a---

1–

a2 x2–--------------------∫ x

a---

aa2 x2+-----------------∫ x

a---



Multiple choice

1 The derivative of tan 4x with respect to x is:

2 The derivative of −3 tan is equal to:

3 If f(x) = loge(tan 3x), then f ′(x) equals:

4 If y = e−2x tan 5x, then is equal to:

5 The second derivative of 2x6 − 5x4 + 3 is:

6 If g(x) = e2x5, then g′′(x) is equal to:

7 If h(x) = loge 5x3, then h′′(x) is equal to:

8 The second derivative of 9 cos is:

A 4 sec2 4x B 4 sec2 x C sec2 4x D sec2 x E 4

A −9 sec2 B −3 sec2 C sec2

D 9 sec2 E 3 sec2(−x)

A B C

D 3 sec2 3x tan 3x E

A e−2x (sec2 5x − 2 tan 5x) B e−2x (5 sec2 5x − 2 tan 5x) C e−2x (5 sec2 5x − tan 5x)

D −10e−2x sec2 5x E 5e−2x sec2 5x

A 60x4 − 60x2 B 12x5 − 20x3 C 12x4 − 20x2

D 240x3 − 120x E 12x4 − 60x2 + 3

A 20x3e2x5 (5x5 + 2) B 100x8e2x5 C e40x3

D 40x3e2x5 E 10x4e2x5

A 30 B 15x C D – E

A −6 sin B −4 cos C −4 sin

D −6 cos E 4 cos

CHAPTERreview

5A14--- 1

4---

5A–x3---

–x3---

–x3---

–x3---

–x3---

5A1x--- 1

tan 3x-------------- sec2 3x

tan 3x------------------

3sin 3x cos 3x---------------------------------

5Adydx------

5B

5B

5B3x--- 3

x2----- 3

5x4--------

5B2x3

------2x3

------ 2x3

------ 2x3

------

2x3

------ 2x3

------



For questions 9 to 11 consider the function f(x) = x3 − x2 − 6x + 2.

9 f (x) has a local maximum stationary point at:

10 f (x) has a local minimum stationary point at:

11 f (x) has a point of inflection at:

Use the graph at right to answer questions 12 and 13.

12 The graph of g(x) has:

A 2 stationary points

B 1 point of inflection

C 3 stationary points and 2 points of inflection

D 2 stationary points of inflection

E no stationary points of inflection

13 g′′(x) = 0 when x is equal to:

14 If y = e−kx and , then k is equal to:

15 The gradient of y = tan 2x when x = is:

16 The gradient of f(x) = 4x − 3 tan x is equal to zero when x is equal to:

17 If f(x) = Sin−1 then f ′(x) is equal to:

A ( , − ) B (2.22, −16.78) C (− , 0)

D (−1.12, −7.29) E (2, −6)

A (2.34, 7.85) B (− , 0) C (1.79, −6.21)

D (− , 6) E ( , )

A (2, 6) B (2, −6) C (− , −6)

D ( , ) E ( , − )

A −2 and 3 B −4 and −2 C −4 and 3 D −4, −2 and 3 E −4 and 1

A 0 or −1 B 1 or 4 C −1 or −4 D 0 or −4 E 0 or 4

A B C D 8 E 4

A B C D E

A B C

D E

5C13--- 2

27------ 4

3---

5C43---

43--- 1

3--- 2

27------

5C43---

13--- 2

27------ 1

3--- 2

27------

x0

y

5C

5C

5Cd2ydx2-------- 5

dydx------ 4y+ + 0=

5Cπ6---

312--- 8

3---

5C π6--- π

3--- π

4--- π

2--- π

5Dx

10------

1

1 100x2–--------------------------- 1–

1 100x2–--------------------------- 1

100 x2–------------------------

1–

100 x2–------------------------ 10

100 x2+--------------------


C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s 20918 If g(x) = 2 Cos−1 2x then g′(x) is equal to:

19 The derivative of Tan−1 is equal to:

20 The antiderivative of is:

21 The antiderivative of is equal to:

22 The antiderivative of is equal to:

Short answer1 Find the derivative of:

2 Given that f(x) = 3x2 logex, find f ′′(x).

3 If g(x) = (x − 2)(x − 3)(x + 1), find:

4 For the function h(x) = 2x3 −15x2 + 24x − 7 find:a all of its stationary points and their natureb any points of inflection.

5 Sketch the graph of the function y = x4 − 3x3 + 2x2, clearly labelling all stationary points and x-intercepts.

A B C

D E

A B C

D E

A 5 Tan−1 2x + c B 10 Tan−1 2x + c C Tan−1 + c

D Tan−1 2x + c E 5 Tan−1 + c

A 2 Sin−1 3x + c B 2 Sin−1 + c C 2 Cos−1 3x + c

D 2 Cos−1 + c E 6 Sin−1 2x + c

A 4 Sin−1 6x + c B 4 Cos−1 6x + c C Cos−1 + c

D 4 Cos−1 + c E −4 Cos−1 6x + c

a tan e2x b

a g′(x) b g′′(x)

5D1–

4 x2–------------------ 2

1 4x2–--------------------- 4

1 4x2–---------------------

4–4 x2+-------------- 4–

1 4x2–---------------------

5D2x3

------6

4 9x2+----------------- 6

9 4x2+----------------- 3

9 4x2–---------------------

32 9 4x2+( )-------------------------- 3

4 9x2+-----------------

5E5

4 x2+--------------

52---

x2---

52---

x2---

5E2

9 x2–------------------

x3---

x3---

4–

6 x2–------------------ 5E

23--- 6

6x

6-------

5Ax2 1+tan 2x--------------

5B5B

5C

5C



6 If y = Sin−1 , find a and b .

7 If y = Tan−1 4x, find a and b .

8 Differentiate each of the following:

9 Find the antiderivative of .

10 If f ′(x) = and f = −5, find f (x).

11 Find f (x) if f ′(x) = and f (0.1) = 1.

Analysis1 The cost, $C, of producing x litres of a particular aromatic oil on any day is C = 144 + 2x + .

a Find the ‘set up’ cost, i.e. the cost incurred before any oil is produced.b Find the cost of producing 50 litres of oil.

The average cost per litre of oil produced, A, is A = .c Express A as a function of x, A(x).d Find the average cost of producing 100 litres.e Find the minimum average cost and the number of litres produced for the minimum cost.f Show that the rate of change of cost is the same as the average cost for the number of

litres producing the minimum cost.

2 The proportion of the original population of kangaroos remaining, P(t), t years after culling

was introduced onto an island can be modelled by the function: P(t) = Tan−1 , t ≥ 0.

a Sketch the graph of P(t).b What percentage remains

1 year after culling began?

c By what percentage has the population dropped after 4 years?

d When is the population half of its original number?

e Find the rate of change of the population proportion after 4 years.

f When is the rate of decline of the population proportion greatest?

g When is the rate of decline of the population

proportion per year?

a Sin−1(3x + 4) b Cos−1(4 − x) c Tan−1(2x + 5)

5Dx

12------ dy

dx------ d2y

dx2--------

5Ddydx------ d2y

dx2--------

5D

5E4–

81 4x2–------------------------

5E1

16 x2+----------------- 4( )

5E2

9 25x2–------------------------

x2

10------

Cx----

1 2π---–

t2---

testtest

CHAPTERyyourselfourselftestyyourselfourself

5 110π---------


7

VCEVCEcocovverageerage

In thisIn this chachapterpter7A Graphing the gradient

function from the graph of a function

7B Limits and differentiation from first principles

7C The derivative of xn

7D The chain rule7E The derivative of ex

7F The derivative of loge x7G The derivatives of sin x and

cos x7H The product rule7I The quotient rule7J Mixed problems on

differentiation

Differentiation

Area of study

Units 3 & 4 • Calculus

262

M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4

Graphing the gradient function from the graph of a function

A function is

smooth

if there are no ‘sharp’ points on its graph.

The gradient does not exist where the function is not smooth. An example is shown at right.

A function is

continuous

if there are no ‘breaks’, ‘jumps’ or ‘asymptotes’ on its graph. The gradient does not exist where the function is not continuous. An example is shown at right.

The gradient of a function exists wherever the function is

smooth and continuous

. That is, the gradient of a function exists at a point providing only

one

tangent can be drawn at that point. An exampleis shown at right.

Gradient function of straight lines

In general, for straight lines in the form

f

(

x

)

=

mx

+

c

(or

y

=

mx

+

c

), the rule for thegradient is

f

′

(

x

)

=

m

, that is, a horizontal straight line through

y

=

m

.

y

x0 2

f (x) is not smooth at x = 2so gradient does not existat x = 2. That is, gradientexists for R\{2}

f (x)

y

x0–1

There is a break in thegraph at x = 2 (discontinuous),so gradient does not existat x = –1. That is, gradientexists for R\{–1}

f (x)

y

x0

f (x)

The gradient of f (x)exists for x ∈ R asf (x) is smooth andcontinuous.

2

–2

Sketch the graph of the gradient function of the following function.

THINK WRITE

Find the gradient of f (x) using

.

y

x0

f (x)

1

–2

1

mriserun--------=

m12---=

1WORKEDExample

C h a p t e r 7 D i f f e r e n t i a t i o n

263

Gradient function of quadratic functions

The gradient function of a polynomial function is also a polynomial function but thedegree is reduced by 1. That is, the gradient function of

f

(

x

)

=

ax

2

+

bx

+

c

is of theform

y

=

mx

+

c

. So the gradient of a quadratic function is a linear function.

THINK WRITE

Sketch the graph of y = to represent the gradient function.

Note: The domain of f ′(x) is R as f (x) is smooth and continuous.

212--- y

x0

1–2 f '(x)

Sketch the graph of the gradient function for the quadratic function shown and state its domain.

THINK WRITE

Find when f ′(x) = 0. f ′(x) = 0 if x = 1.

Find when f ′(x) > 0. f ′(x) > 0 if x > 1.

Find when f ′(x) < 0. f ′(x) < 0 if x < 1.

Sketch f ′(x).

Find where f (x) is smooth and continuous and hence find the domain of f ′(x).

Domain is R.

y

x0

f (x)

1

1

2

3

4 y

x0 1

f '(x)

5

2WORKEDExample

264 M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4

Gradient function of cubic functionsThe gradient function of a cubic is a quadratic function.

Sketch the gradient function of the following function and state its domain.

THINK WRITE

Find when f ′(x) = 0. f ′(x) = 0 if x = −3 and x = 1.

Find when f ′(x) > 0. f ′(x) > 0 if x < −3 and x > 1.

Find when f ′(x) < 0. f ′(x) < 0 if −3 < x < 1.

Sketch the graph of the gradient function.

Find the domain by determining where f (x) is smooth and continuous.

Domain is R.

y

x0–31

f (x)

1

2

3

4 y

x0 1–3

f '(x)

5

3WORKEDExample

For the function f(x) shown, state the domain of the gradient function f ′(x).

THINK WRITE

The function is smooth and continuous everywhere except at x = −1 (discontinuous) and x = 2 (not smooth).

Domain = R\{−1, 2}

y

x0 2–1

f (x)

4WORKEDExample

C h a p t e r 7 D i f f e r e n t i a t i o n 265

Graphing the gradient function from the graph of a function

1 For each straight line function shown below, sketch the graph of its gradientfunction.

a b c

d e

remember1. A function is smooth if there are no sharp points on its graph.2. A function is continuous if the graph can be drawn without lifting pen from

paper.3. The gradient of a function exists where the function is smooth and

continuous.4. The gradient function of a polynomial function is also a polynomial function

but the degree is reduced by 1. For example, the gradient function off (x) = ax2 + bx + c is of the form y = mx + c.

5. Wherever the gradient of a function is zero, its gradient function will have an x-intercept.

6. Wherever the gradient of a function is positive (sloping /), the gradient function graph is above the x-axis.

7. Wherever the gradient of a function is negative (sloping \), the gradient function graph is below the x-axis.

8. The gradient of a horizontal line is 0.9. The gradient of a vertical line is undefined.

remember

7A

SkillSH

EET 7.1WORKEDExample

1

SkillSH

EET 7.2y

x0–1

1

f (x)

y

x0 1

3

f (x)

y

x0–2–1

f (x)

y

x0 2

–5

f (x) y

x0

3

f (x)


2

a The gradient of the line in the graph at right is:

b The graph of the gradient function in the graph at rightis represented by which of the diagrams below?

3 Sketch the graph of the gradient function for each quadratic function shown below.

A 1 B 2 C 0

D E −1

A B C

D E

a b c

d e


y

x0 1

2

f (x)

12---

y

x0 1

2f'(x)

y

x0 2

1f'(x)

y

x0

2 f'(x)

y

x0

1–2 f'(x)

y

x0

–2f'(x)

WORKEDExample

2

y

x0

g(x)y

x0

g(x)

y

x0

g(x)

–2

y

x0

g(x)

3

y

x0

g(x)

–2–1


a The gradient of the function shown in the graph at right is:

b The gradient function for the graph at right isshown by which of the graphs shown below?

5 For each cubic function f (x) graphed below, sketch the gradient function.

A always increasing B always decreasingC decreasing then increasing D increasing then decreasingE constant

A B C

D E

a b c

d e f

g

mmultiple choiceultiple choicey

x0

f (x)

y

x0

f'(x)

y

x0

f'(x) y

x0

f'(x)

y

x

f'(x)

0

y

x0

f'(x)

WORKEDExample

3 y

x0–3 2f (x)

y

x0 41

f (x)y

x0 5

f (x)

y

x0

f (x)Gradient = 0

y

x0

f (x) Gradient = 0

y

x0–3

f (x)

y

x0

2

Gradient = 0

f (x)


6

a The figure at right has a positive gradient where:

A −1 < x < 2

B x < −1 only

C x > 2 only

D x < −1 and x > 2

E x > 0

b The figure above has a negative gradient where:

c The graph of the gradient function for the figure above is:

7 For the functions graphed below, state the domain (where applicable) where thegradient:

i is equal to zero ii is positive iii is negative iv does not exist.

A x > −1 B x < 2 C −1 < x < 2

D x < −1 and x > 2 E x < 0

A B C

D E

a b c

d e f


y

x0

f (x)

–1 2

y

x0 2–1

f '(x)

y

x0 2–1

f '(x)y

x0 2–1

f '(x)

y

x0 2–1

f '(x)

y

x0 2–1

f '(x)

y

x0

1

–1

f (x)y

x0

6

2f (x)

y

x0

5

–1 g(x)

y

x0

–4

4

g(x) y

x0

g (x)

Gradient = 0

y

x0–2 3f (x)


8 Sketch the gradient function for each function in question 7.9 For each function f (x) graphed below, state the domain of the gradient function f ′(x).

(Do not sketch the graph of f ′(x).)

g

j

h i

a b c

d e f

g h i

j

y

x02

1

f (x)

Gradient = 0

y

x0 4–2

2

4

f (x)

y

x0–1 2

g(x)

y

x0

f (x)

WORKEDExample

4

y

x0 8

–3

2

2

f (x)

y

x0

–4

2

f (x)

y

x0

f (x)

y

x0–1 2

f (x)

y

x0 3

f (x)

y

x0

f (x)

y

x0

f (x)

–2

y

x0 2

f (x)y

x0 5

f (x)

y

x0 4–5

f (x)


Limits and differentiation from first principlesThe limit of a function is the value that the function approaches as x approaches a givenvalue.

If the function is continuous at the point in question then the limit exists and can be found by direct substitution.

For example, the limit of f (x) = 3x + 1 as x approaches 1 is denoted as

In this example f (x) is a continuous function at x = 1.(In fact, it is continuous for all values of x.)Therefore, the limit is found by direct substitution, that is,

= 3(1) + 1 = 4

If the function is discontinuous at the point in question then the limit exists if thefunction is approaching the same value from the left as it is from the right.

y

x0 1

4 Continuous

1

f (x)

3x 1+( )x 1→lim

3x 1+( )x 1→lim

For example:

The limit exists at x = 2, as the functionis approaching 3 from the left and fromthe right. Therefore:

For example:

The limit does not exist at x = −2, as thefunction is approaching 3 from the left and5 from the right. That is, if the left-handlimit is not equal to the right-hand limit,then the limit as x → −2 does not exist.

y

x0 2

3Discontinuous

f (x)

f x( )x 2→lim 3=

y

x0

3

5

–2

Discontinuousat x = –2

f (x)

Evaluate the following limits.

a b

THINK WRITE

a Decide whether f (x) is continuous at x = 5. If so, substitute x = 5 into f (x).

a = 52 − 3(5)

Evaluate. = 10

b Decide whether f (x) is continuous at x = 0. If so, substitute x = 0 into f (x).

b =

Evaluate. =

= 2

x2 3x–( )x 5→lim

x2 5x 6+ +x 3+---------------------------

x 0→lim

1 x2 3x–( )x 5→lim

2

1x2 5x 6+ +

x 3+---------------------------

x 0→lim

02 5 0( ) 6+ +0 3+

--------------------------------

263---

5WORKEDExample

C h a p t e r 7 D i f f e r e n t i a t i o n 271The limit of rational expressions can be evaluated, if direct substitution makes the

denominator zero, by first simplifying the expressions and then using directsubstitution.

Differentiation using first principlesConsider the chord (straight line) to the curve below.

The gradient of =

=

As Q moves along the curve towards P, the value of h gets smaller and smaller. Or asQ gets as close as possible to P, h → 0, and becomes a tangent at P. The gradientof the curve at a point P is the gradient of the tangent at that point.

That is, gradient at point P is f ′(x) = h ≠ 0.

Finding the gradient this way is known as differentiation from first principles.Differentiating f (x) gives f ′(x) or f ′(x) is the derivative of f (x).

Differentiating y gives , or is the derivative of y with respect to x.

Evaluate .

THINK WRITE

If the limit can’t be found with f (x) in the given form, factorise the numerator.

=

Simplify by cancelling. = , x ≠ −3

Substitute x = −3. = −3 + 2

Evaluate. = −1

x2 5x 6+ +x 3+---------------------------

x 3–→lim

1x2 5x 6+ +

x 3+---------------------------

x 3–→lim x 3+( ) x 2+( )

x 3+----------------------------------

x 3–→lim

2 x 2+( )x 3–→lim

3

4

6WORKEDExample

PQ

y

x0 x x + h

f (x+h)

f (x+h) – f (x)

f (x)

f (x)Q

Ph

PQriserun--------

f x h+( ) f x( )–h

--------------------------------------

PQ

f x h+( ) f x( )–h

--------------------------------------h 0→lim

dydx------ dy

dx------


Find the gradient of the chord drawn to the curve f (x) = x2 + 2 in the diagram.

THINK WRITE

Find the gradient and simplify.

. Gradient = , h ≠ 0

=

=

=

=

=

= 2 + hThe gradient is 2 + h

PQy

x0

Q [1 + h, f (1 + h)]

P [1, f (1)]

Gradientriserun--------= f x h+( ) f x( )–

h--------------------------------------

f 1 h+( ) f 1( )–h

--------------------------------------

1 h+( )2 2 12 2+( )–+h

------------------------------------------------------

3 2h h2 3–+ +

h------------------------------------

2h h2+

h------------------

h 2 h+( )h

--------------------

7WORKEDExample

Find the gradient of f (x) = x2 − 1 at the point where x = 2 by:a sketching a graph and finding the gradient of the tangent at x = 2b differentiating using first principles.

THINK WRITE

a Sketch the graph of f (x) over a domain which includes the given value of x.

a

Construct a tangent at the given point on the curve. (It is difficult to be accurate.)

Find the gradient of this tangent by

evaluating .

Gradient of tangent at x = 2 is

approximately .

1 y

x0 1 2 3 4 5

12345

–5–4–3–2

678

(3, 6)

(0, –5)3

11

2

3

riserun-------- 11

3------ 32

3---=

8WORKEDExample


Note: Answer a (3 ) is very close to answer b (4).

THINK WRITE

b Find f ′(x) using first principles. b f ′(x) = , h ≠ 0

=

=

=

=

=

= 2x

Evaluate f ′(2) to find the gradient at x = 2. f ′(2) = 2(2)f ′(2) = 4

1f x h+( ) f x( )–

h--------------------------------------

h 0→lim

x h+( )2 1– x2 1–( )–h

-----------------------------------------------------h 0→lim

x2 2xh h2 1– x2 1+–+ +h

--------------------------------------------------------------h 0→lim

2xh h2+h

---------------------h 0→lim

h 2x h+( )h

-----------------------h 0→lim

2x h+( )h 0→lim

2

23---

Use first principles to differentiate g(x) = x2 − x.

THINK WRITE

a Find g(x + h) and simplify. a g(x + h) = (x + h)2 − (x + h)= x2 + 2xh + h2 − x − h

b Find g ′(x) using first principles. b g ′(x) = ,

=

=

=

=

=

= 2x − 1

g x h+( ) g x( )–h

------------------------------------- h 0≠,h 0→lim

x2 2xh h

2x– h– x

2x–( )–+ +

h---------------------------------------------------------------------------

h 0→lim

x2 2xh h

2x– h– x

2x+–+ +

h-----------------------------------------------------------------------

h 0→lim

2xh h2

h–+h

------------------------------h 0→lim

h 2x h 1–+( )h

--------------------------------h 0→lim

2x h 1–+( )h 0→lim

9WORKEDExample


Limits and differentiation from first principles

1 Evaluate the following limits.

2 By first simplifying the rational expression, evaluate the following limits.

a b c

d e f

g h i

j

a b c

d e f

g h i

j

remember1. The limit of a function is the value that y approaches as x approaches a given

value.2. If a function is continuous at a point then a limit exists at that point and can be

found by direct substitution.3. If a function is discontinuous at a point then the limit exists only if the function

is approaching the same value from both left and right. If there is a break in the curve and the point lies within the break then the limit does not exist.

4. The limit of expressions with a denominator can be found.(a) If direct substitution makes the denominator equal to zero then factorise the

numerator, cancel and then use direct substitution.(b) If direct substitution does not make the denominator equal to zero then use

direct substitution.

5. The gradient of a chord or secant is found using .

6. The gradient at a point P on a curve, is f ′(x) = , h ≠ 0.

7. The gradient at a point P on a curve is the gradient of the tangent to the curve at that point.

f x h+( ) f x( )–h

--------------------------------------

f x h+( ) f x( )–h

--------------------------------------h 0→lim

remember

7B

Mathca

d

Limits

WORKEDExample

5 x 4+( )x 2→lim 2 p 3–( )

p 2–→lim 8 3h–( )

h 0→lim

x2 5–( )x 3→lim x2 4x 3–+( )

x 1–→lim x3 5x– 2+( )

x 3→lim

2m2 5m– 3–( )m 4→lim 10 x– x2 x3–+( )

x 2–→lim x2 5x 6+ +

x 2+---------------------------

x 0→lim

x2 2x– 3–x 3–

--------------------------x 1→lim

WORKEDExample

6 x2 3x+x

-----------------x 0→lim x3 4x2 5x+–

x--------------------------------

x 0→lim 3x2 3x–

x 1–--------------------

x 1→lim

2x2 2x+x 1+

---------------------x 1–→lim x2 4–

x 2–--------------

x 2→lim x2 3x 2+ +

x 1+---------------------------

x 1–→lim

x2 5x– 6–x 6–

--------------------------x 6→lim h3 8–

h 2–--------------

h 2→lim x3 27+

x 3+-----------------

x 3–→lim

x2 4x 5–+x 5+

--------------------------x 5–→lim

C h a p t e r 7 D i f f e r e n t i a t i o n 2753 Evaluate the following.

Questions 4 and 5 refer to the following diagram. Consider the chord drawn to thecurve f (x) as shown.

4 Find the gradient of the chord drawn to the curve f(x) = x2 + 1 in thediagram at right.

5

The gradient of the tangent at P in the diagram is:A h B 4 + h C 4hD 4 + h2 E 4

6 a Find the gradient of the chord to the function f (x) = x(x + 2) if thex-coordinates of P and Q are −1 and −1 + h respectively.

b Hence, find the gradient of the function at P.

7 Find the gradient of g(x) = 4 − x2 at the point where x = 2 by:a sketching a graph and finding the gradient of the tangent at x = 2b differentiating using first principles.

8 Find the gradient of h(x) = 2x2 − 6x at x = −1 by using:a a sketch graph to find the gradient of the tangent at x = −1b differentiation from first principles.

9

The gradient of a function f (x) at the point where x = 3 is:

10 Use first principles to differentiate f (x) if:

11 Use first principles to find if:

a b c

d e f

g h i

A B C

D E

a f (x) = 3x + 5 b f (x) = x2 − 3 c f (x) = x2 + 6xd f (x) = (x − 4)(x + 2) e f (x) = 8 − 3x2 f f (x) = x3 + 2

a y = 9 − 4x b y = x2 + 3x c y = x2 − 2x + 7d y = 3x2 + 8x − 5 e y = x3 − 4x f y = 5x − 2x3

g y = h y = −x2 − 2x

3x 4–( )x 3→lim x2 9–

x 3+--------------

x 3–→lim x2 x 6–+

x 2–-----------------------

x 2→lim

x3 4+x 2+--------------

x 1→lim h3 64–

h 4–-----------------

h 4→lim x3 x2 6–+( )

x 2–→lim

x2 3x+x 1–

-----------------x 3→lim x3 1+

x 1–--------------

x 1–→lim x

2 7x 12+ +x 4+

------------------------------x 4–→lim

PQ

WORKEDExample

7

PQy

x0 2

1

f (x) = x2+1

Q (2+h, f (2+h))

P (2, f (2))mmultiple choiceultiple choice

PQ

WORKEDExample

8


f x h+( ) f x( )–h

--------------------------------------h 0→lim f 3 h+( ) f x( )–

h--------------------------------------

h 0→lim f 3 h+( ) f 3( )–

h--------------------------------------

h 0→lim

f 3 h+( ) f 3( )–h

-------------------------------------- f x h+( ) f x( )–h

--------------------------------------

WORKEDExample

9Mathcad

Differentiationfrom firstprinciplesdy

dx------

x2

3-----


The derivative of xn

Instead of using the procedure of differentiating from first principles, rules can beapplied to find derivatives. These rules can be derived from first principles and havebeen looked at in detail in Mathematical Methods Units 1 and 2.

If f (x) = axn then f ′(x) = naxn − 1, where a and n are constants.If f (x) = c then f ′(x) = 0, where c is a constant. (This is because c = x0 and, using the

rule, the derivative of x0 is 0 × x−1 or 0).

For example, if y = x7 then = 7x6. If f(x) = 5x4 then f ′(x) = 20x3.

If f (x) = g(x) + h(x) then f ′(x) = g ′(x) + h ′(x), that is, differentiate each term of afunction separately.

If f (x) = a × g(x), where a is a constant, then f ′(x) = a × g ′(x).

dydx------

Differentiate .

THINK WRITE

Write the equation.

Differentiate each of the three terms separately.

y x4 32--- x2 7+–=

1 y x4 32---x2 7+–=

2 dydx------ 4x4 1– 3

2--- 2( )x2 1–– 0+=

4x3 3x–=

10WORKEDExample

Find the derivative of:

a b .

THINK WRITE

a Write the equation. a

Rewrite and using negative indices.

Differentiate each term.

Write the function in the form originally given.

f x( ) 1x---

1

x-------+= f x( ) x x+

x2-----------------=

1 f x( ) 1x--- 1

x-------+=

21x--- 1

x------- f x( ) x 1– x

12---–+=

3 f ′ x( ) x 1– 1–– 12--- x

12--- 1––

–=

x 2–– x32---–

2-------–=

4 1x2-----– 1

2 x3------------–=

11WORKEDExample


THINK WRITE

b Write the equation. b

Rewrite using indices.

Separate the function into two terms expressed in index form.

Simplify each term.

Differentiate each term.

Simplify f ′(x).

1 f x( ) x x+x2

----------------=

2 x x x12---+

x2--------------=

3xx2----- x

12---

x2-----+=

4 x 1– x32---–+=

5 f ′ x( ) 1x 1– 1––32---x

32---– 1–

–=

x 2––32---x

52---–

–=

61x2-----– 3

2x52---

--------–=

1x2-----–= 3

2 x5------------–

If , find a f ′(x) b f ′(2).

THINK WRITE

a Write the equation. a f (x) = x3 − 2x2 +

Express f (x) so each term is in index form.

= x3 − 2x2 + 8x−1

Differentiate f (x) to obtain the gradient function f ′(x).

f ′(x) = 3x2 − 4x − 8x−2

Simplify f ′(x). = 3x2 − 4x −

b Evaluate f ′(2). b f ′(2) = 3(2)2 − 4(2) −

= 12 − 8 − 2= 2

Therefore the gradient of f (x) is 2 when x = 2.

f x( ) x3 2x2–8x---+=

18x---

2

3

48x2-----

822-----

12WORKEDExample


The derivative of xn

1 Find the derivative of each of the following.

2 Differentiate each of the following.


a y = x6 b y = 3x2 c y = 5x4 d y = x20

e y = −4x3 f y = −5x g y = x3 h y =

i y = 10 j y = 8x5

a f (x) = 4x3 + 5x b g(x) = −5x2 + 6x + 1

c h(x) = 9 + d h(x) = 4 − 3x + 6x2 + x3

e g(x) = 7x11 + 6x5 − 8 f f (x) =

g f(x) = –6x + 3x2 – 4x3 h g (x) = 7x2 − 4x +

i h(x) = (x + 4)(x − 1) j f (x) = (x2 + 2x)(3x − 6)

a b c

d e f

g h i

j k l

m x3 − 4x + x−3 n

remember1. Rules for differentiating axn

(a) If f (x) = axn then f ′(x) = naxn − 1, where a and n are constants.(b) If f (x) = c then f ′(x) = 0, where c is a constant.(c) If f(x) = ag(x), where a is a constant then f ′(x) = ag′(x).

2. If f (x) = g(x) + h(x) then f ′(x) = g′(x) + h′(x).Differentiate each term of a function separately.

3. To evaluate f (a) where a is a constant, replace every x in the f (x) equation with the value of a. For example, if f (x) = 2x2 − 3x + 1, f (2) = 2 × 22 − 3 × 2 + 1 = 3.

remember

7C

Mathca

d

Derivatives12---

x4

3-----

WORKEDExample

10x3

5-----

2x5

5-------- x3

3----- 10+ +

23---

SkillSH

EET 7.3 WORKEDExample

11 2x3----- 3 x x

13---

4x54---

x 2x2–1x--- x2+

x12---–

x23---+ x 3+

x------------ x2 x3+

x----------------

34x------ 2

5x2-------- 2

x------- 3x 2–+

13---

x x14---+

x--------------

C h a p t e r 7 D i f f e r e n t i a t i o n 2794 If f (x) = 2x5 − 10x + 5 find:

5

If f (x) = x2 − 6x then f ′(4) is equal to:

6

The value of f ′(9) if f (x) = x2 + − 10x is:

7 Find g ′(−2) if g(x) = .

8 Find the gradient of the curve y = at the point where a x = 2 and b x = 0.

9 Find the gradient of f (x) = 2x3 − x2 + at the point where x equals:

10 If g(x) = , find a g ′(x) b g ′(1) c g ′(8) d g ′(−8)

11 Show that the derivative of y = k, where k is a constant, is zero.

12 For each of the following:i expand the bracketsii differentiate the expanded expressioniii factorise.

13 Using the results of question 12 give the derivative of (ax + b)n in factorised form. (a, b, n are constants.)

The gradient of a function at a particular point can be found by using the nDerivfunction on the graphics calculator. nDeriv(Y1, x, 2) will calculate the gradient of thefunction entered as Y1 at the point when x = 2.

Or, to find the gradient of f (x) = x3 − 2x2 + at x = 2, follow these steps.

1. Press and 8 to access the nDeriv function.2. Enter the equation as shown and press .3. The answer 2.000 000 5 will appear, as the nDeriv

function gives an approximate answer only.

a f ′(x) b f ′(2)

A 8 B −12 C 12 D 2 E −16

A B 18 C 12 D 8 E 0

a 1 b 4 c 9

a (x + 1)2 b (x + 1)3 c (2x + 1)2

d (2x + 1)3 e (3x + 1)2 f (3x + 1)3

WORKEDE

12



x32---

12--- 1

2---

1x2----- 3x 8–+

5x4-----

1

x-------

x3 4x+

Graphics CalculatorGraphics Calculator tip!tip! Finding the gradient of a function at aparticular point

8x---

MATHENTER

xample


The chain ruleA function which can be expressed as a composition of two simpler functions is calleda composite function. For example, y = (x + 3)2 can be expressed as y = u2 where u = x + 3.

That is, to obtain y from x, the first function to be performed is to add 3 tox (u = x + 3), then this function has to be ‘squared’ (y = u2).

Or if x = 1, to obtain y first calculate 1 + 3 (= 4), then secondly ‘square’ the result,42, giving y = 16.

Composite functions can be differentiated using the chain rule. For example, usingthe previous function, y = (x + 3)2:

Let u = x + 3, so y = u2.

Then and .

But we require and . This is known as the chain rule. It is known

as the chain rule because u provides the ‘link’ between y and x.

Now = 2u × 1

= 2(x + 3) × 1 (replacing u with x + 3)= 2(x + 3)

The chain rule is used when it is necessary to differentiate a ‘function of a function’as above.

dudx------ 1= dy

du------ 2u=

dydx------ dy

dx------ dy

du------ du

dx------×=

dydx------

If y = (3x − 2)3 is expressed as y = un, find:

a b and hence c .

THINK WRITE

a Write the equation. a y = (3x − 2)3

Express y as a function of u. Let y = u3 where u = 3x − 2

Differentiate y with respect to u.

b Express u as a function of x. b u = 3x − 2

Differentiate u with respect to x.

c Find using the chain rule. c =

= 3u2 × 3= 9u2

Replace u as a function of x. =

dydu-------

dudx-------

dydx------

12

3dydu------ 3u2=

1

2dudx------ 3=

1dydx------ dy

dx------ dy

du------ du

dx------×

2 9 3x 2–( )2

13WORKEDExample


A quicker way to apply the chain rule when a function can be expressed in indexform is as follows.

If f (x) = [g(x)]n then f ′(x) = n[g(x)]n − 1 × g '(x). That is, differentiate the bracket andthen what is inside the bracket; ‘outside then inside’.

If find f ′(x).

THINK WRITE

Write the equation.

Express f (x) in index form, that is, as y = [g(x)]n.

Express y as a function of u. Let where u = 2x2 − 3x


Express u as a function of x. u = 2x2 − 3x


Find f ′(x) using the chain rule. =

Replace u as a function of x and simplify.

=

=

=

f x( ) 1

2x2 3x–-------------------------=

1 f x( ) 1

2x2 3x–-------------------------=

2 y 2x2 3x–( )12---–=

3 y u12---–=

4dydu------ 1

2---u

32---–

–=

5

6dudx------ 4x 3–=

7dydx------ f ′ x( )= 1

2---u

32---–

4x 3–( )×–

812--- 4x 3–( ) 2x2 3x–( )

32---–

–

4x 3–( )–

2 2x2 3x–( )32---

-------------------------------

3 4x–

2 2( x2 3x )– 3-------------------------------------

14WORKEDExample

Find the derivative of f (x) = (x2 − 2x)3.

THINK WRITE

Write the equation. f (x) = (x2 − 2x)3

Let g(x) equal what is inside the bracket. g(x) = x2 − 2xFind g′(x). g′(x) = 2x − 2Use the rule f ′(x) = n[g(x)]n − 1 × g′(x) to differentiate f (x).

n = 3, f ′(x) = 3(x2 − 2x)3 − 1 × (2x − 2)= 3(x2 − 2x)2 × [2(x − 1)]= 6(x − 1)(x2 − 2x)2

Simplify f ′(x) as far as possible. = 6(x − 1)[x(x − 2)x(x − 2)]= 6x2(x − 1)(x − 2)2.

1234

5

15WORKEDExample


The chain rule

1 If each of the following functions are expressed in the form y = un, statei u and ii n.

2

If y = (x + 3)5 is expressed as y = u5 then:

3 If each of the following composite functions are expressed as y = un, determine:

i ii and hence iii .

For questions 4, 5 and 6 below, y = is expressed as y = un.

4

=

a y = (5x − 4)3 b y = c

d e y = (5x + 3)−6 f

A u = (x + 3)5 B u = x + 3 C u = xD u = 3 E u = x5

a b c

d e y = f

g y = 3(2x2 + 5x)5 h y = (4x − 3x2)−2 i

j y = 4(5 − 6x)−4

A B C

D E

remember1. A composite function is a function composed of two (or more) functions.

2. Composite functions can be differentiated using the chain rule, .

3. A short way of applying the chain rule is:If f (x) = [g(x)]n then f ′(x) = n[g(x)]n − 1 × g ′(x).

dydx------ dy

du------ du

dx------×=

remember

7D

Mathca

d

Chain rule 3x 1+ y

12x 3+( )4

----------------------=

y1

7 4x–---------------= y 4 3x–( )

43---=


WORKEDExample

13 dydu------ du

dx------ dy

dx------

y 3x 2+( )2= y 7 x–( )3= y1

2x 5–---------------=

y1

4 2x–( )4----------------------= 5x 2+ y

3

3x 2–-------------------=

y x1x---+

6=

x2 3x– 2+


dydu------

12---u u

12---– 1

2 u----------

12---u

32--- 1

2---u

12---


=

6

Using the chain rule, is equal to:

7 Use the chain rule to find the derivative of the following.

8 If , find f ′(x).

9 Use the chain rule to find the derivative of the following. (Hint: Simplify first usingindex notation and the laws of indices.)

a b

10 Find the derivative of:

a f (x) = (x2 + 5x)8 b y = (x3 − 2x)2

c d

11 If f (x) = (2x − 1)6, find f ′(3).

12 If g(x) = (x2 − 3x)−2, find g′(−2).

13 If , find:

14 Find the gradient of the function h(x) = at the point where x = 2.

15 Find the value of f ′(−1) if f (x) = .

A 2x − 3 B x2 − 3x + 2 C x2 − 3x

D x2 − x + 1 E x − 3

A B C

D E

a y = (8x + 3)4 b y = (2x − 5)3

c f (x) = (4 − 3x)5 d y =

e f (x) = f g(x) = (2x3 + x)−2

g h y = (x2 − 3x)−1

a f (3) b f ′(x) c f ′(3) d f ′(x) when x = 2.


dudx------

12--- 3

2---


dydx------

2x 3–u

--------------- 2x 3–

2 x2 3x– 2+---------------------------------- x2 3x– 2+

2 u--------------------------

12--- 2x 3–( ) 1

2--- 2x 3–( ) x2 3x– 2+( )

12---

3x2 4–

x2 4x–( )13---

g x( ) x1x---–

6=

WORKEDExample

14

f x( ) 1

4x 7+-------------------=

y6x 5–

6x 5–-------------------= f x( ) x2 2+( )2

x 2+---------------------=

WORKEDExample

15

f x( ) x3 2x2 7–+( )15---= y 2x

4 3x2– 1+( )

32---=

f x( ) x2 2x– 1+=

3x2 2x+

WorkS

HEET 7.13

5 4x–-------------------


The derivative of ex

If f (x) = ex then using first principles

f ′(x) = , h ≠ 0

=

=

=

=

Note that can be deduced by using a calculator and substituting values of h

close to zero.

hThat is,

Therefore, f ′(x) = ex × 1= ex

If f (x) = ex then f ′(x) = ex.

0.01 1.0050

0.0001 1.000 05

0.000 001 1.000 000

f x h+( ) f x( )–h

--------------------------------------h 0→lim

ex h+

ex–

h----------------------

h 0→lim

exe

he

x–h

---------------------h 0→lim

ex eh 1–( )h

------------------------h 0→lim

ex eh 1–h

--------------h 0→lim

eh 1–h

--------------h 0→lim

eh 1–h

-------------- eh 1–h

--------------h 0→lim 1.=

Differentiate y = e−5x.

THINK WRITE

Write the equation. y = e−5x

Express u as a function of x and find

.

Let u = −5x so

Express y as a function of u and find

.

y = eu so

Find using the chain rule.

Replace u as a function of x.

1

2

dudx------

dudx------ 5–=

3

dydu------

dydu------ eu=

4dydx------ dy

dx------ 5eu–=

5 5e 5x––=

16WORKEDExample

C h a p t e r 7 D i f f e r e n t i a t i o n 285This example shows that if f (x) = ekx then f ′(x) = kekx.

Find the derivative of y = e2x + 1.

THINK WRITE

Write the equation. y = e2x + 1


.

Let u = 2x + 1 so


.

y = eu so



1

2

dudx------

dudx------ 2=

3

dydu------

dydu------ eu=

4dydx------ dy

dx------ e

u 2×=

2eu=

2e2x 1+=5

17WORKEDExample

Differentiate a f (x) = ex(ex − 2) b .

THINK WRITE

a Write the equation. a f (x) = ex(ex − 2)

Expand. = e2x − 2ex

Differentiate. f ′(x) = 2e2x − 2ex

Factorise in order to leave the answer in the form it was given.

= 2ex(ex − 1)

b Write the equation. b

Write each term in the numerator over each term in the denominator.Divide the numerator of each term by its denominator using the laws of indices.

Differentiate each term. f ′(x) = ex + 4e−2x

Write your answer in the form it was given.

f x( ) e2x 2e x––ex

------------------------=

1

2

3

4

1 f x( ) e2x 2e x––ex

------------------------=

2e2x

ex-------= 2e x–

ex----------–

3 e2x x–= 2e x– x––

ex 2e 2x––=

4

5 ex=4

e2x-------+

18WORKEDExample


This example shows that if f (x) = eg(x) then f ′(x) = g ′(x) eg(x).

The following screen shows how a Mathcad file can be used to obtain the result inworked example 19.

Find the derivative of .

THINK WRITE

Write the equation.


.

Let u = x3 − x so = 3x2 − 1


.

y = eu so = eu



y ex3 x–=

1 y ex3 x–=

2

dudx------

dudx------

3

dydu------

dydu------

4dydx------ dy

dx------ e

u 3x2 1–( )×=

3x2 1–( )e

u=

3x2 1–( )e

x3

x–=5

19WORKEDExample

Mathca

d

Derivatives


The derivative of ex

1 Differentiate each of the following using the chain rule.


3The derivative of y = e3x + 2 is equal to:A 3e3x + 2 B (3x + 2)e3x + 2 C 3e3x

D 3xe3x + 2 E 3xe3x


5 Find the derivative of each of the following:

6The derivative of is equal to:

7 If f (x) = 5e9 − 4x, find the exact value of f ′(2).

8 If , give the exact value of g ′(0).

9 Find the exact value of h′(−1) if .

a y = e10x b cd y = e−x e y = 2e3x f y = 4e−5x

g y = −6e−2x h y = 5e0.2x i y = −2e−11x

a y = e6x − 2 b y = e8 − 6x c y = 2e5x + 3

d y = 4e7 − 2x e y = −3e8x + 1 f y = −2e6 − 5x

g y = 10e6 − 9x h y = −5e3x + 4 i y = 6e−7x

j k l

a b c

d e f

g h

a b cd e fg h i

j k l

m n

A B C

D E

remember1. If f (x) = ex, f ′(x) = ex. 2. If f (x) = ekx, f ′(x) = kekx.3. If f (x) = aekx + c, f ′(x) = akekx + c. 4. If f (x) = aeg(x), f ′(x) = g ′(x) × aeg(x).

remember

7EMathcad

Derivatives

WORKEDExample

16 y e13--- x= y e

x4---=

WORKEDExample

17

y 2ex2--- 1+= y 3e

2 x3---–= y 4e

x4--- 5+

–=


WORKEDExample

18 f x( ) 2 ex 1+( )= f x( ) 3e2x ex 1+( )= f x( ) 5 e 4x– 2x+( )=

f x( ) ex 2+( ) e x– 3+( )= f x( ) 3e3x e 6x–+ex

--------------------------= f x( ) 4e7x 2e x––e 2x–

---------------------------=

f x( ) ex e2+= f x( ) 4e5x 2x2 e 3––+=WORKEDExample

19 y ex2 3x+= y ex2 3x– 1+= y ex2 2x–=f x( ) e2 5x–= f x( ) e6 3x– x2+= g x( ) ex3 3x 2–+=h x( ) 3e4x2 7x–= y 5e1 2x– 3x2––= y e 2x 1+( )3=

f x( ) e 4 x–( )4= g x( ) e x 2+( ) 2–= y e 3x 4+=

f x( ) e x 1+( )13---

= h x( ) e x2 3x+( )2=


6ex3 5x–

3x2 5– 6 3x2 5–( )ex3 5x– 3x2 5–( )ex3 5x–

6 x3 5x–( )ex3 5x– 6 3x2 5–( )e3x2 5–

g x( ) 2ex2 3x– 2+=

h x( ) 5ex3 2x+–=


The derivative of loge xThe inverse of the function f (x) = ex is f −1(x) = loge x.If y = loge x then ey = x as shown in chapter 3, Exponential and logarithmic equations.Let x = ey

= ey

But = and ey = x

Therefore, = .

=

That is, if f (x) = loge x then f ′(x) = .

If f (x) = loge kx, where k is a constant then f ′(x) = .

dxdy------

dydx------ 1

dxdy------

---------

dydx------ 1

ey

----

1x---

1x---

Differentiate y = loge 7x.

THINK WRITE

Write the equation. y = loge 7x

Express u as a function of x and find .u = 7x, so

Express y as a function of u and find .y = loge u, so

Find . = × 7

1

2dudx------

dudx------ 7=

3dydu------

dydu------ 1

u---=

4dydx------ dy

dx------ 1

7x------

1x---=

20WORKEDExample

1x---

Find the derivative of y = 2 loge (3x − 4).

THINK WRITE

Write the equation. y = 2 loge (3x − 4)Express u as a function of x. Let u = 3x − 4.

12

21WORKEDExample


This example shows that if f (x) = loge [g(x)] then f ′(x) = .

THINK WRITE


Express y as a function of u. y = 2 loge u



Replace u with 3x − 4.

3dudx------ 3=

4

5dydu------ 2

1u---×=

2u---=

6dydx------ dy

dx------ 2

u--- 3×=

6u---=

76

3x 4–---------------=

Differentiate y = loge (x2 + 4x − 1).

THINK WRITE

Write the equation. y = loge (x2 + 4x − 1)Let u equal the section in brackets. Let u = x2 + 4x − 1


Express y as a function of u. y = loge u



Replace u with what is in the brackets.

12

3dudx------ 2x 4+=

4

5dydu------ 1

u---=

6dydx------ dy

dx------ 1

u--- 2x 4+( )×=

7 2x 4+

x2 4x 1–+--------------------------=

22WORKEDExample

g′ x( )g x( )------------

remember1. If f (x) = loge x, then f ′(x) = .

2. If f (x) = loge kx, then f ′(x) = .

3. If f (x) = loge (g(x)), then f ′(x) = .

1x---

1x---

g′ x( )g x( )------------

remember


The derivative of loge x

1 If y = loge 4x is expressed as y = loge u, then find:

2 Differentiate each of the following.

a y = loge 10x b y = loge 5x

c y = loge (−x) d y = loge (−6x)

e y = 3 loge 4x f y = −6 loge 9x

g y = loge h y = loge

i y = 4 loge j y = −5 loge

3

The derivative of loge 8x is:

4

To differentiate y = loge (3x + 7) using the chain rule:

a ‘u’ would be used to represent

b and are respectively

c Hence is equal to


a y = loge (2x + 5) b y = loge (6x + 1)

c y = loge (3x − 4) d y = loge (8x − 1)

a u b c d using the chain rule.

A 8 B C D E loge 8

A 3x + 7 B 3x C loge x D loge 3x E x

A and 3x + 7 B and 3x C 3 and

D and 3 E 1 and 3

A 3 B C

D E

7F

Mathca

d

Derivative of loge x

dudx------ dy

du------ dy

dx------

WORKEDExample

20

x2---

x3---

x5---

2x3

------–


18---x

8x--- 1

x---


dydu------ du

dx------

1u--- 1

u--- 1

3x------

1u---

dydx------

1x--- 3

3x 7+---------------

13x 7+--------------- 3

x---

WORKEDExample

21

C h a p t e r 7 D i f f e r e n t i a t i o n 291e y = loge (3 − 5x) f y = loge (2 − x)

g y = loge (4 − 7x) h y = 6 loge(5x + 2)

i y = 8 loge(4x − 2) j y = −4 loge(12x + 5)

k y = −7 loge(8 − 9x)

6 Differentiate the following.

7

Using the chain rule the derivative of f (x) = loge (x2 − 5x + 2) would be:

8 Find the gradient of the function f (x) = 6 loge (4 − 3x) when x = −1.

9 If g(x) = 3 loge (3x + 5) find the value of g′(0).

10 Find the exact value of f ′(2) if f (x) = 3x2 + 4 loge (x2 + x).

11 If , find:

a

b the exact gradient when i x = 1 ii x = 2 iii x = 4 iv x = 10.Can you explain this result?

12 If , find:

a f ′(x)

b the exact value of i f ′(1) ii f ′(5) iii f ′(−2).

a y = loge 3x4 b y = loge (x2 + 3)

c y = loge (x2 + 4x) d y = loge (x2 − 3x + 2)

e y = loge (x3 + 2x2 − 7x) f y = loge (x2 − 2x3 + x4)

g y = loge h y = loge

i y = loge j y = loge

k y = loge l f (x) = loge

m f (x) = loge n f (x) = loge (3x − 2)4

o f (x) = loge (5x + 8)−2 p f (x) = loge

A B C

D E

WORKEDExample

22

2x 1+ 3 4x–

x2 2+ x 3+( )14---

5x 2+( )13---

2 3x–( )15---

1x 3+------------

24 3x+---------------


1x2 5x– 2+-------------------------- 5–

x2 5x– 2+-------------------------- 2x 5–

1x 2x 5–( )-----------------------

2x 5–x2 5x– 2+--------------------------

y eloge x=dydx------

f x( ) eloge x2=


The derivatives of sin x and cos xThe derivatives of sin x and cos x can be obtained by differentiation from first prin-ciples but are beyond the requirements of this course.

If f (x) = sin x then f ′(x) = cos x and if f (x) = cos x then f ′(x) = −sin x.

This example shows that if f (x) = sin ax, then f ′(x) = a cos ax. Similarly if f (x) = cos ax, then f ′(x) = −a sin ax.

This example shows that the chain rule can be applied as follows.If f (x) = sin [g(x)] then f ′(x) = g ′(x) cos [g(x)]. If f (x) = cos [g(x)] then f ′(x) = −g ′(x) sin [g(x)].

Find the derivative of y = sin 5x.

THINK WRITE

Write the equation. y = sin 5x

Express u as a function of x and find .Let u = 5x so

Express y as a function of u and find .y = sin u so


Replace u with 5x.

1

2dudx------

dudx------ 5=

3dydu------

dydu------ cos u=

4dydx------ dy

dx------ 5 cos u=

5 5 cos 5x=

23WORKEDExample

Differentiate y = cos (x2 + 2x − 3).

THINK WRITE

Write the equation. y = cos (x2 + 2x − 3)

Express u as a function of x and find .Let u = x2 + 2x − 3 so

Express y as a function of u and find .y = cos u so

Find using the chain rule. = −sin u × (2x + 2)

Replace u with the part in brackets in the rule and simplify.

= −2(x + 1) sin (x2 + 2x − 3)

1

2dudx------

dudx------ 2x 2+=

3dydu------

dydu------ sin u–=

4dydx------ dy

dx------

5

24WORKEDExample


The derivatives of sin x and cos x

1 Find the derivative of each of the following:


3

a The derivative of sin 6x is:

b The derivative of cos 4x is:

c The derivative of sin (−4x) is:

d The derivative of cos (−8x) is:

e The derivative of sin is:

4 If y = sin (4x + 3) is expressed as y = sin u, find:

a y = sin 8x b y = sin (−6x) c y = sin x

d y = sin e y = sin f y = sin

a y = cos 3x b y = cos (−2x) c y = cos

d y = cos 21x e y = cos (−7x) f y = cos

g y = cos h y = cos

A 6 cos 6x B 6 cos x C 6 sin x D −6 cos 6x E cos 6x

A 4 sin 4x B 4 sin x C −4 sin x D 4 cos 4x E −4 sin 4x

A 4 cos (−4x) B 4 cos (–x) C −4 cos 4x D −4 cos (−4x) E −4 sin 4x

A 8 cos (−8x) B 8 sin (−8x) C −8 sin (–8x) D −8 sin (−x) E 8 sin (−x)

A 5 cos B − cos C cos D sin E cos x

a b c using the chain rule.

remember1. If f (x) = sin x then f ′(x) = cos x2. If f (x) = cos x then f ′(x) = −sin x3. If f (x) = sin ax then f ′(x) = acos ax4. If f (x) = cos ax then f ′(x) = −asin ax5. If f (x) = sin [g(x)] then f ′(x) = g ′(x) cos [g(x)]6. If f (x) = cos [g(x)] then f ′(x) = −g ′(x) sin [g(x)]

remember

7GMathcad

Derivativeof sine and

cosine

WORKEDExample

23x3--- x

2---–

2x3

------

x3---

πx4

------

x8---–

2x5

------


16---

x5---

x5--- 1

5---

x5--- 1

5---

x5--- 1

5---

x5--- 1

5---

dydu------ du

dx------ dy

dx------


5 If y = cos (3x + 1) is expressed as y = cos u, find:




9 If f(x) = 3 sin (x2 + x) find f ′(1) (answer correct to 3 decimal places).

10 Find the gradient of the curve g(x) = 2 cos (x3 − 3x) at the point where x = 0.

11 For each of the following functions find:

i f ′(x) and

ii the exact value of f ′ .

a b c using the chain rule.

a y = sin (2x + 3) b y = sin (6 − 7x) c y = sin (5x − 4)

d y = sin e y = sin f y = 5π sin 2π x

g y = −4 sin

a y = cos (3x − 2) b y = cos (4x + 7) c y = cos (8 − x)

d y = cos (6 − 5x) e y = cos f y = cos

g y = 4π cos 10π x h y = −6 cos (−2x)

a cos (x2 − 4x + 3) b sin (10 − 5x + x2)

c sin (ex) d cos (x2 + 7x)

e cos (4x − x2) f sin (x2 + 3x)

g cos (loge x) h sin (e4x)

i sin (loge 3x) j cos

k sin [loge (2x − 1)] l sin (3e2x)m cos (2e3x) n 3 cos (loge 10x)

o 4 sin (x3 + 2x2) p −8 sin

q −2 cos r cos (x2 + 2x) + sin (3x − 9)

s sin (x2 − 4) − 3 cos (8 − 3x) t −5 cos + 6 sin + 4x3

a f(x) = esin x b f(x) = ecos x c f(x) = loge (sin x) d f(x) = loge (cos x)

dydu------ du

dx------ dy

dx------

3x 2+4

--------------- 8 7x–

3---------------

3x8

------

2x 3+3

--------------- 4x 1–

5---------------

WORKEDExample

24

1x---

3x5

------–

x4---

3x 7+10

--------------- 5 4x–

3---------------

π6---


The product ruleAny function which is a product of two simpler functions, for example,

1. f (x) = (x + 2)(x − 5) or 2. f (x) = (x2 − 5x + 6) sin (3x + 5)can be differentiated using the product rule of differentiation.

While example 1 could be expanded and then differentiated, example 2 cannot andtherefore can be differentiated only by using the product rule.

Product ruleIf y = uv then .

Or if f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x).

dydx------ u

dvdx------ v

dudx-------×+×=

If y = (3x − 1)(x2 + 4x + 3) is expressed as y = uv, find:

a u and v b and c using .

THINK WRITE

a Write the equation. a y = (3x − 1)(x2 + 4x + 3)Identify u and v, two functions of x which are multiplied together.

Let u = 3x − 1 and v = x2 + 4x + 3.

b Differentiate u with respect to x. b

Differentiate v with respect to x.

c Apply the product rule to find . c = u × + v ×

= (3x − 1)(2x + 4) + (x2 + 4x + 3) × 3Expand and simplify where possible. = 6x2 + 10x − 4 + 3x2 + 12x + 9

= 9x2 + 22x + 5

dudx-------

dvdx------

dydx------

dydx------ u

dvdx------ v

dudx-------+=

12

1dudx------ 3=

2dvdx------ 2x 4+=

1dydx------ dy

dx------ dv

dx------ du

dx------

2

25WORKEDExample

Find the derivative of y = loge 4x × sin (3x − 2).

THINK WRITE

Write the equation. y = loge 4x × sin (3x − 2)Identify u and v. Let u = loge 4x and let v = sin (3x − 2).

Find and . = 3 cos (3x − 2)

Find using the product rule. = loge 4x × 3 cos (3x − 2) + sin (3x − 2) ×

Simplify wherever possible. = 3 loge 4x × cos (3x − 2) + sin (3x − 2)

12

3dudx------ dv

dx------ du

dx------ 1

x---= dv

dx------

4dydx------ dy

dx------ 1

x---

51x---

26WORKEDExample


The product rule

1 If y = (x + 3)(2x2 − 5x) is expressed as y = u × v, find:a u and v

b and

c using the product rule, .

2 Find the derivative of:a y = 4x3 × loge 6x b g(x) = (3x − 2) loge 2x

3

The derivative of f (x) = x2 sin 2x is:

4 Use the product rule to differentiate each of the following.

5 If f (x) = (2x + 1) loge (x + 3), find the exact value of f ′ (1).

6 Find g′(0) if g(x) = 5e2x cos 4x.

7 Find the value of f ′(−2) if f (x) = (x2 + 2) sin (4 − 3x) (answer correct to 3 decimalplaces).

8 If g(x) = (6x + x2) ex − 3, find the exact value of g′(2).

A f ′(x) = 2x cos 2x B f ′(x) = 4x cos 2xC f ′(x) = 2x sin 2x + x2 cos 2x D f ′(x) = 2x sin 2x + 2x2 cos 2xE f ′(x) = 2x sin x + 2x2 cos x

a y = x cos x b y = 3x sin xc y = (5x − 2) ex d y = e3x (2 − 11x)e y = x5 cos (3x + 1) f y = 2x3 loge 7xg y = e−2x loge (2x − 5) h y = 8 sin 5x loge 5x

i y = 5 cos 2x sin x j y = sin cos x

k f (x) = e4x − 3 loge 6x l f (x) = 4e−5x sin (2 − x)

m f (x) = cos 6x n f (x) = −3x

o f (x) = 2x−3 sin (2x + 3) p f (x) = e−2x loge (3x2 + 5)q f (x) = (x2 + e3x)(4 − e−3x) r f (x) = (x3 + 7x2)(x2 − 4x + 1)s f (x) = (x2 − 6)(2 + 3x − x2) t f (x) = πx cos 2πx

remember1. If y = u × v then .

2. If f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x).

dydx------ u

dvdx------×= v

dudx------×+

remember

7H

Mathca

d

Product rule

WORKEDExample

25dudx------ dv

dx------

dydx------ dy

dx------ u

dvdx------ v

dudx------×+×=

WORKEDExample

26


4x3

------

1

x------- x e

WorkS

HEET 7.2


The quotient ruleThe quotient rule is used to differentiate functions which are rational expressions (thatis, one function divided by another). For example,

1. or 2.

Quotient rule

If y = then =

Or if f (x) = then f ′(x) = .

f x( ) x2 6x– 3+5x 2+

--------------------------= f x( ) e3x 8+

cos 6 x–( )--------------------------=

uv---

dydx------

vdudx-------× u

dvdx------×–

v2--------------------------------------

u x( )v x( )----------- v x( )u′ x( ) u x( )v′ x( )–

v x( )[ ]2-----------------------------------------------------

If y = is expressed as y = , find:

a u and v b and c .

THINK WRITE

a Write the equation. a y =

Identify u and v. Let u = 3 − x and v = x2 + 4x.

b Differentiate u with respect to x. b

Differentiate v with respect to x.

c Apply the quotient rule to obtain

.

c =

=

Simplify where possible,

factorising the final answer where appropriate.

3 x–x2 4x+------------------

uv---

dudx-------

dvdx------

dydx------

13 x–

x2 4x+-----------------

2

1dudx------ 1–=

2dvdx------ 2x 4+=

1

dydx------

dydx------

vdudx------× u

dvdx------×–

v2-------------------------------------

x2 4x+( ) −1 3 x–( ) 2x 4+( )–×

x2 4x+( )

2-------------------------------------------------------------------------------

2dydx------ –x2 4x– 12 2x 2x2–+( )–

x2 4x+( )2----------------------------------------------------------------=

–x2 4x– 12– 2x– 2x2+x2 4x+( )2

------------------------------------------------------------=

x2 6x– 12–x2 4x+( )2

-----------------------------=

x2 6x– 12–x2 x 4+( )2

-----------------------------=

27WORKEDExample


Compare the solution in the above worked example with the computer generated solution below.

Find the derivative of .

THINK WRITE

Write the equation.

Identify u(x) and v(x). Let u(x) = 2e3x.

Let v(x) = cos (2x − 3).

Differentiate u(x) and v(x) with respect to x.

u ′(x) = 6e3x

v ′(x) = −2 sin (2x − 3)

Apply the quotient rule to obtain f ′(x).

f ′(x) =

=

Simplify where possible. =

=

f x( ) 2e3x

cos 2x 3–( )------------------------------=

1 f x( ) 2e3x

cos 2x 3–( )------------------------------=

2

3

4v x( )u′ x( ) u x( )v′ x( )–

v x( )[ ]2-----------------------------------------------------

cos 2x 3–( ) 6e3x× 2e3x 2 sin 2x 3–( )–[ ]–cos 2x 3–( )[ ]2

-----------------------------------------------------------------------------------------------------------

56e3x cos 2x 3–( ) 4e3x sin 2x 3–( )+

cos 2x 3–( )[ ]2-------------------------------------------------------------------------------------------

2e3x 3 cos 2x 3–( ) 2 sin 2x 3–( )+[ ]cos2 2x 3–( )

--------------------------------------------------------------------------------------------

28WORKEDExample

Mathca

d

Quotient rule

remember1. If then .

2. If then .

yuv---= dy

dx------

vdudx------ u

dvdx------×–×

v2-------------------------------------=

f x( ) u x( )v x( )-----------= f ′ x( ) v x( )u′ x( ) u x( )v′ x( )–

v x( )[ ]2-----------------------------------------------------=

remember


The quotient rule

1 If y = is expressed as y = , find:

a u and v

b and

c using the quotient rule, = .

2 If is expressed as f(x) = , find:

a u(x) and v(x)

b u ′(x) and v ′(x)

c f ′(x) using the quotient rule.


4

If then h′(x) equals:

a b c

d e f

g h i

j k l

m n o

p q r

s t u

A B C

D E

7IMathcad

Quotientrule

WORKEDExample

27

x 3+x 7+------------ u

v---

dudx------ dv

dx------

dydx------ dy

dx------

vdudx------× u

dvdx------×–

v2-------------------------------------

f x( ) x2 2x+5 x–

-----------------= u x( )v x( )-----------

WORKEDExample

28 2xx2 4x–----------------- x2 7x 6+ +

3x 2+---------------------------

cos xex

------------

4x 7–10 x–--------------- 5 x2–

x32---

-------------- e2x

x-------

3x2

loge 4x------------------

sin 4xcos 2x---------------- loge x 1+( )

x2 2+-----------------------------

e3x 2+

cos 2x----------------

loge 5x 2–( )sin 3x 2–( )--------------------------------

4xx2 3x 2–+--------------------------

2x3 7x+e5x

--------------------- x2 5–

x--------------

sin xcos x-------------

e 3x–

3x 8+---------------

4 loge 8x

x2 2x–----------------------- sin x

x--------------

2 cos 3 2x–( )x2

----------------------------------- 3e2 7x–

x 3+----------------- e

2x

2x 3–---------------


h x( ) 8 3x2–x

-----------------=

9x2 8–x2

----------------- 8 9x2–x2

----------------- 3x2– 8+x2

----------------------

3x2– 8–x2

---------------------- 3x2– 8+x

----------------------


5

The derivative of is:

6

If then g ′(x) is equal to:

7 If find when x = 0.

8 Find the gradient of the function at the point where x = −1.

9 Find the exact value of g ′(5) if .

A B

C D

E

A B

C D

E

Derivation of thequotient rule

Use the product rule and chain rule with y = to obtain and hence confirm the

formula for the quotient rule.


f x( ) sin 4x4x 1+---------------=

f ′ x( ) 4 4x 1+( ) cosx 4 sin 4x–4x 1+( )2

---------------------------------------------------------------= f ′ x( ) 4x 1+( ) cos 4x 4 sin 4x–4x 1+( )2

-----------------------------------------------------------------=

f ′ x( ) 4 4x 1+( ) cos 4x 4 sin 4x–4x 1+

--------------------------------------------------------------------= f ′ x( ) 4 4x 1+( ) cos 4x 4 sin 4x–4x 1+( )2

--------------------------------------------------------------------=

f ′ x( ) 4 sin 4x 4 4x 1+( ) cos 4x–4x 1+( )2

-------------------------------------------------------------------=


g x( ) cos 3x 2–( )ex

-----------------------------=

3ex sin 3x 2–( ) ex cos 3x 2–( )–e2x

----------------------------------------------------------------------------------- –ex sin 3x 2–( ) ex cos 3x 2–( )–e2x

-----------------------------------------------------------------------------------

ex cos 3x 2–( ) 3ex cos 3x 2–( )+e2x

------------------------------------------------------------------------------------ –3ex sin 3x 2–( ) ex cos 3x 2–( )–ex

--------------------------------------------------------------------------------------

–3ex sin 3x 2–( ) ex cos 3x 2–( )–e2x

--------------------------------------------------------------------------------------

ycos 2x

e3x----------------= dy

dx------

f x( ) 2x 3x2–loge 3x 4+( )--------------------------------=

g x( )4 loge 2x

3x-----------------------=

uv--- dy

dx------


Mixed problems on differentiationProblems on differentiation may involve any combination of chain, product andquotient rules.

For each of the following decide which rule of differentiation, that is, chain, product or quotient rule, would be useful to find the derivative.

a b (x2 − 5x)6 c (x2 + 2x − 3) cos 2x d ex2 + 3x

THINK WRITE

a Write the equation. a

It is of the form , that is, a rational function.

Quotient rule

b Write the equation. b y = (x2 − 5x)6

It is a composite function of the form u6, where u = x2 − 5x.

Chain rule

c Write the equation. c y = (x2 + 2x − 3) cos 2xIt is of the form u × v, that is, the product of two functions.

Product rule

d Write the equation. d y = ex2 + 3x

It is a composite function of the form eu, where u = x2 + 3x.

Chain rule

loge x

sin x---------------

1 yloge x

sinx--------------=

2uv---

12

12

12

29WORKEDExample

Find the derivative of y = 2x3 cos (x2 + x).

Continued over page

THINK WRITE

Write the equation. y = 2x3 cos (x2 + x)Decide which rule to use and identify u and v to apply the rule.

u = 2x3 and v = cos (x2 + x) so use the product rule.


12

3dudx------ 6x2=

30WORKEDExample


THINK WRITE

v is a composite function, so differentiate v with respect to x using the chain rule.

v = cos (x2 + x)

Let w = x2 + x.

= 2x + 1

v = cos w

= −sin w

=

So = (−sin w)(2x + 1)

= −(2x + 1) sin (x2 + x)

Apply the product rule

to find . = u × + v ×

= 2x3[−(2x + 1)] sin (x2 + x) + cos (x2 + x) × 6x2

Simplify where possible. = −2x3(2x + 1) sin (x2 + x) + 6x2 cos (x2 + x)

= 2x2[3 cos (x2 + x) − x(2x + 1) sin (x2 + x)]

4

dwdx-------

dvdw-------

dvdx------ dv

dw------- dw

dx-------×

dvdx------

5

dydx------

dydx------ dv

dx------ du

dx------

6

remember1. Chain rule

(a)

(b) A short way to apply the chain rule is:

If then .

2. Product rule

(a) If y = uv then .

(b) If f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x).

3. Quotient rule

(a) If then .

(b) If then .

dydx------ dy

du------ du

dx------×=

f x( ) g x( )[ ]n= f ′ x( ) n g x( )[ ]n 1– g′ x( )×=

dydx------ u

dvdx------×= v

dudx------×+

yuv---= dy

dx------

vdudx------× u

dvdx------×–

v2-------------------------------------=

f x( ) u x( )v x( )-----------= f ′ x( ) v x( )u′ x( ) u x( )v′ x( )–

v x( )[ ]2-----------------------------------------------------=

remember


Mixed problems on differentiation

1 For each function given below, state which rule of differentiation would be used to findthe derivative, that is, chain (C), product (P) or quotient (Q).

2 Using the appropriate rule find the derivative of each function in question 1.

3 Find the derivative of each of the following. (Note that more than one rule will need tobe applied in some cases.)

a f (x) = loge 8x b f (x) = 3x sin x c

d e g(x) = e5x sin x f

g h(x) = cos (x2 − 4x) h f (x) = e−x loge 5x i g(x) = loge (sin x)

j f (x) = sin2 x k l

m g(x) = ecos x n f (x) = tan x

a y = e−5x cos (4x − 7) b

c y = loge (x + 1)3 d y = cos (x2 − 6x)

e f (x) = ex cos 2x f

g h y = loge (sin 3x)

i j f (x) = (x − 1)(x2 + 5x + 3)

k g(x) = ex(x2 + 3) l

m f (x) = e−4x cos (4x − 3) n y = cos2 3x

o y = loge (cos 3x) p

q r f (x) = [loge (5x − 1)]4

s t y = sin

u f (x) = 3x5 cos (2x + 1) v

w y = ex sin x x f (x) = 3 cos2 x + e−7x − x3

y z h (x) = cos3 x

7JMathcad

Derivatives

WORKEDExample

29

g x( ) 3x 7+4x2

---------------=

h x( ) 4xcosx-----------= g x( ) x2 9x 8–+

loge x--------------------------=

h x( ) x 2–ex

-----------= f x( ) loge x=

WORKEDExample

30

yx 2–

3x 1+-------------------=

f x( ) sin 2xcos 2x----------------=

f x( ) 1sin x------------=

y 4e3x2 5x– 2+=

g x( ) 2x 3+( )5

x3 5–----------------------=

f x( ) sin x4

x2--------------=

f x( ) cos 2xsin 2x----------------=

yloge x

x-----------------=

x 3+x 2–------------

g x( )loge x

32---

x2-----------------=

x( ) 3 sin 6x loge 5x2( ) 4ex2---

–+=


• The gradient of a function exists wherever the graph of the function is smooth and continuous.

• If the gradient of a function, f (x), is zero at x = a, then its gradient function, f ′(x), will have an x-intercept at x = a.

• When the gradient of a function is positive, the graph of the gradient function is above the x-axis and when the gradient of a function is negative, the graph of the gradient function is below the x-axis.

• A polynomial function has a gradient function which is also a polynomial function, but its degree is reduced by one.

• The gradient of a chord (secant) or the average rate of change is given by:

.

• A limit is the value that y approaches as x approaches a given value.• A limit exists if the function is approaching the same value from both left and right.• The gradient of the tangent to a curve at a point P is the gradient of the curve at P

and is given by .

• For a function y = f (x), its derivative is expressed as either or f ′(x).

• Summary of derivatives:

f (x) f ′(x)

c 0

axn naxn − 1

[g(x)]n ng′(x)[g(x)]n − 1

ex ex

ekx kekx

eg(x) g′(x)eg(x)

loge x

loge kx

loge [g(x)]

sin x cos x

sin ax a cos ax

cos x −sin x

cos ax −a sin ax

summary

f x h+( ) f x( )–h

--------------------------------------

f x h+( ) f x( )–h

--------------------------------------h 0→lim

dydx------

• The chain rule of differentiation is:

• The product rule of differentiation states:

1. if y = u × v, then

2. if f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x).

• The quotient rule of differentiation states:

1. if then .

2. if then

f ′(x) =

dydx------ dy

du------ du

dx------×=

dydx------ u

dvdx------×= v

dudx------×+

yuv---= dy

dx------

vdudx------× u

dvdx------×–

v2-------------------------------------=

f x( ) u x( )v x( )-----------=

v x( )u′ x( ) u x( )v′ x( )–v x( )[ ]2

-----------------------------------------------------

1x---

1x---

g′ x( )g x( )------------


Multiple choice

1 The graph of f (x) is shown below.

The graph of its gradient function is:

2 For the function g(x) graphed below

the gradient function g′(x) is defined over the domain:

3 The value of is:

4 is:

A B C

D E

A R B R\{1} C R\{4} D R\{1, 4} E [1, 4]

A undefined B 0 C 1 D 4 E 6

A 22 B −22 C 6 D undefined E −6

CHAPTERreview

7Ay

x0 3

4

f (x)

y

x0 3

y

x0 3

y

x0 3

y

x0 3

y

x0

4

7Ay

x0 1 4

7Bx2 2x– 8–

x 4–--------------------------

x 4→lim

7Bx3 5x–x 1+

-----------------x 3–→lim


5 The derivative of f (x) = 4x3 − x2 + 3x is:

6 The derivative of is:

7 The derivative of (2x + 5)6 is:


9 If y = 5e−6x then is equal to:

10 If y = e4x + 7 then is:

11 The derivative of loge (3x − 2) is:

12 The derivative of 2 loge (x2 + x) is:

13 If y = cos 8x then is:

14 If y = 2 sin (2x + 3) then is equal to:

A 12x2 − 2x + 3 B 4x2 − 2x + 3 C 12x2 − 2x

D 12x2 − x + 3 E 4x2 − 2x

A B C D E

A 6(2x + 5)5 B 12x(2x + 5)5 C 6x(2x + 5)5

D 12(2x + 5)5 E 12(2x + 5)4

A B C

D E

A −30e−6x B −6e−6x C 5e−6x − 1 D −30e−6x − 1 E 5e−7x

A 4e4x + 6 B e4x + 6 C e4x + 7 D 4e4x + 7 E 4e3x + 7

A B C D E

A B C

D E

A 8 sin 8x B sin 8x C −8 sin 8x D −8x sin 8x E −sin 8x

A 4x cos (2x + 3) B −4 cos (2x + 3) C 4 cos (2x + 3)D 4 cos 2x E 2 cos (2x + 3)

7C

7C g x( ) 1x2----- 2 x–=

1x---– 2

x-------– 2

x3-----– 2

x-------– 2

x3-----– 1

x-------– 2

x---– 2

x-------– 2

x3-----– 1

x32---

-----–

7D

7D1

4x 9–-------------------

2 4x 9–2–

4x 9–( )3--------------------------

4

4x 9–( )32---

----------------------

4 4x 9–2

4x 9–---------------

7Edydx------

7Edydx------

7F1

3x 2–--------------- 1

3x------ 1

x--- 1

3 3x 2–( )----------------------- 3

3x 2–---------------

7F2 2x 1+( )

x2 x+----------------------- 2 2x 1+( )

x-----------------------

2x 1+x2 x+---------------

2xx2 x+-------------- 4x

x2 x+--------------

7Gdydx------

7Gdydx------

C h a p t e r 7 D i f f e r e n t i a t i o n 30715 If f (x) = x2 e2x then f ′(x) is equal to:

16 If g(x) = 2x loge 3x then g′(x) must be:



19 If g(x) = (x2 + 3x − 7)5 then g′(x) is equal to:

20 The derivative of sin x cos x is:

Short answer1 The graph of a cubic function is shown below.

Sketch the graph of its gradient function.

2 Find .

3 a Find the derivative of f (x) = x3 + 2x using first principles.b Hence find the gradient at the point where x = 1.

A 2xe2x + 2x2e2x B 2xe2x C 4xe2x

D 2xe2x − 2x2e2x E 2xe2x + x2e2x

A 2 loge 3x + B 2 loge 3x + 2 C 2 loge 3x + 6x

D 2 loge 3x − E 2 loge 3x + 6x loge 3x

A B C

D 4x − 5 E

A B C

D E

A 5(x2 + 3x − 7)4 B (2x + 3)(x2 + 3x − 7)4 C 5(2x + 3)4

D 5(2x + 3)(x2 + 3x − 7)4 E (x2 + 3x − 7)4

A 2 sin x cos x B sin2 x − cos2 x C sin2 x + cos2 x

D cos2 x − sin2 x E −sin2 x − cos2 x

7H

7H23---

23---

7I2x 1+x 2–

---------------

4x 5–x 2–( )2

------------------- 3–x 2–( )2

------------------- 4x 3–x 2–( )2

-------------------

5–x 2–( )2

-------------------

7Ie4x

x2-------

x 2–( )e4x

x3------------------------

2 1 2x–( )x3

----------------------- 2 2x 1–( )e4x

x3-------------------------------

x2e4x 2e4x–x4

----------------------------- 2e4x

x3----------

7J

7J

7Ay

x0 21 3–1–4

f (x)

7Bh3 2h2 4h+ +

h---------------------------------

h 0→lim

7B


4 a Find the gradient function if g(x) = .

b Find the gradient of g(x) when x = 3.

5 If , find:

a h′(x)b i h′(−1) ii h′(2).

6 If , find the gradient when a x = 2 and b x = 1.

7 Find the derivative of .

8 Differentiate f(x) = e2x − 1.

9 a Find f ′(x) if f(x) = .b What is the value of x when f ′(x) = 0?

10 Find if y = loge (2x3 − 4).

11 The tangent to the curve f(x) = loge (ax − 1) when x = 2, has a gradient of 1.

a Find the value of a.b Find the equation of the tangent at the point when x = 2.

12 Find f ′(x) if f(x) = 3 sin 2x and hence find f ′(2π).

13 If y = 3x2 loge 6x, find .

14 Find f ′(x) if f(x) = .

15 Differentiate esin 2x.

AnalysisA section of a roller-coaster ride follows part of the curve with the equation y = (x3 + 30x2) as shown on the right.a For what values of x (domain) is the gradient:

i zero?ii positive?iii negative?

b Sketch the gradient function.c Use the graph of the gradient function to find the value of x

where the gradient is steepest over the domain [−25, 10].

d Find .

e Find the gradient where x equals:i −25ii −10iii 10.

f Does this verify your answer to part c? Briefly explain.g What is the highest point reached by the roller-coaster? (Give your answer in metres.)

7Cx3

3----- 4x–

7C h x( ) 3x4

2-------- x3

4----- 3x–+=

7D y 4x 1+( )32---=

7D x2 4+

7E7E e

x–2

7Fdydx------

7F

7G7H

dydx------

7Icos x

x------------

7J

y

x0 12–20–28

1200---------

dydx------

testtest

CHAPTERyyourselfourself

testyyourselfourself

7

9VCEVCEcocovverageerageArea of studyUnit 2 • Calculus

In thisIn this chachapterpter9A Rates of change9B Sketching graphs

containing stationary points

9C Solving maximum and minimum problems

9D Applications of antidifferentiation

Applicationsofdifferentiation

396


Rates of change

The rate of change of a function refers to the rate at which its gradient is changing. For linearfunctions the gradient is constant; however, the gradient for other functions such as quadraticor cubic polynomials is continually changing.

Differentiation provides uswith a tool to describe the gradient

of a function and hence determine itsrate of change at any particular point.

In essence, while average rates ofchange can be determined from the

original function, differentiation of thisfunction provides a new function that

describes the instantaneous rate of change.

Note

: The term

instantaneous rate ofchange

is often referred to as

rate of change

.

The rate of change ofvelocity with respect to

time is acceleration.

➙

If f (x) = x2 − 2x + 4 find:a the average rate of change between x = 2 and x = 4b a new function that describes the rate of changec the instantaneous rate of change when x = 4.

THINK WRITE

a Write the function. a f(x) = x2 − 2x + 4

Average rate of change = . Average rate of change =

=

= 4

b Differentiate f (x). b f ′(x) = 2x − 2

c Substitute x = 4 into f ′(x). c f ′(4)= 2(4) − 2= 6

So the rate of change when x = 4 is 6.

1

2change in f x( )

change in x------------------------------------ f 4( ) f 2( )–

4 2–-----------------------------

12 4–2

---------------

1WORKEDExample

The rate of change of position with respect totime is velocity.

➙

C h a p t e r 9 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n 397

It is worth noting that there are two common ways of writing the derivative as a func-tion. For example, the derivative of the function P(x) = x2 + 5x − 7 may be written as

P′(x) = 2x + 5 or as .

A javelin is thrown so that its height, h metres above theground, is given by the rule:h(t) = 20t − 5t2 + 2, where t represents time in seconds.

a Find the rate of change of the height at any time, t.b Find the rate of change of the height when i t = 1 ii t = 2 iii t = 3.c Briefly explain why the rate of change is initially positive,

then zero, and then negative over the first 3 seconds.d Find the rate of change of the height when the javelin first

reaches a height of 17 metres.

THINK WRITE

a Write the rule. a h(t) = 20t − 5t2 + 2Differentiate h(t). h′(t) = 20 − 10t

b Evaluate h′(1). b h′(1) = 20 − 10(1)= 10 m/s

Evaluate h′(2). h′(2) = 20 − 10(2)= 0 m/s

Evaluate h′(3). h′(3) = 20 − 10(3)= −10 m/s

c For rates of change:Positive means increasing.Zero means neither increasing nor decreasing.Negative means decreasing.

c The javelin travels upward during the first 2 seconds.When t = 2 seconds the javelin has reached its maximum height.When t > 2 seconds the javelin is travelling downward.

d Find the time at which a height of 17 m occurs, by substituting h = 17 into h(t).

d 20t − 5t2 + 2 = 17

Make RHS = 0. −5t2 + 20t − 15 = 0

Solve for t.Note: the quadratic formula could also be used to solve for t.

t2 − 4t + 3 = 0(t − 1)(t − 3) = 0

t = 1 or 3

The first time it reaches 17 m is the smaller value of t.

It first reaches 17 m when t = 1 s.

Evaluate h′(1). h′(1) = 20 − 5(1)2

= 15 m/sRate of change of height is 15 m/s.

12

1

2

3

1

2

3

4

5

2WORKEDExample

dPdx------- 2x 5+=


The shockwave from a nuclear blast spreads out at ground level in a circular manner.a Write down a relationship

between the area of ground,A km2, over which theshockwave passes and itsradius, r km.

b Find the rate of change ofA with respect to r.

c Find the rate of change ofA when the radius is 2 km.

d What is the rate of changeof A when the area coveredis 314 km2?

THINK WRITE

a State the formula for the area of a circle. a A(r) = π r2

b Differentiate A(r). b A′(r) = 2π r

c Substitute r = 2 into A′(r).Note: The units for the rate of change of A (km2) with respect to r (km) are km2 per km or km2/km.

c A′(2) = 2(3.14)(2)= 12.56

Rate of change of A when the radius is 2 km is 12.56 km2/km.

d Substitute A = 314 into the area function A(r) and solve for r.

d A(r) = π r2

314 = 3.14r2

r2 =

= 100r = 10 since r > 0

Find the rate of change when r = 10. A′(10) = 2π(10)= 62.8

Rate of change of A when area is 314 km2 is 62.8 km2/km.

1

3143.14----------

2

3WORKEDExample

r kmArea A km2

rememberAverage rates of change are calculated using the original function, while differentiation of this function is needed in order to calculate instantaneous rates of change at specific points.

remember


Rates of change

1 If f (x) = x2 + 5x + 15 find:a the average rate of change between x = 3 and x = 5b a new function that describes the rate of changec the (instantaneous) rate of change when x = 5.

2 A balloon is inflated so that its volume, V cm3, at any time, t seconds later is:

V = − t3 + 24t2, t ∈ [0, 10]

a What is the volume of the balloon when: i t = 0? ii t = 10?

b Hence, find the average rate of change between t = 0 and t = 10.c Find the rate of change of volume when

i t = 0 ii t = 5 iii t = 10.

3

The average rate of change between x = 1 and x = 3 for the function y = x2 + 3x + 5 is:

4

The instantaneous rate of change of the function f (x) = x3 − 3x2 + 4x, when x = −2 is:

5

If the rate of change of a function is described by = 2x2 − 7x, then the functioncould be:

6 In a baseball game the ball is hitso that its height above theground, h metres, is

h(t) = 1 + 18t − 3t2

t seconds after being struck.a Find the rate of change, h′(t).b Calculate the rate of change

of height after: i 2 secondsii 3 secondsiii 4 seconds.

c What happens when t = 3 seconds?

d Find the rate of change ofheight when the ball firstreaches a height of 16 metres.

A 1 B 9 C 5 D 3 E 7

A 2 B −2 C 28 D 3 E 12

A y = 6x3 − 14x B y = x3 − 7x C y = x3 − x2 + 5

D y = x3 − x2 + 2 E 2x2 − 7x + 5

9AWORKEDExample

1 SkillSH

EET 9.1

Mathcad

Gradientbetween two

points ona graph

85---

EXCEL Spreadsheet

Gradientbetween two

points ona graph



mmultiple choiceultiple choicedydx------

23--- 2

3--- 7

2---

72---

WORKEDExample

2


7 The position, x metres, of a lift (above ground level) at any time, t seconds,is given by: x(t) = −2t2 + 40ta Find the rate of change of displacement (velocity) at any time, t.b Find the rate of change when:

i t = 5 ii t = 9 iii t = 11.c What happened between t = 9 and t = 11?d When and where is the rate of change zero?

8 The number of seats, N, occupied in a soccer stadium t hours after the gatesare opened is given by:

N = 500t2 + 3500t, t ∈ [0, 5]a Find N when:

i t = 1 and ii t = 3.b What is the average rate of change between t = 1 and t = 3?c Find the instantaneous rate when:

i t = 0 ii t = 1 iii t = 3 iv t = 4.d Why is the rate increasing in the first 4 hours?

9 The weight, W kg, of a foal at any time, t weeks, after birth is given by:W = 80 + 12t − t2 where 0 < t < 20

a What is the weight of the foal at birth?b Find an expression for the rate of change of weight at any time, t.c Find the rate of change after:

i 5 weeks ii 10 weeks iii 15 weeks.d Is the rate of change of the foal’s weight increasing or decreasing?e When does the foal weigh 200 kg?

SkillSH

EET 9.2

310------

C h a p t e r 9 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n 40110 The weekly profit, P (hundreds of dollars), of a factory is given by ,

where n is the number of employees.

a Find .

b Hence, find the rate of change of profit, in dollars per employee, if the number ofemployees is:

i 4ii 16iii 25.

c Find n when the rate of change is zero.

11 Gas is escaping from a cylinder so that its volume, V cm3, t seconds after the leakstarts, is described by V = 2000 − 20t − t2.a Find the rate of change after:

i 10 secondsii 50 secondsiii 100 seconds.

b Is the rate of change ever positive? Why?

12 Assume an oil spill from an oil tanker is circular and remains that way.a Write down a relationship between the area of the spill, A m2, and the radius,

r metres.b Find the rate of change of A with respect to the radius, r.c Find the rate of change of A when the radius is:

i 10 mii 50 miii 100 m.

d Is the area increasing more rapidly as the radius increases? Why?

13 A spherical balloon is being inflated.a Express the volume of the balloon, V m3, as a function of the radius, r metres.b Find the rate of change of V with respect to r.c Find the rate of change when the radius is:

i 0.1 mii 0.2 miii 0.3 m.

P 4.5n n32---

–=

dPdn-------

1100---------

WORKEDExample

3

r


14 A rectangular fish tank has a square base with its height being equal to half its baselength.a Express the length and width of the base in terms of its height, h.b Hence, express the volume, V m3, in terms of the height, h, only.c Find the rate of change of V when:

i h = 1 mii h = 2 miii h = 3 m.

15 For the triangular package shown find:a x in terms of hb the volume, V, as a function of h onlyc the rate of change of V when

i h = 0.5 mii h = 1 m.

16 A new estate is to be established on the side of a hill.

Regulations will not allow houses to be built on slopes where the gradient is greaterthan 0.45. If the equation of the cross-section of the hill is

y = −0.000 02x3 + 0.006x2 find:

a the gradient of the slope

b the gradient of the slope when x equals:i 160ii 100iii 40iv 20

c the values of x where the gradient is 0.45d the range of heights for which houses cannot be built on the hill.

17 A bushfire burns out A hectares of land, t hours after it started according to the ruleA = 90t2 − 3t3

a At what rate, in hectares per hour, is the fire spreading at any time, t?b What is the rate when t equals:

i 0 ii 4 iii 8 iv 10 v 12 vi 16 vii 20?c Briefly explain how the rate of burning changes during the first 20 hours.d Why isn’t there a negative rate of change in the first 20 hours?e What happens after 20 hours?f After how long is the rate of change equal to 756 hectares per hour?

30°h

x

6

200

y

x

80

dydx------


Sketching graphs containing stationary points

The derivative of a function gives its gradient function — that is, it gives the gradient ofa tangent to the curve for any specified value of the independent variable. When thederivative equals zero the tangent is horizontal. The point, or points, on the curve wherethis occurs are called stationary points.

In other words, a function f (x) has stationary points when f ′(x) = 0.Stationary points can take the form of:

1. A local minimum turning point2. A local maximum turning point, or3. A point of inflection.

Local minimum turning pointJust to the left of a, the gradient is negative; that is, ifx < a, but close to a, then f ′(x) < 0.At the point where x = a the gradient is zero; that is, atx = a, f ′(x) = 0.Just to the right of a the gradient is positive; that is, ifx > a, but close to a, f ′(x) > 0.

In other words, for a stationary point at x = a, if thegradient changes from negative to positive as wemove from left to right in the vicinity of a, it is a local minimum.

Local maximum turning pointIf x < a, but close to a, then f ′(x) > 0.At x = a, f ′(x) = 0.If x > a, but close to a, f ′(x) < 0.

In other words, for a stationary point at x = a. If thegradient changes from positive to negative as we movefrom left to right in the vicinity of a, it is a local maximum.

The term local maximum or local minimum impliesthat the function has a maximum or minimum in thevicinity of x = a. This is important because some functions can have more than onestationary point.

Point of inflection

If x < a, but close to a, then f ′(x) > 0. If x < a, but close to a, f ′(x) < 0.At x = a, f ′(x) = 0. At x = a, f ′(x) = 0.If x > a, but close to a, f ′(x) > 0. If x > a, but close to a, f ′(x) < 0.In other words, for a stationary point at x = a, if the gradient remains positive or

negative in the vicinity of a it is a point of inflection.

y

x0 a

f'(x) < 0 f '(x) > 0

f '(a) = 0

y

x0 a

f'(x) > 0f '(x) < 0

f '(a) = 0

y

x0 a

f'(x) < 0

f '(x) < 0

f '(a) = 0

y

x0 a

f'(x) > 0

f '(x) > 0

f '(a) = 0


If f (x) = x3 − 6x2 − 15x, find:a the value(s) of x where the gradient is zero b the stationary point(s).

THINK WRITE

a Write the function. a f(x) = x3 − 6x2 − 15xDifferentiate f (x) to find the gradient function f ′(x).

f ′(x) = 3x2 − 12x − 15

Solve f ′(x) = 0 to find the x-values of each stationary point.

For stationary points: f ′(x) = 03x2 − 12x − 15 = 03(x2 − 4x − 5) = 0

x2 − 4x − 5 = 0(x − 5)(x + 1) = 0

x = 5 or x = −1

b Substitute each value of x into f (x) to find the corresponding y-values.

b f (5) = (5)3 − 6(5)2 − 15(5)= −100

f (−1) = (–1)3 − 6(−1)2 − 15(−1)= 8

Write the coordinates of each stationary point.

Stationary points occur at (5, −100) and (−1, 8).

12

3

1

2

4WORKEDExample

If y = x3 + 2x2 + 3x − 2 find all stationary points and determine their nature.

THINK WRITE

Write the rule. y = x3 + 2x2 + 3x − 2

Differentiate y to find the gradient function . = x2 + 4x + 3

Solve = 0 to find the x-values of

each stationary point.

For stationary points: = 0x2 + 4x + 3 = 0

(x + 3)(x + 1) = 0x = −3 or x = −1

Substitute each value of x into

y = x3 + 2x2 + 3x − 2 to find the

corresponding y-values.

When x = −3,

y = (−3)3 + 2(−3)2 + 3(−3) – 2

y = −2One stationary point is (−3, −2).When x = −1,

y = (−1)3 + 2(−1)2 + 3(−1) − 2

y = −3

The other one is (−1, −3 ).

13---

113---

2dydx------ dy

dx------

3dydx------ dy

dx------

413--- 1

3---

13---

13---

13---

5WORKEDExample


THINK WRITE

Determine the nature of the stationary

point at x = −3 by evaluating to the left

and right, say at x = −4 and at x = −2.

Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = −3, the stationary point (−3, −2) is a local maximum.

When x = −4

= (−4)2 + 4(−4) + 3

= 3.When x = −2

= (−2)2 + 4(−2) + 3

= −1.

(−3, −2) is a local maximum.Determine the nature of the stationary

point at x = −1 by evaluating to the left

and right, say at x = −2 and at x = 0.

Since the gradient changes from negative to positive as we move from left to right in the vicinity of x = −1, the stationary point

(−1, −3 ) is a local minimum.

When x = −2

= (−2)2 + 4(−2) + 3

= −1.When x = 0

= (0)2 + 4(0) + 3

= 3.

(−1, −3 ) is a local minimum.

5

dydx------

6

dydx------

dydx------

x–4 –3 –2

Zero gradient

NegativegradientPositive

gradient

7

dydx------

8

13---

dydx------

dydx------

x–2 –1 0

Zero gradient

Positivegradient

Negativegradient

13---

Sketch the graph of the function f (x) = 5 + 4x − x2 labelling all intercepts and stationary points.

Continued over page

THINK WRITE

Write the function. f(x) = 5 + 4x − x2

Find the y-intercept by letting x = 0. y-intercept: if x = 0,f (0) = 5 + 4(0) − (0)2

= 5so y-intercept is (0, 5).

12

6WORKEDExample


To find a local maximum or minimum point on a func-tion, enter it in the Y= menu and follow these steps.1. Press [CALC] and select 4:maximum (or

3:minimum).2. Use the arrow key to scroll to the left of a turning point

and press .3. Use the arrow key to scroll to the right of the same

turning point and press twice.

THINK WRITE

Find the x-intercepts by letting f (x) = 0. x-intercepts: if f(x) = 0,5 + 4x − x2 = 0x2 − 4x − 5 = 0

(x + 1)(x − 5) = 0x = −1 or x = 5

so x-intercepts are (−1, 0) and (5, 0).Differentiate f (x) to find the gradient function f ′(x).

f ′(x) = 4 − 2x

Solve f ′(x) = 0 to find the x-value/s of each stationary point.

For stationary points: f ′(x) = 04 − 2x = 0

−2x = −4x = 2

Substitute this value of x intof (x) = 5 + 4x − x2 to find the corresponding y-value.

f (2) = 5 + 4(2) − (2)2

= 9so there is a stationary point at (2, 9).

Determine the nature of the stationary point at x = 2 by evaluating f ′(x) to the left and right, say at x = 1 and x = 3.

Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = 2, the stationary point (2, 9) is a local maximum.

f ′(1) = 4 − 2(1)= 2

f ′(3) = 4 − 2(3)= −2

(2, 9) is a local maximum.Sketch the graph.

3

4

5

6

7

8

x1 2 3

Zero gradient

Negativegradient

Positivegradient

9 y

f (x)

x0

5

2–1

9

5

(2, 9)

Graphics CalculatorGraphics Calculator tip!tip! Finding stationary (turning) points

2nd

ENTER

ENTER


Sketching graphs containing stationary points

1 For each of the following functions determine the value(s) of x where the gradient iszero.a f (x) = x2 + 2x b f (x) = x2 − 8x + 5c f (x) = x3 − 3x2 d f (x) = 2x3 + 6x2 − 18x + 1e y = (x + 6)(x − 2) f y = x2(x − 1)g y = 10 + 4x − x2 h y = x3 − 3x2 + 5x − 2

2 For each function in question 1 determine all of the stationary points.

3 If f (x) = x2 − 8x + 1:a show that it has a stationary point when x = 4b evaluate f ′(3) and f ′(5) c state which type of stationary point it is.

4 For the function f (x) = 5 − x2:a find x when f ′(x) = 0 b state what type of stationary point it has

5 If f (x) = x3 − 4 then:a show that it has a stationary point when x = 0 onlyb find f ′(−1) and f ′(1) c state which type of stationary point it is

6 If f (x) = x3 − x2 − 3x + 5:

a show there are stationary points when x = −1 and x = 3b evaluate f ′(−2), f ′(0) and f ′(4) c state what type of stationary points they are.

7

When x = 1, the curve y = 2x2 − 3x + 1:

8

When x = −1 the function y = x3 − 2x2 − 7x:

A is decreasing B has a local maximum C has a local minimumD is increasing E does not exist

A is decreasing B has a local maximum C has a local minimumD is increasing E does not exist

rememberWhen sketching graphs of polynomial functions, four main features should be indicated:1. the y-intercept (found by calculating y when x = 0)2. the x-intercept(s) (found by solving the equation for x when y = 0)

3. the stationary point(s) (found by solving the equation = 0)

4. the type of stationary point(s) (found by checking the sign of the gradient to the left and right of the stationary point).

dydx------

remember

9B

EXCEL Spreadsheet

Cubicgraphs

Mathcad

Cubicgraphs

WORKEDExample

4a

Mathcad

Quadraticgraphs

EXCEL Spreadsheet

Quadraticgraphs

13---

WORKEDExample

4b

13---




9

The graph below best representing a function with f ′(−2) = 0, f ′(x) < 0 if x < −2, andf ′(x) > 0 if x > −2 is:

10

f ′(1) = f ′(4) = 0 and f ′(x) < 0 if 1 < x < 4 and f ′(x) > 0 if x < 1 and x > 4. The graphbelow which satisfies these conditions is:

11 For each of the following find the stationary points and determine their nature.

12 Sketch graphs of the following functions, labelling all intercepts and stationary points.

A B C

D E

A B C

D E

a y = x2 + 6x + 2 b y = 8x − 2x2 c y = x3 − x2

d y = x3 + x2 − 3 e y = x3 − x2 − 2x f y = (x − 1)3

g y = x3 + 3 h y = x3 − 27x + 5

a f (x) = x2 − 2x − 3 b f (x) = x3 − 3x − 2c f (x) = x3 − 2x2 + x d f (x) = x2(3 − x)e f (x) = x3 + 4x2 + 4x f f (x) = x3 − 4x2 − 11x + 30g f (x) = (x + 2)3 h f (x) = 24 + 10x − 3x2 − x3

i f (x) = x3 − 2x2 − x + 2 j f (x) = 8 − x3


y

f (x)

x0–2

y

f (x) x0–2

y

f (x)

x0–2

yf (x)

x0–2

f (x)y

x0

–2


y

x0 41

f (x)

x0 41

f (x)

y y

x0 41

f (x)

y

x0 1 4

f (x)

y

x0 1 4

f (x)

WORKEDExample

5 12--- 1

3--- 1

2---

WORKEDExample

6

SkillSH

EET 9.3

SkillSH

EET 9.4


Solving maximum and minimum problemsThere are many practical situations where it is necessary to determine the maximum orminimum value of a function. For quadratic functions, differentiation makes this a rela-tively simple task because as we saw in the previous section, setting the derivativeequal to zero allows us to solve an equation to obtain the value(s) of x for which thelocal maximum or minimum values (turning points) occur.

When solving maximum or minimum problems it should be verified that it is in facta maximum or minimum by checking the sign of the derivative to left and right of theturning point.

In the case of cubic and higher order polynomials, the local maximum (or minimum) may or may not be the highest (or lowest) value of the function in a given domain.

An example where the local maximum, found by solving f ′(x) = 0, is not the largest value of f(x) in the domain [a, b] is shown. Here, B is the point where f(x) is greatest in this domain, andis called the absolute maximum for the interval.

Case 1. The function is known

f (x)y

x0 a b

Absolutemaximum inthe interval[a, b]

Localmaximum

A baseball fielder throws the ball so that the equation of its path is:y = 1.5 + x − 0.02x2

where x (metres) is the horizontal distance travelled by the ball and y (metres) is the vertical height reached.a Find the value of x for which the maximum height is reached (verify that it is a

maximum).b Find the maximum height reached.

Continued over page

THINK WRITE

a Write the equation of the path. a y = 1.5 + x − 0.02x2

Find the derivative . = 1 − 0.04x

Solve the equation = 0 to find the

value of x for which height is a maximum.

For stationary points: = 0

1 − 0.04x = 0−0.04x = −1

x = 25Determine the nature of this stationary

point at x = 25 by evaluating to the

left and right, say, at x = 24 and at x = 26.

When x = 24

= 1 − 0.04(24)

= 0.04.When x = 26

= 1 − 0.04(26)

= −0.04.

1

2dydx------ dy

dx------

3dydx------ dy

dx------

4

dydx------

dydx------

Zero gradientNegativegradient

Positivegradient

dydx------

7WORKEDExample


Case 2. The rule for the function is not givenIf the rule is not given directly then the following steps should be followed:1. Draw a diagram if necessary and write an equation linking the given information.2. Identify the quantity to be maximised or minimised.3. Express this quantity as a function of one variable only (often this will be x).4. Differentiate, set the derivative equal to zero, and solve.5. Determine, in the case of more than one value, which one represents the maximum

or minimum value.6. For some functions, a maximum or minimum may occur at the extreme points of the

domain so check these also.7. Answer the question that is being asked.8. Sketch a graph of the function if it helps to answer the question, noting any restric-

tions on the domain.

THINK WRITE

Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = 25, the stationary point is a local maximum.

The stationary point is a local maximum.

b Substitute x = 25 into y = 1.5 + x − 0.02x2 to find the corresponding y-value (maximum height).

b When x = 25,y = 1.5 + 25 − 0.02(25)2

= 14So the maximum height reached is 14 m.

5

A farmer wishes to fence off a rectangular paddock on a straight stretch of river so that only 3 sides of fencing are required. Find the largest possible area of the paddock if 240 metres of fencing is available.

THINK WRITE

Draw a diagram to represent the situation, using labels to represent the variables for length and width and write an equation involving the given information.

Let w = widthl = length

P = perimeter

P = l + 2w = 240 [1]

Write a rule for the area, A, of the paddock in terms of length, l, and width, w.

A = l × w [2]

1

Fence Fence

Fence

w

l

w

River

2

8WORKEDExample


THINK WRITE

Express the length, l, of the rectangle in terms of the width, w, using equation [1].

l + 2w = 240l = 240 − 2w [3]

Express the quantity, A, as a function of one variable, w, by substituting [3] into [2].

Substituting [3] into [2]:A(w) = (240 − 2w)w

= 240w − 2w2

Solve A ′(w) = 0. A ′(w) = 240 − 4wFor stationary points: A ′(w) = 0240 − 4w = 0

240 = 4ww = 60

Test to see if the stationary point atw = 60 will produce a maximum or minimum value for the area by evaluating A ′(w) to the left and right, say, at w = 59 and at w = 61.

When w = 59A ′(59) = 240 − 4(59)

= 4.When w = 61A ′(61) = 240 − 4(61)

= −4.

Since the gradient changes from positive to negative as we move from left to right in the vicinity of w = 60, the stationary point is a local maximum.

Find the maximum area of the paddock by substituting w = 60 into the function for area.

The stationary point is a local maximum.The area of the paddock is a maximum when w = 60.

A(60) = (240 − 2 × 60) × 60= 7200 m2

3

4

5

6

7 Zero gradient

Negativegradient

Positivegradient

8

rememberDefining a function and setting its derivative equal to zero to form an equation helps to tell us when a local maximum or minimum occurs. The solution(s) must then be substituted into the original function to find the actual maximum or minimum value(s).

remember


Solving maximum and minimum problems

1 A golfer hits the ball so that theequation of its path is:

y = 1.2 + x − 0.025x2

where x (metres) is the horizontaldistance travelled by the ball andy (metres) is the vertical heightreached.a Find the value of x for which

the maximum height is reached(and verify that it is amaximum).

b Find the maximum heightreached.

2 If the volume of water, V litres, in a family’s hot water tank t minutes after the showeris turned on is given by the rule V = 200 − 1.2t2 + 0.08t3, where 0 ≤ t ≤ 15:a find the time when the volume is minimum (that is, the length of time the shower

is on)b verify that it is a minimum by checking the sign of the derivativec find the minimum volumed find the value of t when the tank is full again.

3 A ball is thrown into the air so that its height, h metres, above the ground at time tseconds after being thrown is given by the function:

h(t) = 1 + 15t − 5t2

a Find the greatest height reached by the ball and the value of t for which it occurs.b Verify that it is a maximum.

4 A gardener wishes to fence off a rectangular vegetable patch against her back fence sothat only 3 sides of new fencing are required. Find the largest possible area of thevegetable patch if she has 16 m of fencing material available.

5 The sum of two numbers is 16.a By letting one number be x, find an expression for the other number.b Find an expression for the product of the two numbers, P.c Hence, find the numbers if P is a maximum.d Verify that it is a maximum.

6 The rectangle at right has a perimeter of 20 cm.a If the width is x cm, find an expression for the

length.b Write an expression for the area, A, in terms of x

only.c Find the value of x required for maximum area.d Find the dimensions of the rectangle for maximum area.e Hence, find the maximum area.

9C

Mathca

d

Quadratic graphs

Mathca

d

Cubic graphs

EXCEL

Spreadsheet

Quadratic graphs

EXCEL

Spreadsheet

Cubic graphs

GCpro

gram

Maximum

GCpro

gram

Minimum

WORKEDExample

7

WORKEDExample

8

Width = x

Length

C h a p t e r 9 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n 4137 A farmer wishes to create a rectangular pen to contain as much area as possible using

60 metres of fencing.a Write expressions for the dimensions (length and width) of the pen.b Hence, find the maximum area.

8 The cost of producing a particular toaster is $(250 + 1.2n2) where n is the number pro-duced each day. If the toasters are sold for $60 each:a write an expression for the profit, P, dollarsb find how many toasters should be produced each day for maximum profitc hence, find the maximum daily profit possible.

9 A company’s income each week is $(800 + 1000n − 20n2) where n is the number ofemployees. The company spends $760 per employee for wages and materials.a Write an expression for the company weekly profit, P dollars.b Determine the number of employees required for maximum profit and hence cal-

culate the maximum weekly profit.

10 The sum of two numbers is 10. Find the numbers if the sum of their squares is to be aminimum.

11 A square has four equal squares cut out of the corners asshown at right. It is then folded to form an open rectangularbox.a What is the range of possible values for x?b In terms of x find expressions for the

i height, ii length, and iii width of the box.c Write an expression for the volume, V (in terms of x

only).d Find the maximum possible volume of the box.

12 The base and sides of a shirt box are to be made from a rectangular sheet of cardboardmeasuring 50 cm × 40 cm. Find:a the dimensions of the box required for maximum volumeb the maximum volume.(Give answers correct to 2 decimal places.)

13 The volume of the square-based box shown at right is 256 cm3.a Find h in terms of l.If the box has an open top find:b the surface area, A, in terms of l onlyc the dimensions of the box if the surface area is to be a

minimumd the minimum area. (Hint: = l –1.)

14 A closed, square-based box of volume 1000 cm3 is to be constructed using theminimum amount of metal sheet possible. Find its dimensions.

15 The cost of flying an aircraft on a 900 km journey is 1600 + v2 dollars per hour,where v is the speed of the aircraft in km/h. Find:a the cost, C dollars, of the journey if v = 300 km/hb the cost, C dollars, of the journey in terms of v. (Hint: time = distance ÷ speed.)c the most economical speed and minimum cost.

12 cm

12 cm

x

h

l

l

1l---

WorkS

HEET 9.1

1100---------


When is a maximum not a maximum?Often finding a maximum (or minimum) of a function is achieved by setting the derivative equal to zero and solving for the independent variable (such as x). This value of x is then substituted into the original function to find the maximum (or minimum) value of the function.

One danger of this method is that it assumes the local maximum or minimum isthe absolute maximum or minimum (see page 409).

Consider finding the maximum value of the function f(x) = − 2x2 + 3x + 2between x = 0 and x = 6:

f ′(x) = x2 – 4x + 3f ′(x) = (x − 1)(x + 3)

So x = 1 or x = 3 for a local maximum or minimum.A computer generated graph of f(x) reveals that the absolute maximum of f(x) in

the interval [0, 6] is not at x = 1 or x = 3!

The absolute maximum of f(x) obviously occurs at x = 6, and equals

f(6) = − 2(32) + 3(3) + 2

= 201 Find the absolute maximum of f(x) = x3 + 5x2 − 8x − 12 between x = 0 and x = 3.2 Find the minimum of f(x) = −x3 + 4x2 + 11x − 30 in the interval [1, 5].3 The temperature T°C of a pottery classroom x minutes after the class has started

is described by the function T(x) = 0.000 08x(x + 2)2 + 21.Sketch a graph of the temperature during a 50-minute class and determine when

the classroom is hottest, and what the temperature is then.

x3

3-----

y

x02–2 4 6

10

5

–5

–10

f (x) = x3

–3 – 2x2 + 3x + 2

63

3-----


Applications of antidifferentiationWe will now consider the situation where the derivative of a function (the gradientfunction) is known but the original function is unknown. Finding the original functionrequires a process called antidifferentiation.

If f ′(x) = xn, n ∈ N, then

f (x) =

where c represents a constant.

This can be verified by differentiating .

The result is xn.Similarly, if f ′(x) = axn, a ∈ R, n ∈ N, then

f (x) =

We saw previously that an alternative expression for the derivative was .

Likewise, if , then

y =

The derivative of a function is also known as the gradient function and describes the rate of change of the function.

The pieces of information used to find the value of the constant that is generated following antidifferentiation are called boundary conditions.

1n 1+------------xn 1+ c+

1n 1+------------xn 1+ c+

an 1+------------xn 1+ c+

dydx------

dydx------ xn=

1n 1+------------xn 1+ c+

Find the rule for the function f(x) if f ′(x) = 3 + 4x − x2 and f(0) = 7.

THINK WRITE

Write the given expression. f ′(x) = 3 + 4x − x2

Antidifferentiate f ′(x) to obtain the general rule for f (x).

f (x) = 3x + 2x2 − + c

Substitute x = 0 and f (x) = 7 into f (x) and solve to find the value of the constant, c.

7 = 3(0) + 2(0)2 − + cc = 7

Write the rule for f (x). f (x) = 3x + 2x2 − + 7

1

2x3

3-----

30( )3

3----------

4x3

3-----

9WORKEDExample


The Mathcad file ‘Antidifferentiation’ may be used to check your antiderivatives, andtakes into account boundary conditions if there are any.

The rate of change of the volume, V cm3, of a balloon at any time, t seconds, after it is inflated beyond 6 litres is given by:

t ∈ [0, 3]

a Express V as a function of t. b What is the volume of the balloon when t = 1?

THINK WRITE

a Write the given expression. a = 3t 2 − 8t + 1

Find the general rule for volume by antidifferentiation.

V(t) = t3 − 4t2 + t + c

Find the value of the constant, c, by substituting t = 0 and V = 6 into V(t).

6 = (0)3 − 4(0)2 + (0) + cc = 6

Write the rule for V(t). V(t) = t3 − 4t2 + t + 6

b Substitute t = 1 into the volume function V(t).

b V(1) = (1)3 − 4(1)2 + (1) + 6= 4

So the volume of the balloon at t = 1 is 4 litres.

dVdt------- 3t2 8t– 1+=

1dVdt-------

2

3

4

10WORKEDExample

rememberWhen finding the antiderivative of a function:For each term in the function, increase the power of the variable by one and divide by the resulting power. Add a constant.

remember

Mathca

d

Antidifferentiation

C h a p t e r 9 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

417

Applications of antidifferentiation

1

Find the rule for the function

f

(

x

) if

f

′

(

x

)

=

3

x

2

−

2

x

and

f

(2)

=

0.

2

If

f

′

(

x

)

=

3

+

5

x

−

2

x

2

and the

y

-intercept is 7, find

f

(

x

).

3

The

y

-intercept of a curve is 10 and . Find the value of

y

when

x

=

3.

4

If the gradient function of a curve that passes through the point (2, 2) is

f

′

(

x

)

=

2

x

−

5,then the function

f

(

x

) is:

5

If

f

′

(

x

)

=

4

x

+

1 and the

y

-intercept is

−

3 then

f

(

x

) equals:

6

A curve passes through the point (2, 1) and has a gradient function

f

′

(

x

)

=

x

(3

x

−

5).The function must be:

7

The velocity (

v

) of an aircraft is changing as it accelerates. Its acceleration (rate ofchange of velocity) at any time,

t

, after it begins accelerating from rest along a runway

is given by

where

v

is in km/h and

t

is in seconds.

a

Express

v

as a function of

t

.

b

Find the velocity after 5 seconds.

A

x

2

−

5

x

+

8

B

x

2

−

5

x

−

1

C

x

2

−

5

D

x

2

−

5

x

E

x

2

−

2

A

x

2

+

2

x

−

3

B

2

x

2

+

x

−

1

C

2

x

2

+

x

−

3

D

2

x

2

+

2

x

− 1 E x2 + x

A f (x) = x3 − 3x2 + 5 B f (x) = x3 − x2 + 2 C f (x) = 3x2 − 5x − 1

D f (x) = x3 − x2 + 3 E f (x) = x4 − x3 + 9

9DWWORKEDORKEDEExamplexample

9

Mathcad

Antidifferentiation

dydx------ x 1+( ) x 3–( )=




52---

52--- 3

4--- 5

2---

WWORKEDORKEDEExamplexample

10dvdt------ 6t2 4t– 5+=

Ch 09 MM 1&2 YR 11 Page 417 Friday, June 29, 2001 11:28 AM


h

8 The rate of change of position (velocity) of a particle travelling in a straight line is

given by , where x is in metres and t is in seconds.

If the particle starts at x = 1, find its position when t = 3.

9 The rate of increase of volume per unit increase in depth for a particular container isgiven by:

where V cm3 is the volume and the depth is h cm.a If V = 0 when h = 0, express V as a function of h.b Find the volume at a height of 7 cm.

10 The weekly rate of change of profit with respect to the number of employees, n, in afactory is:

where P is in thousands of dollars.a Find the number of employees for maximum profit (assume P = 0 when n = 0).b Hence find the maximum profit.

11 The rate of deflection from the horizontal of a 2 m long diving board when a 70 kgperson is x metres from its fixed end is:

a What is the deflection, y, when x = 0.b Find the equation that measures the deflection at any point on the board.c Find the maximum deflection. (Be careful.)

12 The rate of change of height of a hot

air balloon is given by = 4t − 1

where h is the height above the groundin metres after t seconds.a Write h as a function of t.b Find the height after

4 seconds.c How long does it take the

balloon to reach a height of60 metres?

dxdt------ t2 6t– 2+=

dVdh------- 2 h 5+( )2=

dPdn------- 3.182

34--- n–=

dydx------ –0.06 x 1+( )2 0.06+=

x0Deflection

y

WorkS

HEET 9.2 dhdt------

h


Rates of change• Average rate of change = .

• The derivative of a function, f ′(x) (or ), is needed in order to calculate the

(instantaneous) rate of change at a particular point. The rate of change of a function, f (x), at x = a is given by f ′(a).

Sketching graphs containing stationary points• Stationary points occur when f ′(x) = 0.• Three types of stationary point exist and, by testing the sign of

the gradient to the left and right of a stationary point, the nature (type) can be determined.1. Local maximum turning points. ( f ′(x) changes from + to −

moving left to right.)

2. Local minimum turning points. ( f ′(x) changes from − to + moving left to right.)

3. Points of inflection. ( f ′(x) remains the same sign on both sides moving left to right.)

Solving maximum and minimum problems• By solving the equation f ′(x) = 0, and substituting the solutions into the original

function, the maximum or minimum value of a quantity may be found. When the function is not provided it is necessary to formulate a rule in terms of one variable using the information given. Drawing a diagram to represent the situation is often useful.

• Always test to determine if a stationary point is a maximum or a minimum by checking the sign of the gradient to the left and right of the point.

• Check whether or not the local maximum or minimum is the absolute maximum or minimum. The absolute maximum or minimum may be the value of the function at one end of a specified interval.

Applications of antidifferentiation• When the derivative of a function is known, antidifferentiation can provide the

original function. Since the original function may have contained a ‘constant’, this must be allowed for, and can be found using the boundary conditions provided in the question.

summarychange in ychange in x---------------------------

dydx------

y

x0

Local maximum

y

x0

Local minimum

y

x0

Point ofinflection

f (x)y

x0 a b

Absolutemaximum inthe interval[a, b]

Localmaximum


Multiple choice

1 The rate of change of f (x) = 2x3 − 5x2 + 7 when x = 2 is:

2 If V = −3t2 + 7t + 50 then the average rate of change between t = 1 and t = 4 is:

3 If f (x) = 5 + 15x + 6x2 − x3 then the gradient is zero when x equals:

4 The curve y = x2 − 10x + 21 has:

5 When x = −2, the graph of y = 2x2 + 3x − 5:

6 For a particular function g(x), g(1) = 0 and g ′(x) < 0 if x ≥ 1. The graph which could represent g(x) is:

7 The maximum value of f (x) = −2x2 + 8x is:

8 The local minimum value of h(x) = x3 + 6x2 − 28x − 3 occurs when x equals:

A −4 B 7 C −36 D 0 E 4

A −10 B −10 C −6 D −8 E 0

A 1 or −5 B 1 or 5 C −1 or 5 D −1 or −5 E 0 and −1

A a local maximum at (5, 0) B a local minimum at (5, −4)C a point of inflection at (5, 0) D a local maximum at (5, −4)E a point of inflection at (5, −4)

A is increasing B has a local maximumC has a point of inflection D has a local minimumE is decreasing

A B C

D E

A 40 B 0 C 4 D −24 E 8

A 2 B –4 C 0 D −3 E 1

CHAPTERreview

9A

9A 23---

9B

9B

9B

9Bg(x)

y

x0 1

g(x)y

x0

1

y

x0 1

g(x)

g(x)

y

x0 1

g(x)

y

x0

1

9B

9B13---

C h a p t e r 9 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n 4219 The function g(x) = (x + 3)3 has:

10 A curve with a local maximum and a local minimum is:

11 The antiderivative of 12x + 3 is:

12 If the gradient of a curve is and its y-intercept is −3, then its rule is:

Short answer1 If the position of a particle moving in a straight line is given by the rule

x(t) = −2t2 + 8t + 3, where x is in centimetres and t is in seconds, find:a the initial position of the particleb the rate of change of displacement (that is, the velocity) at any time, tc the rate of change when t = 4d when and where the velocity is zeroe whether the particle is moving to the left or to the right when t = 3f the distance travelled in the first 3 seconds.

2 For the function f (x) = x3 − 3x + 2:a find the y-interceptb find the x-interceptsc find the stationary points and state their typed sketch the graph of f (x).

3 If the volume of liquid in a vat, V litres, during a manufacturing process is given by V = 6t − t2, where t ∈ [0, 6], find:a the rate of change 2 hours after the

vat starts to fill upb when the vat has a maximum volume.

4 If a piece of wire is 80 cm long:a find the area of the largest rectangle

that can be formed by the wireb determine whether a circle would give

a larger area.

5 Find the maximum possible volume of a fully enclosed cylindrical water tank given that the total internal surface area of the tank is 600 π square units.

A a local maximum when x = −3 B a point of inflection when x = −3C a local minimum where x = −3 D a local minimum where x = 3E a point of inflection where x = 3

A y = x3 + 2x2 − 7x + 1 B y = x2 − 3x + 1 C y = x3 + 7D y = (x − 2)3 E y = x2 + 6x

A 6x2 + 3x B 24x2 + 3x + c C 24x2 + 3xD 6x2 + 3x + c E 6x2

A y = x3 + x2 − 10x − 3 B y = x3 + x2 − 5x − 3 C y = x3 + 3x2 − 10x − 3

D y = x3 + x2 − 10x − 10 E y = x4 − 10x2

9B

9C

9D

9Ddydx------ x 2–( ) x 5+( )=

13--- 3

2--- 1

3--- 3

2---

13--- 3

2--- 1

4---

9A

9B

9C

9C

9C


6 The rate of increase of height, h metres, of an ascending helicopter at any time, t minutes after

it takes off is .

a Find an expression for the height at any time.b Find the height 6 minutes after takeoff.c Find the maximum height reached in the first 9 minutes.

7 A particle travels such that its velocity at any time, t, is given by v = 2t + 1.a Given that velocity represents the rate of change of position, x, write down the relationship

between v and x.b If x = 3 when t = 2 write an expression for x in terms of t.c Find the position of the particle when t = 10.

Analysis1 A ball is thrown vertically up so that its

height above the ground, h metres, at any time, t seconds, after leaving the thrower’s hand is given by the function h(t) = t − t2 + 2.a Find the height of the ball as it leaves

the thrower’s hand.b Find when and where the ball reaches

its greatest height.c Find when the ball returns to the same

level that it left the thrower’s hand.d If the ball isn’t hit, find when the ball

hits the ground to the nearest thousandth of a second.

e Hence state the domain and range of h(t).f Sketch the graph of h versus t.

2 A piece of wire of length 100 cm is to be cut so that one piece is used to form a square, while the other is used to form a circle. If the edge length of the square is x cm, find, in terms of x,a the radius of the circleb the area of the circle, andc the total area of the two shapes.Show that, when x = 14, the total area is minimum.

h

9D dhdt------ t2 14t– 45+=

9D

9B

83--- 8

9---

9Ctesttest



9

6In this chapter6A Substitution where the

derivative is present in the integrand

6B Linear substitution6C Antiderivatives involving

trigonometric identities6D Antidifferentiation using

partial fractions6E Definite integrals6F Applications of integration6G Volumes of solids of

revolution6H Approximate evaluation

of definite integrals and areas

VCEcoverageArea of studyUnits 3 & 4 • Calculus

Integralcalculus



Integration techniques and applicationsYou will have seen in your Maths Methods course and elsewhere that some functionscan be antidifferentiated (integrated) using standard rules. These common results areshown in the table below where the function f(x) has an antiderivative F(x).

In this chapter you will learn how to find antiderivatives of more complex functionsusing various techniques.

Technique 1: Substitution where the derivative is present in the integrand

Since , n ≠ −1, as an application of the chain rule,

then it follows that:

Since ; f(x) ≠ 0

then it follows that .

The method relies on the derivative, or multiple of the derivative, being present andrecognisable. Then, the appropriate substitutions may be made according to the aboverules.

f(x) F(x)

axn

logekx + c

ekx

sin kx

cos kx

sec2kx

, x ∈ (–a, a)

, x ∈ (–a, a)

axn 1+

n 1+--------------- c+

1x---

ekx

k------- c+

cos kx–k

------------------ c+

sin kxk

-------------- c+

tan kxk

-------------- c+

1

a2 x2–-------------------- Sin 1– x

a--- c+

1–

a2 x2–-------------------- Cos 1– x

a--- c+

aa2 x2+----------------- Tan 1– x

a--- c+

d f x( )[ ]n 1+

dx---------------------------- n 1+( ) f ′ x( ) f x( )[ ]n=

f ′ x( ) f x( )[ ]n dx∫ f x( )[ ]n 1+

n 1+------------------------- c n 1.–≠,+=

d loge f x( )[ ]dx

----------------------------- f ′ x( )f x( )------------=

f ′ x( )f x( )------------∫ dx loge f x( ) c+=


C h a p t e r 6 I n t e g r a l c a l c u l u s 213

Find the antiderivative of the following expressions.

a (x + 3)7 b 4x(2x2 + 1)4 c

THINK WRITE

a Recognise that the derivative ofx + 3 is 1. Let u = x + 3.

a Let u = x + 3.

Find .

Make dx the subject. or dx = du

Substitute for x + 3 and dx. So ∫(x + 3)7 dx = ∫u7 du

Antidifferentiate with respect to u. =

Replace u with x + 3 and state answer in terms of x.

=

b Recognise that 4x is the derivative of2x2 + 1. Let u = 2x2 + 1.

b Let u = 2x2 + 1.

Find .

Make dx the subject. or

Substitute u for 2x2 + 1 and for dx. So ∫4x(2x2 + 1)4 dx

= ∫4x u4

Simplify the integrand by cancelling out the 4x.

= ∫u4 du


Replace u with 2x2 + 1. = + c

c Recognise that 3x2 + 1 is the derivative

of x3 + x. Let u = x3 + x.

c Let u = x3 + x.

Find .

3x2 1+

x3 x+-------------------

1

2dudx------ du

dx------ 1=

3

4

5u8

8----- c+

6x 3+( )8

8------------------- c+

1

2dudx------ du

dx------ 4x=

3dx4x------ du=

4du4x------

du4x------

5

6u5

5----- c+

72x2 1+( )5

5-------------------------

1

2dudx------ du

dx------ 3x2 1+=

1WORKEDExample

Continued over page



THINK WRITE


Substitute u for x3 + x and for

dx.

So

=

Cancel out 3x2 + 1. =

Express the integrand in index form. =


Replace u with x3 + x. =

Express in root notation. =

3 dxdu

3x2 1+------------------=

4du

3x2 1+------------------ ∫3x2 1+

x3 x+------------------dx

∫3x2 1+

u------------------ du

3x2 1+------------------×

5 ∫du

u-------

6 ∫u12---–du

7 2u12---

c+

8 2 x3 x+( )12---

c+

9 2 x3 x+ c+

Antidifferentiate the following functions with respect to x.

a b

THINK WRITE

a Express in integral notation. a

Recognise that x + 3 is half of the derivative of x2 + 6x.Let u = x2 + 6x. Let u = x2 + 6x.

Find .


Substitute u for x2 + 6x and for dx. So =

Factorise 2x + 6. =

Cancel out x + 3 and express u in index form on the numerator.

=

f x( ) x 3+x2 6x+( )3

--------------------------= f x( ) x2 1–( ) cos 3x x3–( )=

1 ∫ x 3+x2 6x+( )3

-------------------------dx

2

3

4dudx------ du

dx------ 2x 6+=

5 dxdu

2x 6+---------------=

6du

2x 6+--------------- ∫ x 3+

x2 6x+( )3-------------------------dx ∫x 3+

u3------------ du

2x 6+---------------×

7 ∫x 3+u3

------------ du2 x 3+( )--------------------×

8 ∫12---u 3– du

2WORKEDExample



THINK WRITE


Replace u with x2 + 6x. =

Express the answer with a positive index number. (Optional.)

=

b Express in integral notation. b

Recognise that x2 − 1 is a multiple of the derivative of 3x − x3.Let u = 3x − x3. Let u = 3x − x3.

Find .


Substitute u for 3x − x3 and for dx.

So

=

Factorise 3 − 3x2. =

=

Cancel out x2 − 1. = du


Replace u with 3x − x3. =

9u 2–

4-------– c+

10x2 6x+( ) 2–

4---------------------------– c+

111

4 x2 x+( )2------------------------- c+–

1 ∫ x2 1–( ) cos 3x x3–( ) dx

2

3

4dudx------ du

dx------ 3 3x2–=

5 dxdu

3 3x2–-----------------=

6du

3 3x2–----------------- ∫ x2 1–( ) cos 3x x2–( ) dx

∫ x2 1–( ) cos udu

3 3x2–-----------------×

7 ∫ x2 1–( ) cos udu

3 1 x2–( )----------------------×

∫ x2 1–( ) cos udu

3 x2 1–( )–-------------------------×

8 ∫ cos u–3

----------------

9sin – u3

--------------- c+

10sin– 3x x3–( )

3-------------------------------- c+

Evaluate the following indefinite integrals.

a ∫ cos x sin4x dx b c d ∫ sin2x cos3x dx

THINK WRITEa Recognise that cos x is the derivative of sin x. a

Let u = sin x. Let u = sin x.

Tan 1– x2---

4 x2+------------------dx∫ ∫ loge4x

x----------------dx

1

2

3WORKEDExample

Continued over page



THINK WRITE

Find .


Substitute u for sin x and for dx. So∫cos x sin4x dx = ∫Cancel out cos x. = ∫Antidifferentiate with respect to u. = u5 + c

Replace u with sin x. = sin5x + c

b Recognise that is half of the

derivative of Tan−1 .

b

Let u = Tan−1 . Let u = Tan−1 .

Find .


Substitute u for Tan−1 and

for dx.

So dx

=

Cancel out 4 + x2. =


Replace u with Tan−1 .=

c Recognise that is the derivative of loge4x. c

Let u = loge4x. Let u = loge4x.

Find .

Make dx the subject. or dx = x du

3dudx------ du

dx------ cos x=

4 dxdu

cos x------------=

5du

cos x------------ cos x( )u4 du

cos x------------

6 u4 du

715---

815---

11

4 x2+--------------

x2---

2x2--- x

2---

3dudx------

dudx------ 2

4 x2+--------------=

4 dx4 x2+( )du

2--------------------------=

5x2--- 4 x2+( )du

2--------------------------

Tan 1– x2---

4 x2+-----------------∫

u4 x2+--------------∫ ×

4 x2+( )du2

--------------------------

6u2--- du∫

7u2

4----- c+

8x2---

Tan 1– x2---

2

4------------------------- c+

11x---

2

3dudx------ du

dx------ 1

x---=

4



THINK WRITE

Substitute u for loge4x and x du for dx in the integral.

So

=

Cancel out x. =

Antidifferentiate with respect to u. = u2 + c

Replace u by loge4x. = (loge4x)2 + c

d Express cos3x as cos x cos2x. d ∫ sin2x cos3x dx

= ∫ sin2x cos x cos2x dx

Express cos x cos2x as cos x (1 − sin2x) (using the identity sin2x + cos2x = 1).

= ∫ sin2x cos x (1 − sin2x) dx

Let u = sin x as its derivative is a factor of the new form of the function.

Let u = sin x.

Find .


Substitute u for sin x and for dx. So ∫ sin2x cos3x dx

= ∫ u2 cos x (1 − u2)

Cancel out cos x. = ∫ u2(1 − u2) du

Expand the integrand. = ∫ (u2 − u4) du

Antidifferentiate with respect to u. = u3 − u5 + c

Replace u by sin x. = sin3x − sin5x + c

5 ∫loge4x

x----------------dx

ux--- x du×∫

6 u du∫7

12---

812---

1

2

3

4dudx------ du

dx------ cos x=

5 dxdu

cos x------------=

6du

cos x------------

ducos x------------

7

8

913--- 1

5---

1013--- 1

5---

If f ′(x) = 4xex2 and f(0) = 5, find f(x).THINK WRITE

Express f(x) in integral notation. f(x) = ∫ 4xex2 dx

Recognise that 4x is twice the derivative of x2.Let u = x2. Let u = x2.

Find .

1

2

3

4dudx------ du

dx------ 2x=

4WORKEDExample

Continued over page



Substitution where thederivative is present in the integrand

1 Find the antiderivative for each of the following expressions.

a 2x(x2 + 3)4 b 2x(6 − x2)−3

c 3x2(x3 − 2)5 d 2(x + 2)(x2 + 4x)−3

e f

g 3x2(x3 − 5)2 h

i 4x3ex4 j (2x + 3) sin(x2 + 3x − 2)

k (3x2 + 5) cos(x3 + 5x) l cos x sin3x

THINK WRITE


Substitute u for x2 and for dx. So f(x) = ∫ 4xeu

Cancel out 2x. = ∫ 2eu du

Antidifferentiate with respect to u. = 2eu + cReplace u by x2. f(x) = 2ex2

+ cSubstitute x = 0 and f(0) = 5. f(0) = 2e0 + c = 5Solve for c. 2 + c = 5

c = 3State the function f(x). Therefore f(x) = 2ex2

+ 3.

5 dxdu2x------=

6du2x------

du2x------

7

8

9

10

11

12

remember1. Since

then .

2. Since

then .

d f x( )[ ]n 1+

dx---------------------------- n 1+( ) f ′ x( ) f x( )[ ]n n 1–≠,=

f ′ x( ) f x( )[ ]n dx∫ f x( )[ ]n 1+

n 1+------------------------- c n 1–≠,+=

d loge f x( )dx

-------------------------- f ′ x( )f x( )------------=

f ′ x( )f x( )------------∫ dx loge f x( ) c+=

remember

6A

Mathca

d

Anti-differentiation

WORKEDExample

1 & 2

2x 5+( ) x2 5x+2x 3–

x2 3x–( )4------------------------

3x2 4x+

x3 2x2+------------------------



2

Given that the derivative of (x2 + 5x)4 is 4(2x + 5)(x2 + 5x)3, then the antiderivative of8(2x + 5)(x2 + 5x)3 is:

3

a The integral dx can be found by making the substitution ‘u’ equal to:

b After the appropriate substitution the integral becomes:

c Hence the antiderivative of is:

4 Antidifferentiate each of the following expressions with respect to x.

m −sin4x cos x n

o sec2x tan3x p

A 2(x2 + 5x)4 + c B (x2 + 5x)4 + c C 4(x2 + 5x)4 + c

D 2(x2 + 5x)2 + c E (x2 + 5x)2 + c

A x2 B x C D x2 + 3 E 2x

A B C

D E

A B C

D E

a 6x2(x3 − 2)5 b x(4 − x2)3

c x2(x3 − 1)7 d (x + 3)(x2 + 6x − 2)4

e (x + 1)(x2 + 2x + 3)−4 f

g h

i (6x − 3)ex2 − x + 3 j x2ex + 2

k (x + 1) sin(x2 + 2x − 3) l (x2 − 2) cos(6x − x3)

m sin 2x cos42x n cos 3x sin23x

o p

logex

x-------------

Sin 1– x( )2

1 x2–-------------------------


12---

12---


x

x2 3+------------------∫

x

u12--- du∫ 1

2--- u

12---–∫ du 1

2--- u 3+( )

12---–∫ du

u 3+( )12--- du∫ 2 u

12---– du∫

xx2 3+--------------

23--- x2 3+( )

32---

c+ 4 x2 3+( )12---

c+ 23--- x2 6+( )

32---

c+

x2 6+( )12---

c+ x2 3+( )12---

c+

WORKEDExample

2

4x 6+

x2 3x+----------------------

2x 5–x2 5x– 2+( )6

---------------------------------- x2 1–( ) 4 3x– x3+

loge3x

2x----------------

(4x 2) loge (x– 2 x)–

x2 x–-----------------------------------------------------



5 Evaluate the following indefinite integrals.

6 Find the antiderivative for each of the following expressions.

7 If and f(2) = 1 find f(x).

8 If and f(0) = 3 find f(x).

9 If g(1) = −2 and then find g(x).

10 If and g′(x) = 16 sin x cos3x then find g(x).

a b

c d

e f

g h

i j

k l

m n

o

a b

c sin x sec3x d

e f

g h

i j

k sin3x cos2x l cos3x sin4x

m

WORKEDExample

3 x x2 1+( )52--- dx∫ x 1 x2– dx∫

ex 3 2ex+( )4 dx∫ sin xcos3x------------- dx∫

x2 sin x3 dx∫ sin x ecos x dx∫ cos x loge sin x( )

sin x------------------------------------------- dx∫ e3x 1 e3x–( )2 dx∫2– Cos 1–

x3---

9 x2–--------------------------- dx∫ 2x 1+( ) x x2 3–+ dx∫x 2+( ) cos x2 4x+( ) dx∫ e x 1+

x 1+---------------- dx∫

Sin 1– 4x

1 16x2–------------------------ dx∫ Tan 1– x

1 x2+------------------- dx∫

x1 4x2–----------------- dx∫

cos x

1 3 sin x+------------------------------ sec2x 2 tan x+

e2x

e2x 3–( )2-----------------------

sec2x5 tan x–( )3

----------------------------4

xlogex----------------

logex( )3

x--------------------

etan x

cos2x-------------

ex e x––

ex e x–+---------------------- sin x cos x–

sin x cos x+------------------------------

loge tan x( )sin x cos x---------------------------

WORKEDExample

4

f ′ x( ) x

x2 5+------------------=

f ′ x( ) e x

x--------=

g′ x( )4 logex2

x--------------------=

gπ4---

0=



Technique 2: Linear substitution

For antiderivatives of the form where g(x) is a linear function,

that is, of the type mx + c, and f(x) is not the derivative of g(x), the substitution u = g(x)is often successful in finding the integral. Examples of this type of integral are:

1. . In this example f(x) = 1 and g(x) = 4x + 1 with n = . By letting

u = 4x + 1, and consequently dx = du, the integral becomes which can be

readily antidifferentiated.

2. . In this example f(x) = 4x and g(x) = x − 3 with n = 4. By letting

u = x − 3, the function f(x) can be written in terms of u, that is, u = x − 3, thus

4x = 4(u + 3) and further, dx = du. The integral becomes which

can be readily antidifferentiated.The worked examples below illustrate how the use of the substitution u = g(x)

simplifies integrals of the type .

f x( ) g x( )[ ]n dx n 0≠,∫

4x 1+ dx∫ 12---

14---

14--- u du∫

4x x 3–( )4 dx∫4 u 3+( ) u du×∫

f x( ) g x( )[ ]n dx∫

i Using the appropriate substitution, express the following integrals in terms of u only.ii Evaluate the integrals as functions of x.

a b

THINK WRITE

a i Let u = x − 2. a i Let u = x − 2 and x = u + 2.

Find .

Make dx the subject. dx = du

Substitute u for x − 2, u + 2 for x and du for dx.

So

=

Expand the integrand. =

ii Antidifferentiate with respect to u. ii =

Replace u with x − 2. =

x x 2–( )52--- dx∫ x2

x 1+---------------- dx∫

1

2dudx------ du

dx------ 1=

3

4 x x 2–( )52--- dx∫

u 2+( )u52--- du∫

5 u72---

2u52---

+ dx∫

129---u

92--- 4

7---u

72---

c+ +

229--- x 2–( )

92--- 4

7--- x 2–( )

72---

c+ +

5WORKEDExample

Continued over page



THINK WRITE

Take out the factor of

.=

Simplify the other factor.=

=

b i Express x + 1 in index form. b i

=

Let u = x + 1. Let u = x + 1.

Find . = 1

Make dx the subject. or dx = du

Express x in terms of u. x = u − 1

Hence express x2 in terms of u. x2 = u2 − 2u + 1

Substitute u for x + 1, u2 − 2u + 1

for x2 and du for dx.So

=


ii Antidifferentiate with respect to u. ii =

Replace u with x + 1. =

Take out as a factor. =

Simplify the other factor. =

=

=

3

2 x 2–( )72---

2 x 2–( )72--- x 2–

9----------- 4

7---+

c+

42 x 2–( )

72--- 7x – 14 36+

63------------------------------

c+

2 x 2–( )72--- 7x 22+

63------------------

c+

1x2

x 1+---------------- dx∫

x2 x 1+( )12---– dx∫

2

3dudx------ du

dx------

4

5

6

7 x2

x 1+( )12---– xd∫

u2 2u– 1+( )u12---– du∫

8 u32---

2u12---

– u12---–

+ du∫

1 25---u

52--- 4

3---u

32---

– 2u12---

c+ +

2 25--- x 1+( )

52--- 4

3--- x 1+( )

32---

– 2 x 1+( )12---

c+ +

3 2 x 1+( )12---

2 x 1+( )12--- x 1+( )2

5------------------- 2 x 1+( )

3--------------------– 1+ c+

4 2 x 1+( )12--- x2 2x 1+ +( )

5-------------------------------- 2x– 2–( )

3------------------------ 1+ + c+

2 x 1+( )12--- 3x2 6x 3 10x– 10– 15+ + +

15---------------------------------------------------------------------- c+

2 x 1+( )12--- 3x2 4x– 8+

15------------------------------ c+



Note: Recall that the logarithm of a negative number cannot be found.

a Find the antiderivative of . b State the domain of the antiderivative.

THINK WRITE

a Since e2x = (ex)2, it can be antidifferentiated the same as a linear function by letting u = ex + 1.

a Let u = e x + 1.

Find . = ex

Make dx the subject. or dx =

Express ex in terms of u. and ex = u − 1

Substitute u for ex + 1 and for dx.So

=

Cancel out ex. =

Substitute u − 1 for the remaining ex. =

Simplify the rational expression. =

Antidifferentiate with respect to u. = u − logeu + c

Replace u with ex + 1. = ex + 1 − loge(ex + 1) + c

b ex + 1 > 0 for all values of x as ex > 0 for all x. The function loge f(x) exists wherever f(x) > 0.

b For loge(ex + 1) to exist ex + 1 > 0, which

it is for all x.

State the domain. Therefore the domain of the integral is R.

e2x

ex 1+--------------

1

2dudx------ du

dx------

3duex------

4

5du

ex

------e2x

ex 1+-------------- dx∫

exex

u---------- du

ex------×∫

6ex

u----- du∫

7u 1–

u------------ du∫

8 1 1u---–

du∫9

10

1

2

6WORKEDExample

rememberFor antiderivatives of the form , make the substitution

u = g(x) and so [g(x)]n dx, n ≠ 0 becomes g′(x) un du, n ≠ 0. This technique can be used for the specific case where g = mx + c since g′(x) = m. The function f(x) needs to be transformed in terms of the variable u as well.

f x( ) g x( )[ ]n dx n 0≠,∫remember



Linear substitution

1 By making the appropriate substitution for u:i express the following integrals in terms of uii evaluate the integrals as functions of x.

2

a The integral can be found by letting u equal:

b The integral then becomes:

3

a Using the appropriate substitution, becomes:

a b

c d

e f

g h

i j

k l

m n

o p

A B C x + 2 D 4x E 2x

A B C

D E

A B C

D E

6BWORKEDExample

5

Mathca

d


4x 3–----------- dx∫ 2

3x 5+--------------- dx∫

4x 1+ dx∫ 3 2x– dx∫x x 1+( )3 dx∫ 4x x 3–( )4 dx∫2x 2x 1+( )4 dx∫ 3x 1 3x–( )5 dx∫6x 3x 2–( )

34--- dx∫ x 2x 7+( )

13--- dx∫

x x 3+ dx∫ x 3x 4– dx∫x 2+( ) x 4–( )

32--- dx∫ x 3–( ) 2x 1+( )

52--- dx∫

2x

x 6–---------------- dx∫ 3x

8 x–---------------- dx∫


4x x 2+ dx∫x 2+ x

u52--- du∫ 2u

12---

4u12---–

– du∫ 2u

12--- du∫

4u12---

2u32---

– du∫ 4u

32---

8u12---

– du∫


x2

x 1–---------------- dx∫

u du∫ u32---

2u12---

u12---–

+ + du∫ u

12---

2u12---–

+ du∫

u52---

2u32---

u12---

+ + du∫ u

32--- du∫


C h a p t e r 6 I n t e g r a l c a l c u l u s 225b The result of the integration is:

4 Find the antiderivative of each of the following expressions.

5 a If and f(1) = −2, find f(x).

b State the domain of f(x).

6 a If and f(0) = 1, find f(x).


7 a Given that and g(2) = 0, find g(x).

b State the domain of g(x).

8 a Given that g(0) = 2 – loge2 and , find g(x).


A B

C D

E

a x2(x − 4)4 b x2(5 − x)3 c

d e f

g h i

j k l

m n o

p q r

s t u

v

23--- x 1–( )

32---

c+ 23--- x 1–( )

32---

4 x 1–( )12---

c+ +

25--- x 1–( )

52---

c+ 25--- x 1–( )

52--- 4

3--- x 1–( )

32---

2 x 1–( )12---

c+ + +

27--- x 1–( )

72--- 4

5--- x 1–( )

52--- 4

3--- x 1–( )

32---

c+ + +

WORKEDExample

6a x2

x 1–

x2 3 x– x2 x 2+( )43---

x2 1 x–( )34---

x 1+( )2 x 2– x 3–( )2 x 1+ex

ex 1+--------------

x2

x 1+---------------- 2x2

3 x–---------------- x3 x 1–

x3

x 4+---------------- 2x3

1 x–---------------- x 3+

x 2–( )2-------------------

2x 1–x 1+( )3

------------------- 4xx 2+( )2

------------------- x2

x 1–( )2-------------------

x 3+( )2

x 2+------------------- x 2–( )2

2 x–------------------- e2x

ex 2+--------------

e3x

ex 1–--------------

WORKEDExample

6b

f ′ x( ) 5 x–( )12---

– 10 5 x–( )12---–

+=

f ′ x( ) 5 x 1+( )32---

2---------------------- 3 x 1+( )

12---

– x 1+( )12---–

2---------------------+=

g′ x( ) 2x 1+x 1–( )2

-------------------=

g′ x( ) e2x

ex 1+--------------=



Technique 3: Antiderivatives involving trigonometric identities

Different trigonometric identities can be used to antidifferentiate sinnx and cosnx; n ∈ J+ depending on whether n is even or odd. Functions involving tan2ax are also discussed.

Even powers of sin x or cos xThe double-angle trigonometric identities can be used to antidifferentiate even powersof sin x or cos x. The first identity is:

cos 2x = 1 − 2 sin2x

= 2 cos2x − 1

Therefore sin2x = (1 − cos 2x)

or cos2x = (1 + cos 2x)

The second identity is: sin 2x = 2 sin x cos x

or sin x cos x = sin 2x

These may be expressed in the following general forms:

sin2ax = (1 – cos 2ax) Identity 1

cos2ax = (1 + cos 2ax) Identity 2

sin ax cos ax = sin 2ax Identity 3

12---

12---

12---

12---

12---

12---

Find the antiderivative of the following expressions.

a sin2 b 2cos2

THINK WRITE


Use identity 1 to change sin2 . =

Take the factor of to the front of the integral. =


Simplify the answer. =


x2--- x

4---

1 sin2 x2--- dx∫

2x2--- 1

2--- 1 cos x–( ) dx∫

312---

12--- 1 cos x–( ) dx∫

412--- x sin x–( ) c+

5x2---

12--- sin x c+–

1 2cos2 x4--- dx∫

7WORKEDExample



Odd powers of sin x or cos xFor integrals involving odd powers of sin x or cos x the identity:

sin2x + cos2x = 1can be used so that the ‘derivative method’ of substitution then becomes applicable.The following worked example illustrates the use of this identity whenever there is anodd-powered trigonometric function in the integrand.

THINK WRITE

Use identity 2 to change cos2 . =

Simplify the integral. =


2x4--- 2 1

2---

1 cosx2---+

dx∫3 1 cos

x2---+

dx∫4 x 2sin

x2--- c+ +

Evaluate the following indefinite integrals as functions of x.

a b

THINK WRITE

a Use identity 3 to change sin x cos x.Note: The integral could be antidifferentiated using technique 1 since the derivative of sin x is cos x.

a

=


b Express sin2 cos2 as a perfect square. b

=

Use identity 3 to change sin cos . =

Square the identity. =


Use identity 1 to change sin2x. =


Simplify the answer. =

sin x cos x dx∫ 4 sin2 x2--- cos2 x

2--- dx∫

1 sin x cos x dx∫12--- sin 2x dx∫

214--- cos 2x c+–

1x2--- x

2--- 4 sin2 x

2--- cos2 x

2--- dx∫

4 sinx2--- cos

x2---

2 dx∫

2x2--- x

2--- 4 1

2--- sin x( )2 dx∫

3 4 14--- sin2x( ) dx∫

4 sin2x dx∫5

12--- 1 cos 2x–( ) dx∫

612--- x 1

2---sin 2x–( ) c+

7x2--- 1

4---sin 2x c+–

8WORKEDExample



Find the antiderivative of the following expressions.a cos3x b cos x sin 2x c cos42x sin32x

THINK WRITE


Factorise cos3x as cos x cos2x. =

Use the identity: (1 − sin2x) for cos2x. =

Let u = sin x so the derivative method can be applied.

Let u = sin x.

Find .


Substitute u for sin x and for dx. So

=

Cancel out cos x. =

Antidifferentiate with respect to u. = u − u3 + c

Replace u with sin x. = sin x − sin3x + c


Use identity 3 in reverse to expresssin 2x as 2 sin x cos x. =

Simplify the integrand. =

Let u = cos x so that the derivative method can be applied.

Let u = cos x.

Find .


Substitute u for cos x and for dx.So

=

1 cos3x dx∫2 cos x cos2x dx∫3 cos x 1 sin2x–( ) dx∫4

5dudx------ du

dx------ cos x=

6 dxdu

cos x------------=

7du

cos x------------ cos x 1 sin2x–( ) dx∫

cos x 1 u2–( ) ducos x------------∫

8 1 u2–( ) du∫9

13---

1013---

1 cos x sin 2x dx∫2

cos x 2 sin x cos x( ) dx∫3 2sin x cos2x dx∫4

5dudx------ du

dx------ sin x–=

6 dxdusin x–

--------------=

7dusin x–

--------------2 sin x cos2x dx∫

2 sin x u2( ) dusin x–

--------------∫

9WORKEDExample



Using the identity sec2x = 1 + tan2xThe identity sec2ax = 1 + tan2ax is used to antidifferentiate expressions involvingtan2ax + c where c is a constant since the antiderivative of sec2x is tan x.

Otherwise, expressions of the form tannx sec2x can be antidifferentiated using the‘derivative method’ of exercise 6A.

THINK WRITE

Cancel out sin x. =

Antidifferentiate with respect to u. = − u3 + c

Replace u with cos x. = − cos3x + c

c Express in integral notation. c

Factorise sin32x as sin 2x sin22x. =

Use the identity 1 − cos22x for sin22x. =

Let u = cos 2x so that the derivative method can be applied.

Let u = cos 2x.

Find .


Substitute u for cos 2x and

for dx.So

Cancel out sin 2x. =


Antidifferentiate with respect to u. = ( u7 − u5) + c

Simplify the result. = u7 − u5 + c

Replace u with cos 2x. = cos72x − cos52x + c

8 2u2 du–∫9

23---

1023---

1 cos42x sin32x dx∫2 cos42x sin 2x sin22x dx∫3 cos42x sin 2x 1 cos22x–( ) dx∫4

5dudx------ du

dx------ 2 sin 2x–=

6 dxdu

2 sin 2x–----------------------=

7du

2 sin 2x–----------------------

u4 sin 2x 1 u2–( ) du2sin 2x–

----------------------∫8

12---u4 1 u2–( ) du–∫

912--- u6 u4–( ) du∫

1012--- 1

7--- 1

5---

11114------ 1

10------

12114------ 1

10------



Find an antiderivative for each of the following expressions.

a b

THINK WRITE

a Express 2 + tan2x as 1 + sec2x using the identity. a

=

Antidifferentiate by rule. There is no need to add c as one antiderivative only is required.

= x + tan x

b Let u = tan 3x so that the derivative method can be applied.

b Let u = tan 3x.

Find .


Substitute u for tan 3x and for dx. So 3 tan2 3x sec2 3x dx

=

Cancel out 3sec23x. =

Antidifferentiate with respect to u. = u3

Replace u with tan 3x. = tan33x

2 tan2x+( ) dx∫ 3 tan23x sec23x dx∫

1 2 tan2x+( ) dx∫1 sec2x+( ) dx∫

2

1

2dudx------ du

dx------ 3 sec23x=

3 dxdu

3 sec23x---------------------=

4du

3 sec23x-------------------- ∫

3 u2 sec23xdu

3sec23x--------------------∫

5 u2 du∫6

13---

713---

10WORKEDExample

remember1. Trigonometric identities can be used to antidifferentiate odd and even powers

of sin x and cos x. These identities are:

sin2ax = (1 − cos 2ax)

cos2ax = (1 + cos 2ax)

sin ax cos ax = sin 2ax

2. The identity sec2ax = 1 + tan2ax is used to antidifferentiate expressions involving tan2ax + c where c is a constant.

12---

12---

12---

remember



Antiderivatives involving trigonometric identities


2 Evaluate the following indefinite integrals as functions of x.

3If a is a constant, then,

a is equal to:

b is equal to:

c is equal to:

a cos2x b sin22x c 2 cos24x d 4 sin23x

e cos25x f sin26x g cos2 h sin2

i 3 cos2 j 2 sin2 k cos2 l sin2

a b

c d

e f

g h

i j

k l

A 2x − sin 2ax + c B x − 2asin 2ax + c C

D E

A B C

D E

A a cos ax − 3a cos3ax + c B a sin ax − 3 cos3ax + c

C D

E

6CWORKEDExample

7

Mathcad

Anti-differentiationx

2--- x

3---

x6--- x

4--- 2x

3------ 3x

2------

WORKEDExample

8 2 sin x cos dx∫ 4 sin 2x cos 2x dx∫3x 3x cossin dx∫ 2 sin 4x cos 4x dx–∫

sin2x cos2x dx∫ sin22x cos22x dx∫2 sin24x cos24x dx∫ 2 sin23x cos23x dx∫6 sin2 x

2--- cos2 x

2--- dx∫ 4 sin2 x

3--- cos2 x

3--- dx∫

sin25x2

------ cos25x2

------ dx∫ 2 sin24x3

------ cos24x3

------ dx–∫mmultiple choiceultiple choice

sin2ax dx∫x2--- sin 2ax

4a------------------– c+

x 12---sinax

2------ c+– x

2---

1a--- sinax

2------ c+–

sin2ax cos2ax dx∫x8--- sin 4ax

32a------------------ c+– x

2--- sin ax

4a-------------- c+– x

4--- cos ax

8a--------------- c+–

xsin 4ax

16a------------------ c+– x

8--- cos 4ax

16a------------------- c+ +

cos3ax dx∫cos4ax

4a---------------- c+ sin4ax

4a--------------- c+

13a------ 3 sin ax sin3ax–( ) c+



4 Find an antiderivative of each of the following expressions.

5 Use the appropriate identities to antidifferentiate the following expressions.


7 Find the following integrals.

8 Find an antiderivative for each of the following expressions:

9 Find the following integrals where n ∈ J+.

a sin3x b cos32x c 6 sin34x d 4 cos3 3x

e sin37x f cos36x g h

i j k l

a sin x cos 2x b cos 2x cos 4x c sin 3x cos 6x

d cos 4x cos 8x e f

a sin x cos4x b sin 2x cos32x c

d cos 3x sin43x e f

a b

c d

e f

g h

i j

k l

m n

a 1 + tan22x b c tan2x sec2x

d tan3x sec2x e 4 tan52x sec22x f

g tan2x sec4x h 6 tan22x sec42x i

j 3 tan33x sec43x k l 12 tan56x sec66x

a b c

d e

WORKEDExample

9a3 sin3 x

2--- 2 cos3 x

3---

sin33x2

------ cos35x2

------ sin33x4

------ cos34x3

------

WORKEDExample

9bsin

x2--- cos x cos

x3--- cos

2x3

------

sinx2--- cos5 x

2---

cosx5--- sin6 x

5--- cos

2x3

------ sin72x3

------

WORKEDExample

9c cos2x sin3x dx∫ sin2x cos3x dx∫cos22x sin32x dx∫ sin23x cos33x dx∫cos2 x

2---sin3 x

2--- dx∫ sin23x

2------cos33x

2------ dx∫

4 cos2 x3--- sin3 x

3--- dx∫ 6– sin25x

4------ cos35x

4------ dx∫

sin3x cos4 dx∫ cos32x sin42x dx∫2sin32x cos52x dx∫ 2 cos33x sin63x dx–∫4 sin3 x

2--- cos6 x

2--- dx∫ cos33x

2------ sin73x

2------ dx∫

WORKEDExample

10 1 tan2 x3---+

8 tan4 x2--- sec2 x

2---

2 tan2 x2--- sec4 x

2---

tan4 x5--- sec4 x

5---

sin x cosnx dx∫ cos x sinnx dx∫ sec2x tannx dx∫sin3x cosnx dx∫ cos3x sinnx dx∫



10 If f ′(x) = 6 sin x cos2x and , find f(x).

11 If f ′(x) = 4 sin22x cos22x and , find f(x).

12 Find g(x) if g′(x) = sin3 cos4 and g(0) = .

The graph of a function and the graphs of its antiderivatives

Given f(x) = , what do the graphs of its

antiderivatives look like?

Using a graphics calculator, press , enter

Y1= 2(cos(X/2))2, move down to , press

and select 9:fnInt(. Complete to obtain fnInt(Y1,X,0,X).

(Remember that to insert the symbol Y1, press , select and

1:Function. Then select 1:Y1 and press (and similarly for any Y variable).

As the given function is trigonometric, press

and select 7: ZTrig.

(Since the numeric integral is repeatedly applied for every X-value on the screen, the antiderivative graph can take some time to plot. You can speed it up considerably by changing the value of Xres in the WINDOW settings to 5.)

1 Which is the graph of f(x) = and which is the graph of the antiderivative?

The antiderivative graph in the second screen is the line that cuts 0 at x = 0, since the integral from 0 to 0 of any function is 0. To see another

antiderivative graph, go to , press , select 9 and complete 9: fnInt(Y1,X,1,X) and

then press .

2 Generate another two antiderivative graphs on your calculator. Sketch the function and the four antiderivative graphs. Describe any relationships you can find.

3 Choose another function and investigate the relationship between the graph of the function and the graphs of its antiderivatives.

fπ3---

0=

fπ4---

π=

x2--- x

2--- 4

35------–

2cos2x2---

Y=

Y2 =

MATH

VARS Y-VARS

ENTER

ZOOM

2cos2x2---

Y3 = MATH

GRAPH



Technique 4: Antidifferentiation using partial fractions

Recall that rational expressions, in particular those with denominators that can beexpressed with linear factors, can be transformed into partial fractions. A summary oftwo common transformations is shown in the table below. These transformations areuseful when the degree of the numerator is less than the degree of the denominator;otherwise long division is generally required before antidifferentiation can be performed.

We have seen how this procedure simplifies the sketching of graphs of rational func-tions. Similarly, expressing rational functions as partial fractions enables them to beantidifferentiated quite easily. However, it is preferable to use a substitution method, ifit is applicable, as the partial-fraction technique can be tedious.

Rational expression Equivalent partial fraction

where f(x) is a linear function


f x( )ax b+( ) cx d+( )

---------------------------------------- Aax b+( )

-------------------- Bcx d+( )

--------------------+

f x( )ax b+( )2

---------------------- Aax b+( )2

---------------------- Bax b+( )

--------------------+

Find a, b and c if ax(x − 2) + bx(x + 1) + c(x + 1)(x − 2) = 2x − 4.THINK WRITE

Let x = 0 so that c can be evaluated. Let x = 0, −2c = −4Solve the equation for c. c = 2Let x = 2 so that b can be evaluated. Let x = 2, 6b = 0Solve the equation for b. b = 0Let x = −1 so that a can be evaluated. Let x = −1, 3a = −6Solve the equation for a. a = −2State the solution. Therefore a = −2, b = 0 and c = 2.

123

4

5

6

7

11WORKEDExample

For each of the following rational expressions:i express as partial fractions ii antidifferentiate the result.

a b

THINK WRITE

a i Express the rational expression as two separate fractions with denominators (x + 2) and (x − 3) respectively.

a i

Express the partial fractions with the original common denominator.

=

x 7+x 2+( ) x 3–( )

----------------------------------- 2x 3–x2 3x– 4–---------------------------

1x 7+

x 2+( ) x 3–( )---------------------------------- = a

x 2+( )----------------- b

x 3–( )----------------+

2a x 3–( ) b x 2+( )+

x 2+( ) x 3–( )----------------------------------------------

12WORKEDExample



THINK WRITE

Equate the numerator on the left-hand side with the right-hand side.

so x + 7 = a(x − 3) + b(x + 2)

Let x = −2 so that a can be evaluated. Let x = −2, and thus 5 = −5a

Solve for a. a = −1

Let x = 3 so that b can be evaluated. Let x = 3, and thus 10 = 5b

Solve for b. b = 2

Rewrite the rational expression as partial fractions.

Therefore

ii Express the integral in partial fraction form.

ii

=

Antidifferentiate by rule. = −loge(x + 2) + 2 loge(x − 3) + c (x > 3)

Simplify using log laws.= loge + c

b i Factorise the denominator. b i =

Express the partial fractions with denominators (x − 4) and (x + 1) respectively.

=

Express the right-hand side with the original common denominator.

=

Equate the numerators. So 2x − 3 = a(x + 1) + b(x − 4)

Let x = 4 to evaluate a. Let x = 4, 5 = 5a

Solve for a. a = 1

Let x = −1 to evaluate b. Let x = −1, −5 = −5b

Solve for b. b = 1

Rewrite the rational expression as partial fractions.

Therefore =

ii Express the integral in its partial fraction form.

ii

=

Antidifferentiate by rule. = loge(x − 4) + loge(x + 1) + c (x > 4)

Simplify using log laws. = loge[(x − 4)(x + 1)] + c (x > 4)

or loge(x2 − 3x − 4) + c (x > 4)

3

4

5

6

7

8x 7+

x 2+( ) x 3–( )---------------------------------- 1–

x 2+------------ 2

x 3–-----------+=

1x 7+

x 2+( ) x 3–( )---------------------------------- dx∫

1–x 2+------------ 2

x 3–-----------+

dx∫2

3 x 3–( )2

x 2+-------------------

12x 3–

x2 3x– 4–-------------------------- 2x 3–

x 4–( ) x 1+( )----------------------------------

2a

x 4–----------- b

x 1+------------+

3a x 1+( ) b x 4–( )+

x 4–( ) x 1+( )----------------------------------------------

4

5

6

7

8

92x 3–

x2 3x– 4–-------------------------- 1

x 4–----------- 1

x 1+------------+

12x 3–

x2 3x– 4–-------------------------- dx∫

1x 4–----------- 1

x 1+------------+

dx∫2

3



Find the following integrals.

a b

THINK WRITE

a Factorise the denominator of the integrand. a =

Express into partial fractions with denominators (1 − x) and (1 + x).

=


=

Equate the numerators. so 2 = a(1 + x) + b(1 − x)

Let x = 1 to find a. Let x = 1, 2 = 2a

Solve for a. a = 1

Let x = −1 to find b. Let x = −1, 2 = 2b

Solve for b. b = 1

Express the integrand in its partial fraction form.

Therefore

=

Antidifferentiate by rule. = −loge(1 − x) + loge(1 + x) + c,(−1 < x < 1)

Simplify using log laws. = loge (−1 < x < 1)

b The degree of the numerator is the same as the degree of the denominator and hence the denominator should divide the numerator using long division.

b

Expand the denominator.

Divide the denominator into the numerator.

Using long division:1

x2 + 5x + 4 ) x2 + 6x − 1

x2 + 5x + 4

x − 5The division yields 1 with remainder(x − 5).

21 x2–-------------- dx∫ x2 6x 1–+

x 4+( ) x 1+( )----------------------------------- dx∫

12

1 x2–-------------- 2

1 x–( ) 1 x+( )----------------------------------

2a

1 x–----------- b

1 x+------------+

3a 1 x+( ) b 1 x–( )+

1 x2–----------------------------------------------

4

5

6

7

8

92

1 x2–-------------- dx∫

11 x–----------- 1

1 x+------------+

dx∫10

111 x+1 x–------------

c+

1

2x2 6x 1–+x 4+( ) x 1+( )

---------------------------------- x2 6x 1–+x2 5x 4+ +---------------------------=

3

13WORKEDExample



THINK WRITE

Rewrite the rational expression using the result of the division.

Therefore

Express as partial

fractions with denominators (x + 4) and (x + 1).

Now

Rewrite the partial fractions with the original common denominator.

=

Equate the numerators. and thus x − 5 = a(x + 4) + b(x + 1)

Let x = −1 to find a. Let x = −1, −6 = 3a

Solve for a. a = −2

Let x = −4 to find b. Let x = −4, −9 = −3b

Solve for b. b = 3

Express the original integrand in its partial fraction form.

Therefore,

=

Antidifferentiate by rule. = x − 2loge(x + 4) + 3loge(x + 1) + c,(x > −1).

4

x2 6x 1–+x 4+( ) x 1+( )

---------------------------------- 1 x 5–x 4+( ) x 1+( )

----------------------------------+=

5x 5–

x 4+( ) x 1+( )---------------------------------- x 5–

x 4+( ) x 1+( )---------------------------------- a

x 4+------------ b

x 1+------------+=

6a x 4+( ) b x 1+( )+

x 4+( ) x 1+( )-----------------------------------------------

7

8

9

10

11

12x2 6x 1–+x 4+( ) x 1+( )

---------------------------------- dx∫1 2–

x 4+------------ 3

x 1+------------+ +

dx∫13

rememberRational polynomials can be antidifferentiated by rewriting the expressions as partial fractions or by long division. If the numerator is of degree less than the denominator then use partial fractions; otherwise rewrite the expression by long division. Two common partial fraction transformations are shown below.




f x( )ax b+( ) cx d+( )

---------------------------------------- Aax b+( )

-------------------- Bcx d+( )

--------------------+

f x( )ax b+( )2

---------------------- Aax b+( )2

---------------------- Bax b+( )

--------------------+

remember



Antidifferentiation usingpartial fractions

1 Find the values of a, b and c in the following identities.a ax + b(x − 1) = 3x − 2b a(x + 2) + b(x − 3) = x − 8c a(x − 4) + b = 3x − 2d a(3x + 1) + b(x − 2) = 5x + 4e a(2 − 3x) + b(x + 5) = 9x + 11f a(x + 2) + bx = 2x − 10g a + b(x + 2) + c(x + 2)(x + 3) = x2 + 4x − 2h a(x + 2)(x − 3) + bx(x − 3) + cx(x + 2) = 3x2 − x + 6

2 Express each of the following rational expressions as partial fractions.

3 Find the antiderivative of each rational expression in question 2.

4

a If = , then:

b Hence dx is equal to:

5

The antiderivative of is equal to:

a b c

d e f

g h i

j k l

m n

A a = 2, b = 3 B a = −2, b = −3 C a = 3, b = 2D a = −2, b = 3 E a = 1, b = −1

A 2loge(x + 6) − 3loge(4 − x) + c B −2loge(x + 6) − 3loge(4 − x) + cC 3loge(x + 6) + 2loge(4 − x) + c D 3loge(x + 6) − 2loge(4 − x) + c

E

A 2loge(x + 3) − loge(x − 2) + c B 2loge

C 2loge D loge(x + 3) − 2loge(x − 2) + c

E loge(x + 1) − 2loge(x − 6) + c

6DWORKEDExample

11

Mathca

d

Partialfractions

WORKEDExample

12i 1x 1+( ) x 2+( )

---------------------------------- 12x 2–( ) x 2+( )

---------------------------------- 6xx 3+( ) x 1–( )

----------------------------------

3xx 2–( ) x 1+( )

---------------------------------- x 3+x 2+( ) x 3+( )

---------------------------------- x 20+x 4–( ) x 4+( )

----------------------------------

4x 5+x 2+( )2

------------------- 5x 26–x 5–( )2

------------------- x 4+x x 2–( )-------------------

7x 4–x 2–( ) x 3+( )

---------------------------------- 8x 10–2x 1+( ) x 3–( )

------------------------------------- 9x 11–3x 2–( ) x 1+( )

-------------------------------------

11 3x–2 x–( ) x 3+( )

---------------------------------- 12 2x–1 x–( ) 3 x–( )

---------------------------------

Mathca

d


WORKEDExample

12ii mmultiple choiceultiple choice5x 10+

24 2x– x2–----------------------------- a

x 6+------------ b

4 x–-----------+

5x 10+24 2x– x2–-----------------------------∫

loge x 6+( )4 x–

---------------------------


x2 x 6–+-----------------------–

x 1+x 6–------------ c+

x 3+x 2–------------ c+


C h a p t e r 6 I n t e g r a l c a l c u l u s 2396 Antidifferentiate each of the following rational polynomials by first expressing them

as partial fractions.

7 By first simplifying the rational expression using long division, find the antiderivativeof each of the following expressions.

8 Evaluate the following integrals in terms of x.

9 a If and f(2) = 3loge2, find f(x).


10 a Find g(x) if and g(4) = 4 − loge5.


a b c

d e f

g h i

j k l

m n

a b c

d e f

g h i

j k l

a b c

d e f

g h i

j

Challenge

k l

m n

WORKEDExample

13a3x 10+x2 2x+------------------ 5x 4–

x2 x– 2–----------------------- x 3+

x2 3x 2+ +---------------------------

6x 1–x2 5x– 6–-------------------------- 5x 7–

x2 4x– 3+-------------------------- x 16+

x2 7x 6+ +---------------------------

7x 9+x2 9–--------------- 7x 1+

x2 1–--------------- 5x

2x2 3x– 2–-----------------------------

16 2x–3x2 7x 6–+------------------------------ x 4+

2x2 5x– 2+------------------------------ 4

4 x2–--------------

3x 4–16 x2–----------------- x 13+

5 4x x2–+--------------------------

WORKEDExample

13bx 1–x 5+------------ x 3+

x 2–------------ x2 1–

x2 3x+-----------------

x2 2x 4+ +x2 4x–

--------------------------- x2 x–x 3+( ) x 1+( )

----------------------------------x2 x 4+ +x2 2x– 3–--------------------------

x2 3x 2–+x2 4–

-------------------------- x3 4x2 x–+x 2+( ) x 1+( )

----------------------------------x3 4x 13–+x2 4x– 5–

-----------------------------

2x3 x2 5–+x2 1–

----------------------------- x2 x– 2+x2 2x 1+ +--------------------------- 2x2 9x– 7+

x2 6x– 9+------------------------------

4 x–x x 2+( )--------------------dx∫ 9x 8+

x 3–( ) x 4+( )----------------------------------dx∫ 5 x 1+( )

x2 25–--------------------dx∫

x2 3+x2 9–--------------dx∫ x2 3x 4–+

x 4–( ) x 2+( )----------------------------------dx∫ x2 4x 1+ +

x2 6x 7–+---------------------------dx∫

x3 x2 4x–+x2 4x– 4+-----------------------------dx∫ 4x2 6x 4–+

2x2

x– 6–------------------------------dx∫ x 1+

x2 4+--------------dx∫

4x 2–x2 9+---------------dx∫

5x2 2x 17+ +x 1–( ) x 2+( ) x 3–( )

---------------------------------------------------dx∫ x2 18x 5+ +x 1+( ) x 2–( ) x 3+( )

---------------------------------------------------dx∫x2 8x 9+ +

x 1–( ) x 2+( )2------------------------------------dx∫ x2 5x 1+ +

x2 1+( ) 2 x–( )------------------------------------dx∫

f ′ x( ) 6x2 1–--------------=

g′ x( ) x2 1+x2 2x– 3–--------------------------=



Definite integrals

The quantity is called the ‘indefinite integral of the function f(x)’. However,

is called the ‘definite integral of the function f(x)’ and is evaluated using

the result that:

=

= F(b) − F(a)where F(x) is an antiderivative of f(x).

The definite integral can be found only if the integrand, f(x), exists for

all values of x in the interval [a, b]; that is, a ≤ x ≤ b.

When using substitution to evaluate definite integrals there is no need to return to anexpression in terms of x providing the terminals are expressed in terms of u. In fact it ismathematically incorrect to show the integral in terms of u but with terminals in termsof x. Therefore when using a substitution, u = f(x), the terminals should also beadjusted in terms of u.

f x( ) dx∫f x( ) dx

a

b

∫

f x( ) dxa

b

∫ F x( )[ ]ab

f x( ) dxa

b

∫

For each of the following integrals, state:i the domain of the integrand ii whether the integral exists.

a b

THINK WRITE

a i For the integrand to exist, must be greater than 0.

a i The integrand exists if .

Solve the inequation for x. x2 < 9–3 < x < 3

State the domain. The domain is (−3, 3).

ii The integral exists for all values of x between the terminals −2 and 2.

ii The integral exists.

b i The integrand does not exist for x = −3 and 1, as these values make the denominator equal to zero.

b i x ≠ –3, 1

State the domain. Domain is R\{–3, 1}.

ii The integral does not exist for all values of x between the terminals 0 and 4 (as 1 lies in the interval).

ii The integral does not exist.

∫2

–2

1

9 x2–------------------ dx ∫

4

0

2x 1–( ) x 3+( )

----------------------------------- dx

1 9 x2– 9 x2– 0>

2

3

1

2

14WORKEDExample



Use an appropriate substitution to express each of the following definite integrals in termsof u, with the terminals of the integral correctly adjusted.

a b

THINK WRITE

a Antidifferentiate the integrand by letting u = x2 − 1 so the derivative method can be applied.

a Let u = x2 − 1.

Find .

Express dx in terms of du. or

Adjust the terminals by finding u when x = 2 and x = 3.

When x = 2, u = 22 − 1= 3

When x = 3, u = 32 − 1= 8

Rewrite the integral.

Therefore the integral is


b Antidifferentiate the integrand by using the linear substitution u = x − 2.

b Let u = x − 2.

Find .

Express dx in terms of du. or dx = du

Express x in terms of u. x = u + 2

Adjust the terminals by finding u when x = 3 and x = 6.

When x = 3, u = 3 − 2= 1

When x = 6, u = 6 − 2= 4

Rewrite the integral.

Therefore the integral is


∫3

2

xx2 1–-------------- dx ∫

6

3

x

x 2–---------------- dx

1

2dudx------ du

dx------ 2x=

3 dxdu2x------=

4

5 xu--- du

2x------×

3

8

∫6 ∫

8

3

12u------ du

1

2dudx------ du

dx------ 1=

3

4

5

6 ∫4

1

u 2+

u12---

------------ du

7 ∫4

1

u12---

2u12---–

+ du

15WORKEDExample



Evaluate the following definite integrals.

a b

THINK WRITE

a Write the integral. a

Factorise the denominator of the integrand.

Consider: =

Express in partial fraction form with denominators x + 1 and x + 4.

=


=

Equate the numerators. x − 2 = a(x + 4) + b(x + 1)

Let x = −1 to find a. Let x = −1, −3 = 3aa = −1

Let x = −4 to find b. Let x = −4, −6 = −3bb = 2

Rewrite the integral in partial fraction form.

So

=

Antidifferentiate the integrand. = [−loge(x + 1) + 2loge(x + 4)]20

Evaluate the integral. = [−loge3 + 2loge6] − [−loge1 + 2loge4]= −loge3 + 2loge6 − 2loge4

Simplify using log laws. = 2loge1.5 − loge3= loge2.25 − loge3= loge0.75(or approx. − 2.88)

b Write the integral. b

Let u = 1 + sin x to antidifferentiate. Let u = 1 + sin x

Find .

Express dx in terms of du. or

∫2

0

x 2–x2 5x 4+ +---------------------------- dx ∫

π2---

0cos x 1 sin x+ dx

1 ∫2

0

x 2–x2 5x 4+ +--------------------------- dx

2x 2–

x2 5x 4+ +--------------------------- x 2–

x 1+( ) x 4+( )----------------------------------

3a

x 1+------------ b

x 4+------------+

4a x 4+( ) b x 1+( )+

x2 5x 4+ +-----------------------------------------------

5

6

7

8 ∫2

0

x 2–x2 5x 4+ +--------------------------- dx

∫2

0

1–x 1+------------ 2

x 4+------------+ dx

9

10

11

1 cos x 1 sin x+ dx0

π2---

∫2

3dudx------ du

dx------ cos x=

4 dxdu

cos x------------=

16WORKEDExample



THINK WRITE

Change terminals by finding u when

x = 0 and x = .

When x = 0, u = 1 + sin 0= 1

When , u = 1 + sin

= 1 + 1= 2

Simplify the integrand. So

=

=

Antidifferentiate the integrand. =

Evaluate the integral. = × – ×

=

or

5

π2---

xπ2---=

π2---

6 cos x 1 sin x+ dx0

π2---

∫cos x( )u

12--- ducos x------------

1

2

∫u

12--- du

1

2

∫7 2

3---u

32---

1

2

8 23--- 2

32--- 2

3--- 1

32---

4 23

---------- 23---–

4 2 2–3

-------------------

By using the substitution x = sin θ, evaluate .

THINK WRITE

Let x = sin θ. Let x = sin θ.

Find .

Make dx the subject. or dx = cos θ dθ

Change the terminals by finding θ when x = and x = 0.

When x = , = sin θ

θ =

When x = 0, 0 = sin θθ = 0

1 x2– dx0

12---

∫

1

2dxdθ------ dx

dθ------ cos θ=

3

412---

12--- 1

2---

π6---

17WORKEDExample

Continued over page



To find the value of a definite integral, press and select 9:fnInt(. Then type in the integrand, the function variable, the lower terminal and the upper terminal. Press

to evaluate the integral.

Alternatively, if the function is already in Y1, press , select 9:fnInt(, complete

9: fnInt(Y1,X,0, 2) and press . (Remember that to insert the symbol Y1, press

, select Y–VARS and 1:Function, then 1:Y1 (similarly for any Y variable).

1 The screen shows both methods for (Worked example 16a).

2 To estimate cos2 dx, press , select 9:fnInt( and complete by entering

2(cos(X ÷2))2,X,0,π) and pressing .

THINK WRITE

Simplify the integrand.

=

=

=

Replace cos2θ by its identity

(1 + cos 2θ).

=

=

Antidifferentiate the integrand.

Evaluate the integral.

=

=

=

=

5 1 x2– dx0

12---

∫1 sin2θ– cos θ dθ

0

π6---

∫cos θ cos θ dθ

0

π6---

∫cos2θ dθ

0

π6---

∫6

12---

12--- 1 cos 2θ+( ) dθ

0

π6---

∫12--- 1 cos 2θ+( )

0

π6---

∫ dθ

7

8

12--- θ 1

2--- sin 2θ+

0

π6---

12---

π6--- 1

2--- sin

π3---+

0 12--- sin 0+( )–

12---

π6---

12--- 3

2-------

+

π12------ 3

8-------+

Graphics CalculatorGraphics Calculator tip!tip! Finding the numeric integral at the HOME screen

MATH

ENTER

MATH

ENTER

VARS

x 2–

x2 5x 4+ +

---------------------------0

2

∫ dx

20

π

∫ x2--- MATH

ENTER



A handy trick to use, if the answer is a simple fraction, is to press , select

1: Frac and press — but it doesn’t work in this case.If the answer could possibly be a fractional multiple of π, first try dividing by π then

pressing , selecting 1: Frac and pressing . In this case, the answer is just π itself. (Don’t expect this trick to always work!)

Definite integrals

1 For each of the following definite integrals , state i the maximal domain

of the integrand f(x) and ii whether the integral exists.

a b c

d e f

g h i

j k l

m n

MATH

�

ENTER

MATH

�

ENTER

remember1. =

= F(b) − F(a), where F(x) is an antiderivative of f(x).

2. The definite integral can be found only if the integrand, f(x), exists

for all values of x in the interval [a, b]; that is, a ≤ x ≤ b.

f x( ) dxa

b

∫ F x( )[ ]ab

f x( ) dxa

b

∫

remember

6EWORKEDExample

14

f x( ) dxa

b

∫

∫2

1

19 x2–-------------- dx ∫

1

0

1–

4 x2–------------------ dx ∫

5

3

dx

16 x2–---------------------

∫2

1–

dx1 x2+-------------- ∫

4

1

2x--- dx

1

2

∫ dxx x 1+( )--------------------

∫1

–1

4x 10+x2 5x 6+ +--------------------------- dx ∫

2

0

1x 1–( )2

------------------- dx ∫3

1

3x 2+x2 8x– 12+----------------------------- dx

∫3

2-------

0

dx4x2 9+------------------ ∫

0

–1

dx

1 9x2–--------------------- 2x 1–( )

32--- dx

12---

2

∫

∫2

0x

1x 2–-----------+

dx ex e x–+( )2 dx0

3

∫



2 Evaluate the integrals in question 1 provided that the integrand, f(x), exists for allvalues within the domain of the integral.

3

The definite integral dx can be evaluated after substituting u = x3 + 1.

a The integral will then be equal to:

b The value of the integral is:

4

a can be evaluated by first making the substitution:

b The integral will then be equal to:

c When evaluated, the integral is equal to:

5 By choosing an appropriate substitution for u, express the following integrals in termsof u. (Do not forget to change the terminals.)

6 Evaluate each of the integrals in question 5.

A B C

D E

A 11 B C 9 D 12 E

A u = sin x B u = cos x CD u = cot x E u = 1 + sin x

A B C D E

A 2 B C −2 D E

a b c

d e f

g h i

j k l

m n

Mathca

d

Integratormmultiple choiceultiple choice

2x2 x3 1+0

2

∫

∫2

0

2 u3

---------- du ∫9

0

3 u 1+2

------------------- du ∫9

1

2 u3

---------- du

∫2

02x2 u du ∫

7

0

4u32---

9-------- du

59--- 8 2

9----------

49--- 10 10 1–


∫π2---

0

cos x

1 sin x+------------------------- dx

u 1 sin x+=

u12---–du

0

1

∫ u12---

du1

0

∫ u12---– du

0

2

∫ u12---– du

1

2

∫ 1 u+ du0

1

∫

2 2 2– 2 223---

WORKEDExample

15

x2 2 x3+( ) dx0

2

∫ ∫ 2

1

4xx2 3– 2---------------- dx x x2 1+ dx

0

1

∫x 1–( ) x2 2x– dx

2

4

∫ x x 1– dx1

2

∫ ∫3

0

x2

x 1+---------------- dx

∫3

1

logex

x------------- dx sin x ecos x dxπ

3---

π2---

∫ x 1 x–( )10 dx0

1

∫cos x sin x dx

0

π2---

∫ tan3x sec2x dx0

π4---

∫ x sin x2 dx0

π2---

∫cos3x dxπ

2---

π

∫ ex

ex 1+------------------ dx

0

1

∫


C h a p t e r 6 I n t e g r a l c a l c u l u s 2477 Evaluate the following definite integrals.

8 By substituting x = sin θ, evaluate .

9 By substituting x = 2sin θ, evaluate .

10 By making the substitution x = tan θ, evaluate .

11 If , find the value of a.

12 If , find a.

13 If , find a.

14 If , find a.

a b

c d

e f

g h

i j

k l

m n

o p

q r

WORKEDExample

16 4xex2 dx

2–

0

∫ 4x 7+( ) 2x2 7x+ dx0

1

∫2 x 3+ dx

3–

2–

∫ ∫0

–1

x 1+

1 x–---------------- dx

sin x cos4x dx0

π3---

∫ ∫2

0

x 5+x2 4x 3+ +--------------------------- dx

∫1

–1

1

4 x2–------------------ dx 1

x 3–( )2 1+----------------------------

0

3

∫ dx

∫1

–1

1–4 x 1–( )2–---------------------------- dx sin3x cos2x dx

0

π

∫cot x dxπ

4---

π2---

∫ ∫6

5

3x 10–x2 7x– 12+----------------------------- dx

2x2

x2 1+-------------- dx

–1

1

∫ 2 sin 2x cos x dx0

π2---

∫2x 1+( )ex2 x+ dx

0

1

∫ 2 tan2x+( ) dxπ4---

π3---

∫x2

x 1–---------------- dx

2

5

∫ 2x3 x2 2x– 4–+x2 4–

----------------------------------------- dx3

4

∫WORKEDExample

17

1 x2– dx0

1

∫4 x2– dx

0

3

∫dx

1 x2+( )2---------------------

0

1

∫4

1 x2+-------------- dx

0

a

∫ π=

44 x2–-------------- dx

0

a

∫ loge– 3=

3 x 1+ dx1–

a

∫ 6 3=

1

4 x2–------------------ dx

a–

a

∫ π2---=

WorkS

HEET 6.1



You can check your answers by using the Mathcad file ‘Integrator’ found on the MathsQuest CD-ROM.

Applications of integrationIn this section, we shall examine how integration may be used to determine the areaunder a curve and the area between curves.

Areas under curvesYou will already be aware that the area between a curvewhich is above the x-axis and the x-axis itself is as shown inthe diagram at right.

Area =

Further, the area between a curve which is below thex-axis, and the x-axis itself, is as shown in the seconddiagram.

Area = –

=

The modulus is required here since, for a curve segment that lies below the x-axis,the integral associated with that curve segment is a negative number. Area is a positivenumber and in this case the integral is negative.

Mathca

d

Integrator

y

x0 ba

y = f (x)

f x( ) dxa

b

∫

y

x0ba

y = g(x)g x( ) dx

a

b

∫g x( ) dx

a

b

∫


C h a p t e r 6 I n t e g r a l c a l c u l u s 249Similarly, the area between a curve and the y-axis can be

found if the rule for the curve is expressed as a function ofy, that is, x = f (y).

Area = (integral measures to the right of the

y-axis are positive)

or

Area = (integral measures to the left of the

y-axis are negative)

Area =

If the graph crosses the x-axis, then the areas of theregions above and below the x-axis have to be calculatedseparately. In this case the x-intercepts must be determined.In the figure at right a single intercept, c, is shown.

Area =

=

Similarly the shaded region in the figure at right has anarea given by:

Area =

=

y

x0

a

b x = f (y)

f y( ) dya

b

∫

y

x0

a

bx = g (y)g y( ) dy

a

b

∫–

g y( ) dya

b

∫y

x0 bca

y = f (x)

f x( ) dxc

b

∫ f x( ) dxa

c

∫+

f x( ) dxc

b

∫ fa

c

∫ x( ) dx–

y

x0a

bc

x = f (y)

f y( ) dyc

b

∫ f y( ) dya

c

∫+

f y( ) dyc

b

∫ fa

c

∫ y( ) dy–

If y = , find:

a the x-intercepts b the area bounded by the curve, the x-axis and the line x = 3.

THINK WRITE

a For x-intercepts, y = 0, when 2logex = 0. a x-intercepts occur when 2loge x = 0.

Solve for x. That is, x = 1.

b Sketch a graph showing the region required. (A graphics calculator may be used.)

b

2 logex

x-----------------

1

2

1 y

x0 31

y =2 loge x————x

18WORKEDExample

Continued over page



A graphics calculator should be used here to verify the result.

To find the area under a curve between two x-values, first graph the curve by enteringits equation as Y1 in the Y= menu.

Consider y = in worked example 18. Press Y= and type in (2ln(X))÷X at Y1.

Then press .

To find the area bounded by this curve and the

x-axis between x = 1 and x = 3, press [CALC]and select 7: ∫f(x) dx. Type in 1 for the lower value

(press ) and 3 for the upper value (press

). Compare this result to that obtained inworked example 18.

THINK WRITE

Express the area as a definite integral. Area =

Antidifferentiate by letting u = loge x to apply the derivative method.

Let u = loge x.

Find .

Make dx the subject. or dx = x du

Express the terminals in terms of u. When x = 1, u = loge1= 0

When x = 3, u = loge3

Area =



Evaluate the integral. =

= (loge3)2

State the area. The area is (loge3)2 or approximately 1.207 square units.

22 logex

x----------------- dx

1

3

∫3

4dudx------ du

dx------ 1

x---=

5

6

2ux

------x du0

loge3

∫7 2u du

0

loge3

∫8 u2[ ]0

loge3

9 loge3( )2[ ] 02[ ]–

10

Graphics CalculatorGraphics Calculator tip!tip! Finding the numeric integral at the GRAPH screen

2loge x

x------------------

GRAPH

2nd

ENTER

ENTER



The shaded area shown in the figure in worked example 19could also have been calculated relative to the x-axis by sub-tracting the area between the curve and the x-axis from thearea of the rectangle as shown in the figure at right. That is:

Area =

Using symmetry propertiesIn some problems involving area calculations, use of symmetry properties can simplifythe procedure.

Examine the figure at right.a Express the rule as a function of y.b Find the area of the shaded section.THINK WRITE

a Write down the rule. a y = Square both sides of the equation. y2 = x − 1Add 1 to both sides to make x the subject. or x = y2 + 1

b Express the area between the curve and the y-axis in integral notation.

b Area =



= 4

State the area. The area is 4 square units.

y

x0 1

2y = x – 1

1 x 1–

2

3

1 y2 1+( ) dy0

2

∫2 [1

3---y3 y+ ]0

2

3 [83--- 2+ ] [0 0]+–23---

423---

19WORKEDExample

y

x0 1 5

2y = x – 1

5 2×( ) x 1– dx1

5

∫–

Find the area inside the ellipse in the figure at right.THINK WRITE

Write the equation. (The ellipse is symmetrical about the x-axis and y-axis and so finding the shaded area in the figure allows for the total enclosed area to be determined.)

= 1

Express the relation as a function of x for the top half of the ellipse.

= 1 − x2

y2 = 4(1 − x2)

y = is the rule for the top halfof the ellipse.

(y = is the bottom half.)

y

x

y2—4

0 1–1

2

–2

x2 + = 1

1 x2 y2

4-----+

2y2

4-----

2 1 x2–

2 1 x2––

20WORKEDExample

Continued over page



Areas between curvesWhen finding the areas between two curves that intersect, itis necessary to determine where the point of intersectionoccurs. In the figure at right, two functions, f and g, inter-sect at the point P with x-ordinate c.

The area contained within the envelope of the two func-tions bounded by x = a and x = b is given by:

.

THINK WRITE

Write the integral that gives the area in the first quadrant (a quarter of the total area).

Area in the first quadrant =

Express the total area as four times this integral.

Total area of ellipse =

=

To antidifferentiate, let x = sin θ. Let x = sin θ.

Find . = cos θ

Make dx the subject. or dx = cos θ dθExpress the terminals in terms of θ. When x = 0, sin θ = 0

θ = 0When x = 1, sin θ = 1

θ =

Rewrite the integral in terms of θ. Area = cos θ dθ

Simplify the integrand using identities. =

=

=



=

= 2πState the area. The exact area is 2π square units.

3 2 1 x2– dx0

1

∫4 4 2 1 x2– dx

0

1

∫8 1 x2– dx

0

1

∫5

6dxdθ------ dx

dθ------

7

8

π2---

9 8 1 sin2θ–0

π2---

∫10 8 cos2θ dθ

0

π2---

∫8 1

2---

0

π2---

∫ 1 cos 2θ+( ) dθ

4 1 cos 2θ+( ) dθ0

π2---

∫11 4[θ 1

2--- sin 2θ]+

0

π2---

12 4 π2---

12--- sin π+ 0 1

2--- sin 0+[ ]–

4 π2--- 0+

13

yP

x0 bca

f (x)

g (x)



Area = +

Similarly, areas between curves can also be found rela-tive to the y-axis.

Area =

Note that on the interval [a, b], g(y) ≥ f (y) and hence theintegrand is g(y) − f (y) and not f (y) − g(y).

When an area between a curve and the x-axis (or between curves) gives an integrand which cannot be antidifferentiated, it may be possible to express the area relative to the y-axis, creating an integrand which can be antidifferentiated.

g x( ) f x( )–[ ] dxa

c

∫ f x( ) g x( )–[ ] dxc

b

∫y

x0

a

b x = f (y)x = g (y)

g y( ) f y( )–[ ] dya

b

∫

Find the area bounded by the curves y = x2 − 2 and y = 2x + 1.

THINK WRITE

Check on a graphics calculator to see if the curves intersect. If they do, solvex2 − 2 = 2x + 1 to find the x-ordinate of the point or points of intersection for the two curves.

x2 − 2 = 2x + 1x2 − 2x − 3 = 0

(x − 3)(x + 1) = 0x = 3 and x = −1The curves intersect at x = 3 and x = −1.

Express the area as an integral.(Use , as without a graph we cannot always be sure which function is above the other. Here is a valuable use for the graphics calculator.)

Area =




=

=

= 10

State the solution. The area bounded by the two curves is

10 square units.

1

2 x2 2– 2x 1+( )–[ ] dx1–

3

∫

3 x2 2x– 3–( ) dx1–

3

∫4 [1

3---x3 x2– 3x]–

1–

3

5 [ 9 9– 9–( ) 13---– 1– 3+( )]–

9– 123---–

1023---–

23---

623---

21WORKEDExample



Consider using the TI calculator for worked example 21.

1. To graph the two curves with equations y = x2 – 2 and y = 2x + 1 enter Y1= X2 – 2 and Y2= 2X + 1. Then press . Use TRACE to locate the points of intersection. Adjust the WINDOW settings if necessary.

2. To show the area bounded by the two curves, press , position the cursor to the left of the Y1 symbol

and press successively to obtain the ‘shade below’ style. Repeat for Y2 to obtain the ‘shade above’ style. Press . The required area is shown unshaded.

3. To determine the value of the area bounded by the curves on the required interval (in this case, between x = –1 and x = 3), press , select 9 and complete 9: fnInt(Y2–Y1,X,–1,3) and press . Remember, to insert Y1, press and select Y–VARS, 1:Function and 1:Y1 (or 2:Y2 to enter Y2). Note that in this case we are subtracting Y1 from Y2 (seen by viewing the graph). However, if it is entered the opposite way, it only produces the negative of the required answer.

Graphics CalculatorGraphics Calculator tip!tip! Showing and finding the area bounded by two curves

GRAPH

Y=ENTER

GRAPH

MATHENTER

VARS

remember1. The area between a curve f (x), the x-axis and lines x = a and x = b is given by:

Area = where F(x) is the antiderivative of f (x).

2. Area measures can also be evaluated by integration along the y-axis. The area between a curve f (y), the y-axis and lines y = a and y = b is given by:

Area = where F(y) is the antiderivative of f (y).

3. If an area measure is to be evaluated over the interval [a, b] and the curve crosses the x-axis at x = c between a and b, then the integral has to be decomposed into two portions.

Area =

4. The area bounded by two curves f (x) and g(x) where f (x) ≥ g(x) and the lines x = a and x = b is given by:

Area =

5. Where possible use a graphics calculator to draw the function or functions to determine whether the integrals have to be decomposed into portions and to check and verify the correct use of the modulus function.

f x( ) dxa

b

∫ F b( ) F a( )–=

f y( ) dya

b

∫ F b( ) F a( )–=

f x( ) dxa

c

∫ f x( ) dxc

b

∫+ F c( ) F a( )– F b( ) F c( )–+=

f x( ) g x( )–[ ] dxa

b

∫ F b( ) G b( )– F a( )– G a( )+=

remember



Applications of integration

For the following problems, give exact answers wherever possible; otherwise give answersto an appropriate number of decimal places. (Use a graphics calculator to assist with, orverify, any graphing required.)

1 For each of the following curves find:i the x-interceptsii the area between the curve, the x-axis and the given lines.

2 For each of the graphs below:i express the relationship as a function of y (that is, make x the subject of the rule)ii find the magnitude of the shaded area between the curve and the y-axis.

a y = , x = 0 and x = 9 b y = x − , x = 1 and x = 2

c y = , x = 2 and x = 5 d y = , x = 3 and x = 4

e y = , x = 1 and x = f y = cos2x, x = 0 and

g y = 2x cos x2, and x = 0 h y = , x = 0 and x = 1

a b c

d e f

g

6F

WORKEDExample

18

Mathcad

Definiteintegral– graph

x1x2-----

x x 1–3x 2–x2 4–---------------

1

4 x2–------------------ 3 x

π2---=

xπ3---–= ex

2 ex+--------------

WORKEDExample

19

y

x0

2y = x

y

x0

4

1

–1

y = (x – 1)2 y

x0 1–1

y = Sin–1 x–2π

–3π

–2π–

y

x0

2

1

y = logex

y

x0 1–1

y = Cos–1 x

π

–2π

y

x0

y = x3

8

y

x0

y = Tan–1 x–2π

–4π



3 Find the magnitude of the shaded areas on each graph below.

4a The definite integral that correctly gives the area bounded by the curve y = 4x − x2

and the x-axis is:

b The area, in square units, is equal to:

5a Which of the graphs below correctly shows the area bounded by the curve

y2 = x + 1 and the y-axis?

b The definite integral which gives the area bounded by y2 = x + 1 and the y-axis is:

a b c

d e f

A B C

D E

A 10 B 2 C 5 D 8 E −5

A B C

D E

A B C 2

D E

WORKEDExample

20 y

x0 2

y = x2 y

x0

y2 = x

2

y

x0 1–1

y = –––––4 + x2

1–41

y

x0 1 e2

y = loge x

y

x0 3–3

1

–1

+ y2 = 1––9x2

y

x0

y = sin3x

–2π π

1


4x x2–( ) dx0

2

∫ 4x x2–( ) dx0

1

∫ 4x x2–( ) dx4

0

∫4x x2–( ) dx

0

4

∫ 2x2 13---x3–( ) dx

2

0

∫23--- 1

3--- 1

3--- 1

3---


y

x0

2y2 = x + 1

y

x0

1

–1

y2 = x + 1y

x0

–2

y2 = x + 1

y

x0

y2 = x + 1

y

x0

y2 = x + 11

–1–1

y2 1–( ) dy0

1

∫ y2 1–( ) dy1–

0

∫ y2 – 1( ) dy1

0

∫y2 1+( ) dy

0

1

∫ x 1– xd0

1

∫


C h a p t e r 6 I n t e g r a l c a l c u l u s 257c The value of the area, in square units, is equal to

6 Find the area bounded by the graph with equation y = (x − 2)2(x + 1) and the x-axis.

7 Find the area bounded by the graph with equation y2 = x + 4 and the y-axis.

8 a Show that the graphs of f (x) = x2 − 4 and g(x) = 4 − x2 intersect at x = −2 and x = 2.b Find the area bounded by the graphs of f (x) and g(x).

9 a On the same axis sketch the graphs of f (x) = sin x and g(x) = cos x over [0, π].

b Show algebraically that the graphs intersect at .

c Find the area bounded by the curves and the y-axis.

10 a On the same axis sketch the graphs of y = and y = x + 3.b Find the value of x where the graphs intersect.c Hence find the area between the curves from x = −1 to x = 2.

11 Find the area bounded by the curves y = x2 and y = 3x + 4.

12 Find the area enclosed by the curves y = x2 and .

13 Find the area bounded by y = ex and y = e−x and the line y = e.

14 Examine the figure at right.a Find the area enclosed by f (x), g(x) and the y-axis.b Find the shaded area.

15 Find the area of the ellipse with equation .

Hints:1. Use symmetry properties.

2. Antidifferentiate by using the substitution x = asin θ.

16 Find the area between the circle x2 + y2 = 9 and ellipse .Hint: Make use of symmetry properties.

17 a Sketch the curve y = ex + 2.b Find the equation of the tangent at x = −2.c Find the area between the curve, the tangent and the y-axis.

18 a Sketch the graph of .

b Find the area bounded by this curve and the x- and y-axes.

19 a Show algebraically that the line y = x does not meet the curve .

b Find the area enclosed by the curve, the lines y = x and , and the y-axis.

A B 2 C 1 D 5 E 223--- 2

3--- 1

3--- 1

3---

WORKEDExample

21

Mathcad

Areabetweencurves

xπ4---=

9 x–

y x=

y

x0 1

2

1

2

g (x) = ex – 1

f (x) = –––––1 + x2

2

1–e

x2

a2----- y2

b2-----+ 1=

a2 x2–

x2

9----- y2

4-----+ 1=

y1 x–x 1+------------=

y1

1 x2–------------------=

x1

2-------=



Volumes of solids of revolutionIf part of a curve is rotated about the x-axis, or y-axis, a figure called a solid of revol-ution is formed. For example, a solid of revolution is obtained if the shaded region infigure 1 is rotated about the x-axis.

The solid generated (figure 2) is symmetrical about the x-axis and any vertical cross-section is circular, with a radius equal to the value of y at that point. For example, theradius at x = a is f (a).

Any thin vertical slice may be considered to be cylindrical, with radius y and heightδx (figure 3).

The volume of the solid of revolution generated between x = a and x = b is found byallowing the height of each cylinder, δx, to be as small as possible and adding the vol-umes of all of the cylinders formed between x = a and x = b. That is, the volume of atypical strip is equal to πy2 δx.

Therefore the volume of the solid contained from x = a to x = b is the sum of all theinfinitesimal volumes:

V =

=

The value of y must be expressed in terms of x so that the integral can be evaluated.From the figure above y = f (x) and thus the volume of revolution of a curve f (x) from

x = a to x = b is .

Similarly if a curve is rotated about the y-axis, the solidof revolution shown in the figure at right is produced.

The volume of the solid of revolution is likewise

For regions between two curves that are rotated about thex-axis:

y

x0

y = f(x)

ba

y

x0

y = f(x)

ba

y

x0

y = f(x)

δx

ba

y

Figure 1 Figure 2 Figure 3

πy2 δxx a=

x = b

∑δx 0→lim

πa

b

∫ y2 dx

V π f x( )[ ]2 dxa

b

∫=

y

x0

x = f (y)

a

b

V π f y( )[ ]2 dya

b

∫=

y

x0 a b

y = f (x)

y = g (x)V π f x( )[ ]2 g x( )[ ]– 2 dx

a

b

∫=



Consider the volume of the solid of revolution formed inWorked example 22.1. The line with equation y = 2x is rotated about the x-axis

to form a cone. To graph the line, enter Y1 = 2X and press

. Press to locate particular coordinates.

2. To determine the volume of the cone, press , then select 9: fnInt( and insert Y1

2, X, 0, 2) and press

. To insert Y1, press and select Y–VARS, 1:Function and 1:Y1 (similarly any Y variable).

(Note that Y12 provides the integrand, X is the variable,

0 and 2 are the terminals of the integral.)

Can you verify the formula V = for this cone? What is the radius for this cone?

You can try to convert your volume answer to a fraction of . Press and [ ],

then . Select 1: Frac and press . This is also shown in the screen above.

a Sketch the graph of y = 2x and show the region bounded by the graph, the x-axis and the line x = 2.

b Find the volume of the solid of revolution when the region is rotated about thex-axis.

THINK WRITE

a Sketch the graph. aShade the region required.

b State the integral that gives the volume. (The volume generated is bounded by x = 0 and x = 2.)

b V =




=

State the volume. The exact volume generated is cubic units.

1 y

x0 2

x = 2 y = 2x2

1 π 2x( )2 dx0

2

∫

2 π 4x2 dx0

2

∫3 π[4

3---x3]0

2

4 π[323------ 0– ]

32π3

---------

532π

3---------

22WORKEDExample

Graphics CalculatorGraphics Calculator tip!tip! Finding the volume of a solid of revolution

GRAPH TRACE

MATHπ

ENTER VARS

π

13---πr

3

π ÷ 2nd πMATH

�

ENTER



a Sketch the region bounded by the curve y = logex, the x-axis, the y-axis andthe line y = 2.

b Calculate the volume of the solid generated if the region is rotated about they-axis.

THINK WRITEa Sketch the graph. (Use a graphics

calculator if necessary.)a

Shade the region required.

b Write the rule y = logex. b y = logexTake the exponent of both sides to get y as a function of x.

ey = elog x

State the function. ey = xor x = ey

Express the volume in integral notation between y = 0 and y = 2. So V =




=

State the volume. The volume is exactly cubic units

(or approximately 84.19 cubic units).

1 y

x0

2

1

y = logex2

1

2

3

4π ey( )2 dy

0

2

∫5 π e2y dy

0

2

∫6 π[1

2---e2y]0

2

7 π[12---e4 1

2---e0– ]

π2--- e4 1–( )

8π2--- e4 1–( )

23WORKEDExample

remember1. To find the volume of revolution about the x-axis for the function f (x) from

x = a to x = b, evaluate the integral:

V =

2. To find the volume of revolution about the y-axis for the function f (y) fromy = a to y = b, evaluate the integral:

V =

3. To find the volume of revolution about the x-axis for the region between f (x) and g(x) where f (x) ≥ g(x) from x = a to x = b, evaluate the integral:

V =

π f x( )[ ]2 dxa

b

∫

π f y( )[ ]2 dya

b

∫

π f x( )[ ]2 g x( )[ ]2 dx–a

b

∫

remember



Volumes of solidsof revolution

Give exact answers where possible; otherwise use an appropriate number of decimalplaces when giving approximate answers. (Use a graphics calculator to check anygraphing.)

1 a Sketch the graph of the region bounded by the x-axis, the curve y = 3x and the linex = 2.

b Calculate the volume generated by rotating this region about the x-axis.c Verify this result by using the standard volume formula for the solid generated.

2 The region bounded by the graph of and the x-axis is rotated about thex-axis.a Calculate the volume of the solid of revolution generated.b Verify this answer using the standard volume formula.

3 a Sketch the region bounded by the curve , the y-axis and the lines y = 0and y = 2.

b Calculate the volume generated when this region is rotated about the y-axis.

4 Find the volume generated when the area bounded by y = x2 − 1 and the x-axis isrotated about:a the x-axisb the y-axis.

5 For the regions bounded by the x-axis, the following curves, and the given lines:i sketch a graph shading the regionii find the volume generated when the region is rotated about the x-axis.

6 For each region defined in question 5 (a to f only) find the volume generated byrotating it about the y-axis.

7

a The region bounded by the curves y = x2 + 2 and y = 4 − x2 is represented by the graph:

a y = x + 1; x = 0 and x = 2 b y = ; x = 1 and x = 4c y = x2; x = 0 and x = 2 d y2 = 2x + 1; x = 0 and x = 3

e x2 + y2 = 4; x = −1 and x = 1 f y = ; x = 1 and x = 3

g y = cos x; and h y = ex + 1; x = −2 and x = −1

A B

6G

WORKEDExample

22

Mathcad

Solid ofrevolution x

y 16 x2–=

WORKEDExample

23

Mathcad

Solid ofrevolution y

y x 1–=

x

2x---

xπ2---–= x

π2---=


y

x0 2–2

y = x2 + 2

y = 4 – x2

(–1, 3) (1, 3)

y

x0

y = x2 + 2

y = 4 – x2



b The volume generated when the region is rotated about the x-axis is equal to:

c The volume generated when the region is rotated about the y-axis is equal to:

8 Find the volume generated when the region bounded by the curves y = x2 and y = −xis rotated about:

9 Find the volume generated when the area bounded by the curve y = sec x, the line and the x- and y-axes is rotated about the x-axis.

10 Find the volume generated by rotating the area bounded by y = e2x, the y-axis and theline y = 2 about the x-axis.

11 The area bounded by the curve y = Tan−1x, the x-axis and the line x = 1 is rotatedabout the y-axis. Find the volume of the solid generated.

12 A model for a container is formed by rotating the area under the curve of between x = −1 and x = 1 about the x-axis. Find the volume of the container.

C D

E

A B

C D

E

A B

C D

E

a the x-axis b the y-axis.

(–2, 2) (2, 2)

y

x0

y = x2 + 2

y = 4 – x2

y

x0 4

y = x2 + 2

y = 4 – x2

y

x

y = x2 + 2y = 4 – x2

(1, –3)

(1, 3)

0

π 2 2x2–( )2 dx0

2

∫ π 2 2y–( )2 dy1–

1

∫π 2 2x2–( )2 dx

0

1

∫ π 2 2x2–( )2 dx1–

1

∫π 6 2x2–( )2 dx

1–

1

∫

π 4 y–( ) dy3

4

∫ π y 2–( ) dy2

3

∫+ π y 2–( ) dy3

4

∫ π 4 y–( ) dy2

3

∫+

π 2 2y–( ) dy2

4

∫ π 2y 2–( ) dy2

4

∫π 2 2x2–( ) dx

2

4

∫

xπ4---=

y 2 x2

6-----–=


C h a p t e r 6 I n t e g r a l c a l c u l u s 26313 For the graph shown at right:

a find the coordinate of Ab find the volume generated when the shaded region is rotated

about the x-axisc find the volume generated when the shaded region is rotated

about the y-axis.

14 What is the volume generated by rotating the ellipse with equation about:

a the x-axis?b the y-axis?

15 Find the volume generated when the region bounded by y = x2 and isrotated about:a the x-axisb the y-axis.

16 Find the volume generated by the rotation of the area bounded by the curves y = x3

and y = x2 about:a the x-axisb the y-axis.

17 A hemispherical bowl of radius 10 cm containswater to a depth of 5 cm. What is thevolume of water in the bowl?

18 A solid sphere of radius 6 cm hasa cylindrical hole of radius 1 cmbored through its centre. Whatis the volume of theremainder of the sphere?

19 Find the volume of a trun-cated cone of height10 cm, a base radius of5 cm and a top radius of2 cm.

20 a Find the equation ofthe circle sketchedbelow.

b Find the volume of atorus (doughnut-shapedfigure) generated byrotating this circle aboutthe x-axis (give your answerin cm3).

y

x0 21

A y = 1–––––4 – x2

x2

4----- y2

9-----+ 1=

y 8x=

y

x0 4

6



Approximate evaluation of definite integrals and areas

When calculating definite integrals or areas that involve integrands which cannot beantidifferentiated using techniques discussed in this chapter, approximation methodscan be used. We shall now look at two useful and simple approximation methods: themidpoint rule and the trapezoidal rule.

The midpoint rule

The definite integral determines the shaded area

under the curve below.It can be approximated by constructing a rectangle with height

equal to the value of y halfway between x = a and x = b.

Area of rectangle =

The estimate for the shaded area is improved by increasingthe number of intervals, that is the number of rectanglesbetween x = a and x = b. In the figure below, the region from x = a and x = b is broken up into n rectangles.

The base width of each rectangle is δx and the height ofeach individual rectangle is obtained from the midpoint rule.The area of each rectangle is given by the product of theheight and the common width δx.

So

where: , the width of each rectangle

n = the number of intervals and hence rectangles usedx0 = axn = b

y

x0 ba

y = f (x)f x( ) dxa

b

∫

y

x0 ba

y = f (x)

–––2

a+b

b a–( ) fa b+

2------------

y

x0 …x0 x1 x2 xn

y = f (n)

(a) (b)

f x( ) dx δx fx0 x1+

2----------------

fx1 x2+

2----------------

. . . fxn 1– xn+

2-----------------------

+ + +a

b

∫ ≈

δxb a–

n------------=

Estimate using the midpoint rule and 4 intervals.

THINK WRITE

State f (x). f (x) = x2 + 2xCalculate δx. δx =

= 1

x2 2x+( ) dx0

4

∫1

2 4 0–4

------------

24WORKEDExample



The trapezoidal ruleThe area under a curve can also be approximated using atrapezium.

The area of the trapezium = .

By increasing the number of intervals between x = a andx = b, that is, the number of trapezia, the estimate becomesmore accurate:

So

≈

Notice here that the terms f (a) and f (b) occur only once and all other terms such asf (x1) and f (x2) occur twice. Thus an approximation to the area is:

Approximate area =

where: n = the number of intervals used

a = x0

b = xn

THINK WRITE

Find x0, x1, x2, x3, x4. x0 = 0x1 = 1x2 = 2x3 = 3x4 = 4

Substitute these values into the midpoint rule.

So

≈ 1[ f (0.5) + f (1.5) + f (2.5) + f (3.5)]

Evaluate the approximation. = 1(1.25 + 5.25 + 11.25 + 19.25)= 37

State the solution. The approximate value of the definite integral is 37.

3

4 x2 2x+( ) dx0

4

∫5

6

y

x0 ba

y = f (x)

b a–2

------------ f a( ) f b( )+[ ]×

y

x0 x0 x1 x2 x3 xn

y = f (x)

(a) (b)

…f x( ) dxa

b

∫b a–

2------------ f a( ) f x1( )+[ ] f x1( ) f x2( )+[ ] f x2( ) f x3( )+[ ] . . . f xn 1–( ) f b( )+[ ]+ + + +{ }×

δx2------ f x0( ) 2 f x1( ) 2 f x2( ) . . . 2 f xn 1–( ) f xn( )+ + + + +[ ]

δxb a–

n------------=



Compare this answer with that in worked example 24. Which is closest to the exactanswer?

Estimate using the trapezoidal rule and four equal intervals.

THINK WRITE

State f (x). f (x) = x2 + 2x

Calculate δx. δx = = 1

Find x0, x1, x2, x3, x4. x0 = 0x1 = 1x2 = 2x3 = 3x4 = 4

Substitute these values into the trapezoidal rule.

So

≈ [f (0) + 2f (1) + 2f (2) + 2f (3) + f (4)]

Evaluate the approximation. = [0 + 2(3) + 2(8) + 2(15) + 24]

= (76)

= 38State the solution. The value of the definite integral is approximately 38.

x2 2x+( ) dx0

4

∫

1

24 0–

4------------

3

4 x2 2x+( ) dx0

4

∫12---

5 12---

12---

6

25WORKEDExample

Estimate the area under the graph of y = x logex from x = 1 to x = 5 using two equalintervals and:a the midpoint rule b the trapezoidal rule.THINK WRITEa State f (x). a f (x) = x logex

Calculate δx. δx = = 2

Find x0, x1, x2. x0 = 1x1 = 3x2 = 5

Substitute the values into the midpoint rule.

So the area =

≈ 2[ f(2) + f(4)]Evaluate the estimate of the area. = 2[2loge2 + 4loge4]

= 4loge2 + 8loge4

State the approximate area. The approximate area is 13.863 square units.

1

25 1–

2------------

3

4 x logex dx1

5

∫5

6

26WORKEDExample



THINK WRITE

b State f (x). b f (x) = xlogex

Calculate δx. δx =

= 2

Find x0, x1, x2. x0 = 1x1 = 3x2 = 5

Substitute these values into the trapezoidal rule.

So the area =

≈ [f (1) + 2f (3) + f (5)]

Evaluate the estimate for the area. = 1[loge1 + 2(3loge3) + 5 loge5]= 6 loge3 + 5 loge5

State the approximate area. The approximate area is 14.639 square units.

1

25 1–

2------------

3

4 xlogex dx1

5

∫22---

5

6

remember1. Some functions cannot be integrated using the

techniques covered in this chapter. Two approxi-mation methods are discussed. The midpoint rule involves subdividing the required area into a finite number of rectangles. The trapezium rule involves subdividing the area into a finite number of trapezia.

2. The midpoint rule:

where: δx = , the width of each rectangle

n = the number of intervals usedx0 = axn = b

3. The trapezoidal rule:


δx =

a = x0

b = xn

y

x0 x0 x1 x2 xn

y = f (n)

(a) (b)…

f x( ) dx ≈ δx fx0 x1+

2----------------

fx1 x2+

2----------------

. . . fxn 1– xn+

2-----------------------

+ + +a

b

∫b a–

n------------

y

x0 x0 x1 x2 x3 xn

y = f (x)

(a) (b)

…

f x( ) dx ≈ δx2------ [ f x0( ) 2 f x1( ) 2 f x2( ) . . . 2 f xn 1–( ) f xn( )]+ + + + +

a

b

∫b a–

n------------

remember




1 Find approximations to the following definite integrals using the midpoint rule withfour equal intervals.

2 Repeat question 1 using the trapezoidal rule.

3 Use the midpoint rule with two equal intervals to estimate the following definite inte-grals.

4 Repeat question 3 using the trapezoidal rule.

5

a Using the midpoint rule and two equal intervals, an estimate for is:

b Compared to the exact answer, the percentage error in answer a is closest to:

6

a Using the trapezoidal rule and four equal intervals, an estimate for

is:

b The percentage error relative to the exact answer is closest to:

7 Find the value of using the midpoint rule and:

a 4 equal intervalsb 8 equal intervals.

a b

c d

a b

c d

A 1.4 B 0.8 C 0.9412 D 0.7906 E 0.863

A 0.7 B 1.9 C 2.5 D 20 E 6.4

A B C D E

A 61 B 21 C 11 D 5 E 1

6HWORKEDExample

24dx

x 2–-----------

3

5

∫ sin x dx0

π

∫logex2 dx

3–

1–

∫ Tan 1– x dx0

4

∫

GCpro

gram

Midpointrule

WORKEDExample

25

Mathca

d

Middleboxes

WORKEDExample

26a

x 2+( ) dx0

2

∫ x2 3–( ) dx1

4

∫x3 x2– 2x+( ) dx

1

2

∫ 16 x2–( ) dx0

4

∫

GCpro

gram

Trapezoidalrule

Mathca

d

Trapezoidalrule

WORKEDExample

26b


11 x2+-------------- dx

0

1

∫


cos x dx–

π2---

π2---

∫

1 2+π2--- π 1 2+( )

4------------------------- π 1 2+( )

8-------------------------

π4---

ex dx1

5

∫



8 Find using the trapezoidal rule and:

a 4 equal intervalsb 8 equal intervals.

9 Estimate the area under the curve y = logex from x = 1 to x = 4 using the midpoint ruleand:a 3 equal intervalsb 6 equal intervals.

10 Estimate the area under the graph of y = Cos−1x from x = −1 to x = 1 using the mid-point rule and 4 equal intervals.

11 Calculate an estimate for the area under the graph of y = 2x between x = 0 and x = 2using the trapezoidal rule and:a 2 equal intervalsb 4 equal intervals.

12 Using the trapezoidal rule with:a 2 equal intervalsb 4 equal intervalsfind the approximate area under the graph of between x = 0 and x = π.

Is there a connection between area and volume of revolution?

The area under a curve can be found directly from integration. Likewise, the volume of revolution can also be found from integration. In both cases it is assumed that the curves have functions which can be readily integrated. This investigation examines the relative size of the area produced by a curveand the x-axis, and the volume of revolution produced. Curves will be restrictedto those of the type f (x) = axn. 1 Consider the function f (x) = axn, where a = , 1, 2 and 4 and n = 1. Find the

area enclosed by the curve, the x-axis and the line x = 1 for each of the four values for a. In what manner is the area dependent on the gradient of the line a?

2 For each of the four curves, find the volume of revolution. In what way is the volume dependent on a?

3 Consider the function f (x) = axn, where n = , 1, 2 and 4 and a = 1. Find the area enclosed by the curve, the x-axis and the line x = 1 for each of the four values for n. In what manner is the area dependent on the value of n?

4 For each of the four curves, find the volume of revolution. In what way is the volume dependent on the value of n?

5 Finally, compare the sizes of the areas found in parts 1 and 3 to the volumes found in parts 2 and 4. In particular, investigate the ratio of the volume of

revolution V to the area A. In what way does the ratio depend on the values

of a and n for the general function f (x) = axn? What happens to the ratio asn → ∞ when a = 1?

6 Write a brief report detailing your findings being careful to illustrate your work with graphs and with calculations.

1 x2+ dx0

4

∫

y sin x=

12---

12---

VA----



Common antiderivatives• The table below lists common antiderivatives.

Substitution where the derivative is present in the integrand

•

•

Linear substitution

The integral may be successfully antidifferentiated using

the substitution u = g(x), provided that g(x) is linear. The function f (x) must be written in terms of y also.

Useful trigonometric identities• Trigonometric identities are used to integrate even and odd powered trigonometric

functions:

sin2ax = (1 − cos 2ax)

cos2ax = (1 + cos 2ax)

sin ax cos ax = sin 2ax

f(x) F(x)

Axn

logekx + c

ekx

sin kx

cos kx

sec2kx

, x ∈ (–a, a)

, x ∈ (–a, a)

summary

axn 1+

n 1+--------------- c+

1x---

ekx

k------- c+

cos kx–k

------------------ c+

sin kxk

-------------- c+

tan kxk

-------------- c+

1

a2 x2–-------------------- Sin 1– x

a--- c+

1–

a2 x2–-------------------- Cos 1– x

a--- c+

aa2 x2+----------------- Tan 1– x

a--- c+

f ′ x( ) f x( )[ ]n dx∫ f x( )[ ]n 1+

n 1+( )------------------------- c+=

f ′ x( )f x( )------------ dx∫ loge f x( ) c+=

f x( ) g x( )[ ]n dx∫ n 0≠,

12---

12---

12---


C h a p t e r 6 I n t e g r a l c a l c u l u s 271Antidifferentiation using partial fractionsMany rational expressions can be antidifferentiated by transforming the expressions into partial fractions. Two common types are shown below.

Definite integrals

• =

= F(b) − F(a), where F(x) is an antiderivative of f(x).

• The definite integral can be found only if the integrand f(x) exists for

all values of x in the interval [a, b]; that is, a ≤ x ≤ b.

Areas under curves

• Area =

• Area =

• Area = +

• Area = +




f x( )ax b+( ) cx d+( )

---------------------------------------- Aax b+--------------- B

cx d+---------------+

f x( )ax b+( )2

---------------------- Aax b+( )2

---------------------- Bax b+---------------+

f x( ) dxa

b

∫ F x( )[ ]ab

f x( ) dxa

b

∫

y

x0 ba

y = f (x) f x( ) dxa

b

∫

y

x0ba

y = g (x)

g x( ) dxa

b

∫

y

x0 bca

y = f (x) f x( ) dxa

c

∫ f x( ) dxc

b

∫

y

x0a

bc

x = f (y)f y( ) dy

a

c

∫ f y( ) dyc

b

∫



Areas between curves

• Area = +

• Area =

Volumes of solids of revolution

• About x-axis: .

• About y-axis:

• Between two functions f (x) and g(x) where f (x) ≥ g(x):


• Approximate measure of dx using the midpoint rule:


δx =

x0 = a

xn = b

• Approximate measure of using the trapezoidal rule:


δx =

x0 = a

xn = b.

yP

x0 bca

f (x)

g (x)

g x( ) f x( )–[ ] dxa

c

∫ f x( ) g x( )–[ ] dxc

b

∫

y

x0

a

b x = f (y)x = g (y)g y( ) f y( )–[ ] dy

a

b

∫

V π f x( )[ ]2 dxa

b

∫=

V π f y( )[ ]2 dya

b

∫=

V π f x( )[ ]2g x( )[ ]– 2 dx

a

b

∫=

f x( )a

b

∫f x( ) dx ≈ δx f

x0 x1+

2----------------

fx1 x2+

2----------------

. . . fxn 1– xn+

2-----------------------

+ + +a

b

∫b a–

n------------

f x( ) dxa

b

∫f x( ) dx

a

b

∫ ≈ δx2------[ f x0( ) 2 f x1( ) 2 f x2( ) . . . 2 f xn 1–( ) f xn( )]+ + + + +

b a–n

------------



Multiple choice

1 The expression is equal to:

2 An antiderivative of is:


4 The antiderivative of x(x + 2)10 is:

5 is equal to:

6 Using an appropriate substitution, is equal to:

A B C D E

A B C D E

A B C D E

A x11 + c B (x + 2)11 + c C

D E x(x + 2)11 + c

A B C

D E –

A B C

D E

CHAPTERreview

6Ax 1–( ) x2 2x–( )5 dx∫u5 du∫ 1

2--- u5 du∫ 2 u5 du∫ 5u4 du∫ u4 du∫

6Asin xcos3x-------------

1cos4x------------- 1–

cos2x------------- 1–

sin2x------------ 1

2 cos2x----------------- 1

4 cos2x-----------------

6A6 sec23x tan43x dx∫2 u4 du∫ u4 du∫ 1

2--- u4 du∫ u2 du∫ 2 u2 du∫

6Bx 2+( )11 11x 2–( )

132--------------------------------------------- c+

x 2+( )11 12x 11–( )132

------------------------------------------------ c+

x 2 x– dx∫ 6B25--- 2 x–( )

52---

2 2 x–( )32---

– c+ 52--- 2 x–( )

52---

3 2 x–( )32---

– c+ x2 2 x–( )32---

c+

15--- 2 x–( )

52---

2 2 x–( )32---

c+ +2

15------ 2 x–( )

32---

3x 4+( ) c+

6Be2x ex 1– dx∫u

32---

u12---

+ du∫ 2u

32---

u12---

+ du∫ u

52---

2u32---

u12---

+ + du∫

u52---

2u32---

+ du∫ u

52---

u12---

+ du∫



7 Using an appropriate substitution, is equal to:

8 If f ′(x) = 4sin2x and , then f (x) is equal to:

9 Using the appropriate substitution, is equal to:


11 Given that , an antiderivative of is:


13 The integral can be evaluated over the largest domain of:

A B C

D E

A B 2x − 1 + sin 2x C 2x + cos 2x

D 2x – cos 2x E 2x − sin 2x

A B C

D E

A x + sec2x + c B 2x + sec2x + c C tan x + cD x + tan x + c E xtan x + c

A B C loge(x2 − 9x + 20)

D E

A B

C D

E loge(x + 1)2 + c

A (–9, 9) B [–3, 3] C (–3, 0)D R E (–3, 3)

6C cos3x sin2x dx∫u4 u2–( ) du∫ u2 cos x dx∫ u5 u3–( ) du∫u3 u5–( ) du∫ u2 u4–( ) dx∫

6C fπ4---

π2---=

4 cos2xπ2--- 2–+

6C sin5x dx∫u2 u4–( ) du∫ u4 2u2–( ) du∫ 2u2 1– u4–( ) du∫–u4 1–( ) du∫ u4 u2– 1+( ) du∫

2 tan2x+( ) dx∫6C

6D1

x2 9x– 20+----------------------------- 1

x 5–----------- 1

x 4–-----------– x 5>,=

1x2 9x– 20+-----------------------------

logex 5–x 4–-----------

logex 4–x 5–-----------

loge x 5–( ) 1x 4–( )2

-------------------– 1x 5–( )2

------------------- 1x 4–( )2

-------------------–

6D2x 3+x 1+( )2

------------------- dx∫2–

x 1+------------ c+

2–x 1+( )2

------------------- 3 loge x 1+( ) c+ +

loge x 1+( ) 1x 1+------------ c+ + 2 loge x 1+( ) 1

x 1+------------– c+

6E1

9 x2–------------------ dx

a

b

∫



14 The value of is:


16 The integral representing the shaded area of this curve is equal to:

A

B

C

D

E

17 The area between the curve y = sin x and the line y = x from x = 0 to x = 1 (see diagram) is approximately equal to:

A 0.04 square unitsB 1.04 square unitsC 0.54 square unitsD 0.84 square unitsE 0.34 square units

18 The shaded area (in square units) on the graph below is equal to:

A loge2 B loge2 C 3 loge2

D loge4 E unable to be calculated.

A cos π B sin (π2) C 0 D 1 E 2

A B 16 C D E 8

6Ex2

x3 1+-------------- dx

1–

1

∫13---

6E2x cos x2 dx0

π

∫

y

x0

y = x2 – 16E

2 x2 1–( ) dx0

1

∫×

x2 1–( ) dx1–

1

∫2 x2 1–( ) dx

1

0

∫×

1 x2–( ) dx0

1

∫x2 1–( ) dx

0

2

∫6F

y

x0

1

1

y = x

y = sin x

–2π

6Fy

x0 4

y = (x – 2)2

163------ 32

3------ 8

3---



Questions 19 and 20 refer to the shaded area in the figure below.

19 The volume generated when the region is rotated about the x-axis is equal to:

20 The volume generated when the region is rotated about the y-axis is equal to:

21 The approximate value of dx using the trapezoidal rule and 3 equal intervals is:

22 The approximate value of the area under the curve y = x2 + 1 from x = −1 to x = 1 (using the midpoint rule with four equal intervals) is:

Short answer

1 Find the antiderivative of:

2 Find the indefinite integral .

3 Find:

A B

C D

E

A B

C D

E

A (6e + 3e2 + 2e3 + e4) B (4e + 4e2 + 2e3 + e4)

C (6e + 6e2 + 4e3 + e4) D (2e + 2e2 + e3 + e4)

E (12e + 12e2 + 8e3 + 3e4)

A 2.625 square units B 1.3125 square units C 2.5 square unitsD 2.75 square units E 1.95 square units

a (cos x) esinx b

a b

y

x0y = 2 – x2

(1, 1)

y = x

6Gπ 4 2x2– x4 x–+( ) dx

0

1

∫ π 4 x4 x–+( ) dx0

1

∫π 4 3x2– x4+( ) dx

0

1

∫ π 4 4x2– x4 x+ +( ) dx0

1

∫π 2 x2– x–( ) dx

0

2

∫6G

π 2 y–( ) dy0

2

∫ π 2 y–( ) dy1

2

∫ π y2 dy0

1

∫+

π y2 dy0

2

∫ π y2 dy1

2

∫ π 2 y–( ) dy0

1

∫+

π 2 y– y2–( ) dy0

2

∫

6Hex

x-----

1

4

∫112------ 1

8---

112------ 1

4--- 1

2---

124------

6H

6A logex( )2

x--------------------

6Bx

x 1+---------------- dx∫

6Ccos22x dx∫ sin2 x

4--- cos2 x

4--- dx∫



4 Find an antiderivative of f (x) where f (x) = .

5 Find f (x) if f ′(x) = sin 2x cos x and f (π) = 1.

6 If f ′(x) = and f (0) = −3, find f (x).(Hint: Use the substitution x = sin θ to antidifferentiate.)

7 Evaluate:

8 a Sketch a graph which shows the region enclosed by the curve y = logex, the line y = 2 and the x- and y-axes.

b Find the area of this region.

9 What is the area bounded by the curve y = x2 + 2 and the line y = 5x − 4?

10 Find the volume generated when the area under the graph of y = ex, between x = −1 and x = 0, is rotated about the x-axis.

11 Find the volume of water in a hemispherical bowl of radius 8 cm if the depth is 3 cm.

12 a Find an approximate value of using four equal intervals and:

i the midpoint rule

ii the trapezoidal rule.

b Which result is closest to the exact answer?

Analysis1 a Find the area of the shaded region on the graph at right.

b What is the volume generated when this region is rotated about the x-axis?

c If the region is rotated about the y-axis, find the approximate volume of the solid generated using the midpoint rule and four equal intervals. (Give your answer correct to 4 decimal places.)

a b

6Dx2 2x– 12–x2 7x– 8–

-----------------------------

6E

6E2 1 x2–

6E1

4 x2+-------------- dx

0

2

∫ x

2 x–---------------- dx

2–

1

∫

6E

6F6G

6G

6H2x2 dx0

4

∫

y

x0

y = tan x

1

–2π–

4π



2 The side view of the right side of a wine glass vessel can be modelled by two curves which join at x = e:

y = 2logex, 0 < x ≤ e (red curve)y = x2 − 2ex + e2 + c, e ≤ x ≤ 5 (blue curve)(All measurements are in centimetres.)a Show that the value of c is 2 and find the height

of the vessel correct to 2 decimal places.The vessel is formed when the region between the curves and the y-axis is rotated about the y-axis.

b Find the volume of wine in the glass when the depth is 2 cm.

c What is the maximum volume of wine that the glass can hold (using maximum height to the nearest mm)?

3 A below-ground skating ramp is to be modelled by the curve

.

This is shown above, where the line y = 4.086 represents ground level. (All measurements are in metres.) (Give all answers correct to 2 decimal places.)a Find the

maximum depth of the ramp.

b Find the area under the curve.

c Find the volume generated if this area is rotated about the x-axis.

d If the ramp is 20 metres long, what is the volume of dirt which must be removed?

y

x0 1 e 5

y2

36 x2–--------------------- 5.98– x 5.98≤ ≤,=

y

x0 6–6

Ground Level4.086

testtest

CHAPTERyyourselfourselftestyyourselfourself

6


VCEVCEcocovverageerageArea of studyUnits 3 & 4 • Calculus

In thisIn this chachapterpter9A Revision of

antidifferentiation9B Integration of ex, sin x and

cos x9C Integration by recognition9D Approximating areas

enclosed by functions9E The fundamental theorem of

integral calculus9F Signed areas9G Further areas9H Areas between two curves9I Further applications of

integration

9

Integration

344


Revision of antidifferentiation

As we have seen, the process of differentiation enables us to find the gradient of a func-tion. The reverse process, antidifferentiation (or integration), will find the function for aparticular gradient.

Integration has wider applications including calculation of areas, volumes, energy,probability and many more quantities in science and business.

Note that

f

(

x

) means differentiate

f

(

x

) with respect to

x

; that is, .

So

f

(

x

) is the antiderivative of

f

′

(

x

), denoted as

f

(

x

)

=

where means

antidifferentiate

, or

integrate

, or

find an indefinite integral

and d

x

indi-

cates that the integration of the function is with respect to

x

.

Since , where

a

and

c

are constants

then

Since

then

Note

: The addition of the constant,

c

, is required when finding general antiderivatives.However, if we have to find an antiderivative, the

c

is to be allocated an actualnumber and for convenience the number chosen is zero. That is, an antiderivativemeans ‘let

c

=

0’, or ‘do not add on the

c

’.

For example, the antiderivative of is . An antideriva-

tive of is .

Properties of integrals

Since is a linear operator, so too is its inverse, . Therefore,

That is, each term can be integrated separately, and

That is, a ‘constant’ factor of the function can be taken to the front of the integral.

ddx------ d

dx------ f x( ) f ′ x( )=

f ′ x( ) dx∫∫

ddx------ ax c+( ) a=

a dx∫ ax c+=

ddx------ axn 1+

n 1+---------------

axn=

axn dx∫ axn 1+

n 1+---------------- c n 1–≠,+=

3x2 4x 5+ + x

3 2x2 5x c+ + +

3x2 4x 5+ + x

3 2x2 5x+ +

ddx------ ∫

f x( ) g x( )±[ ] dx∫ f x( ) dx g x( ) dx∫±∫=

k f x( ) dx∫ k f x( ) dx∫=

C h a p t e r 9 I n t e g r a t i o n 345

Antidifferentiate each of the following, giving answers with positive indices.

a 2x7 b 4x−3 c

THINK WRITE

a Integrate by rule; that is, add 1 to the index and divide by the new index.

a

Simplify.

b Integrate by rule. b

Simplify. = −2x−2 + c

Express the answer with a positive index. = −

c When a square root is involved, replace it with a fractional index.

c =

Bring the x to the numerator and change the sign of the index.

= 3

Integrate by rule. =

Simplify. =

Write the answer in the form it was given. =

3

x-------

1 2x7 dx∫ 2x8

8-------- c+=

2x8

4----- c+=

1 4x 3– dx∫ 4x 2–

2–---------- c+=

2

32

x2

-----

13

x-------dx∫ 3

x12---

----- dx∫

2 x12---– dx∫

33x

12---

12---

-------- c+

4 6x12---

c+

5 6 x c+

1WORKEDExample

Find the following indefinite integral.

THINK WRITE

Expand the expression.

Collect like terms.

Integrate each term separately.

Simplify each term.

x 1–( ) 3x 5+( ) dx∫

1 x 1–( ) 3x 5+( ) dx∫ 3x2 3x– 5x 5–+( ) dx∫=

2 3x2 2x 5–+( ) dx∫=

33x3

3-------- 2x2

2-------- 5x– c+ +=

4 x3 x2 5x– c+ +=

2WORKEDExample


Integration of (ax + b)n

By applying the chain rule for differentiation:

so

or

or

Integration of

Since

then , where x > 0

or .

ddx------ ax b+( )n 1+ a n 1+( ) ax b+( )n=

a n 1+( ) ax b+( )n dx∫ ax b+( )n 1+ c+=

a n 1+( ) ax b+( )n dx∫ ax b+( )n 1+ c+=

ax b+( )n dx∫ ax b+( )n 1+

a n 1+( )----------------------------- c+=

Antidifferentiate 4(5x − 2)3 by rule.

THINK WRITE

Express as an integral and take 4 out as a factor.

Apply the rule where a = 5 and n = 3.

Simplify the antiderivative by cancelling the fraction.

= (5x − 2)4 + c

1 4 5x 2–( )3 dx∫ 4 5x 2–( )3 dx∫=

24 5x 2–( )4

5 4( )------------------------- c+=

315---

3WORKEDExample

1x---

ddx------ loge x

1x---=

1x--- dx∫ loge x c+=

x 1– dx∫ loge x c+=

Antidifferentiate .

THINK WRITE

Take out as a factor.

Integrate by rule.

47x------

147---

47x------ dx∫ 4

7--- 1

x---×

dx∫=

47---

1x--- dx∫=

2 47---loge x c+=

4WORKEDExample


Integration of (ax + b)–1

By applying the chain rule for differentiation:

= ,

where a and b are constants

and × =

=

So =

or =

Note that the a in the fraction is the derivative of the linear function ax + b.

ddx------ loge ax b+( ) a

ax b+---------------

1a--- d

dx------ loge ax b+( ) a

ax b+--------------- 1

a---×

1ax b+---------------

1ax b+--------------- dx∫ 1

a--- loge ax b+( ) c+

ax b+( ) 1– dx∫ 1a--- loge ax b+( ) c+

1a---

1ax2 bx c+ +----------------------------- dx∫ 1

2ax b+------------------ loge ax2 bx c+ +( ) c1+=

Antidifferentiate .

THINK WRITE

Express as an integral and take 5 out as a factor.

Integrate by rule where a = 2.

52x 3+---------------

15

2x 3+--------------- dx∫ 5

12x 3+--------------- dx∫=

252--- loge 2x 3+( ) c+=

5WORKEDExample

Find .

THINK WRITE

Express as separate fractions. =

Simplify each fraction. =

Integrate each term separately by rule. =

Simplify leaving the answer with positive indices.

=

=

6x 5+x2

--------------- dx∫

16x 5+

x2--------------- dx∫ 6x

x2------ 5

x2-----+

dx∫2 6x 1– 5x 2–+( ) dx∫3 6 loge x

5x 1–

1–---------- c+ +

4 6 loge x 5x 1– c+–

6 loge x5x--- c+–

6WORKEDExample


Find the equation of the curve g(x) given that g′(x) = 3 and the curve passes through (1, 2).

THINK WRITE

Write the rule for g ′(x). g ′(x) =

Rewrite g ′(x) in index form. g ′(x) =

Express g(x) in integral notation. g(x) =

Antidifferentiate to obtain a general rule for g(x). =

Simplify. =

g(x) =

Substitute coordinates of the given point into g(x). As g(1) = 2, = 2Find the constant of antidifferentiation, c. 2 + 2 + c = 2

so c = −2

State the rule for g(x) in the form that it is given. g(x) =

=

x 2+

1 3 x 2+

2 3x12---

2+

3 3x12---

2+( ) dx∫4 3x

32--- 3

2---÷ 2x c+ +

53x

32---

1-------- 2

3---× 2x c+ +

2x32---

2x c+ +

6 2 1( )32---

2 1( ) c+ +7

8 2x32---

2x 2–+

2 x3 2x 2–+

7WORKEDExample

If a curve has a stationary point (2, 3), and a gradient of 2x − k, where k is a constant, find:a the value of k b y when x = 1.

THINK WRITE

a The gradient is so write the rule for the gradient. a

Let (as stationary points occur when the

derivative is zero) and substitute the value of x into this equation.

For stationary points,

, so 2x − k = 0

2(2) − k = 0 as x = 2Solve for k. 4 − k = 0 so k = 4

b Rewrite the rule for the gradient function, using the value of k found in a above.

b

Integrate to obtain the general rule for y.

Substitute the coordinates of the given point on the curve to find the value of c.

Since curve passes through (2, 3),3 = 22 − 4(2) + c3 = 4 − 8 + cc = 7

State the rule for y. So y = x2 − 4x + 7Substitute the given value of x and calculate y. When x = 1,

y = (1)2 − 4(1) + 7= 4

1dydx------ dy

dx------ 2x k–=

2dydx------ 0=

dydx------ 0=

3

1dydx------ 2x 4–=

2 y 2x 4–( ) dx∫=x2 4x– c+=

3

4

5

8WORKEDExample


Revision of antidifferentiation

1 Antidifferentiate each of the following, giving answers with positive indices.

2 Find the following indefinite integrals.

a x b x4 c x7 d 3x5 e 5x−2

f −2x4 g −6x−4 h i j

k l m n o

p q r s t

a b

c d

e f

g h

i

remember1.

2.

3.

4. , n ≠ −1

5.

6.

7. , n ≠ −1

8. , where x > 0 or

9. or =

10. Write the answer in the same form as the question unless otherwise stated.

ddx------ f x( ) f ′ x( )=

f x( ) f ′ x( ) dx∫=

a dx∫ ax c+=

axn dx∫ axn 1+

n 1+--------------- c+=

f x( ) g x( )±[ ] dx∫ f x( ) dx g x( ) dx∫±∫=

k f x( ) dx∫ k f x( ) dx∫=

ax b+( )n dx∫ ax b+( )n 1+

a n 1+( )----------------------------- c+=

1x--- dx∫ loge x c+= x 1– dx∫ loge x c+=

1ax b+--------------- dx∫ 1

a--- loge ax b+( ) c+= ax b+( ) 1– dx∫ 1

a--- loge ax b+( ) c+

remember

9AMathcad

Antidifferentiation

WORKEDExample

1

2 xx4

5----- x3

2-----

x 4–

3------- x x

23---

4x34---

x37---–

5x3----- 9

x2----- 10–

x6---------

8

x------- 6–

x x( )---------------

Mathcad

Integrator

WORKEDExample

2 2x 5+( ) dx∫ 3x2 4x 10–+( ) dx∫10x4 6x3 2+ +( ) dx∫ –4x5 x3 6x2 2x+–+( ) dx∫x3 12 x2–+( ) dx∫ x 3+( ) x 7–( ) dx∫

5 x2 2x 1–+( ) dx∫ x2 4+( ) x 7–( ) dx∫x x 1–( ) x 4+( ) dx∫


3

is equal to:

4

is equal to:

5 Antidifferentiate each of the following by rule.

6

is equal to:

7 Antidifferentiate the following.

A B

C D

E

A B

C D

E

a (x + 3)2 b (x − 5)3 c 2(2x + 1)4 d −2(3x − 4)5

e (6x + 5)4 f 3(4x − 1)2 g (4 − x)3 h (7 − x)4

i 4(8 − 3x)4 j −3(8 − 9x)10 k (2x + 3)−2 l (6x + 5)−3

m 6(4x − 7)−4 n (3x − 8)−6 o (6 − 5x)−3 p −10(7 − 5x)−4

A B

C D

E

a b c

d e f

g h i

j k l

m n o

p q r


x2 x 2+ +( ) dx∫x2 dx x 2+ +∫ x2 dx x dx 2 dx∫+∫+∫x2 x+( ) dx 2+∫ x2 x 2+( ) dx∫+

x2 x 2 dx∫+ +


x x 3+( ) dx∫x dx x 3+( ) dx∫∫ x x 3+( ) dx∫x 1+( ) x dx∫ x dx x 3+( ) dx∫+∫

x2 3x+( ) dx∫WORKEDExample

3


3 x 2+( )4 dx∫3 x 2+( )4 dx∫+ 3 dx x 2+( )4 dx∫+∫3 x 2+( )4 dx∫ 3 dx x 2+( )∫× 4 dx∫x 2+( )4 3 dx∫

WORKEDExample

4 3x--- dx∫ 8

x--- dx∫ 6

5x------ dx∫

73x------ dx∫ 4

7x------ dx∫ 1

x 3+------------ dx∫

3x 3+------------ dx∫WORKED

Example

5

2–x 4+------------ dx∫ 6–

x 5+------------ dx∫

43x 2+--------------- dx∫ 8

5x 6+--------------- dx∫ 3

2x 5–--------------- dx∫

5–3 2x+--------------- dx∫ 2–

6 7x+--------------- dx∫ 1

5 x–----------- dx∫

36 11x–------------------ dx∫ 2–

4 3x–--------------- dx∫ 8–

5 2x–--------------- dx∫


is equal to:

9 Find .

10 For the following mixed sets, find:

11 Find the equation of the curve f (x) given that:a f ′(x) = 4x + 1 and the curve passes through (0, 2)b f ′(x) = 5 − 2x and the curve passes through (1, −1)c f ′(x) = x−2 + 3 and the curve passes through (1, 4)d f ′(x) = x + and f (4) = 10

e f ′(x) = and f(8) = −100

f f ′(x) = and f (1) = −5

g f ′(x) = (x + 4)3 and the curve passes through (−2, 5)h f ′(x) = 8(1 − 2x)−5 and f(1) = 3i f ′(x) = (x + 5)−1 and the curve passes through (−4, 2)

j f ′(x) = and f(3) = 7

12 If a curve has a stationary point (1, 5), and a gradient of 8x + k, where k is a constant,find:

13 A curve g(x) has , where k is a constant, and a stationary point(1, 2). Find:

A B

C D

E

a b c

d e f

g h i

j k l

m n

a the value of k b y when x = −2.

a the value of k b g(x) c g(4).


x 5+------------ dx∫

61

x 5+------------ dx∫ 6dx

1x 5+------------ dx∫∫

6dx1

x 5+------------ dx∫+∫

6dx∫x 5+( ) dx∫

---------------------------

6

x 5+( ) dx∫---------------------------

WORKEDExample

62x 7+( )

x-------------------- dx∫

x4 2x

1x---+ +

∫ 3x 1+( )5 dx∫ 3x2 2x 1–+x2

------------------------------ dx∫3

2x 1+--------------- dx∫ 5–

6 10x–------------------ dx∫ 4 2x 5–( )5 dx∫

3 4x 1+( ) 3– dx∫ x 4+( )2

2x------------------- dx∫

x 5–( ) x 3+( )x3

---------------------------------- dx∫x

23 x–-----------+

dx∫ 5x32---

2x– 3x13---–+( ) dx∫

x2 x4+x

---------------- dx∫x2 2x 1–+

x-------------------------- dx∫ 10 x– 2x4+

x3----------------------------- dx∫

WORKEDExample

7 SkillSH

EET 9.1

x

x13---

3x2 50+–

1

x------- 2x–

87 2x–---------------

WORKEDExample

8

g′ x( ) kx x+x2

-------------------=


Integration of ex, sin x and cos xIntegration of the exponential function ex

Since

then

and , where k is a constant

or

Therefore,

ddx------ ex( ) ex=

ex

x ex

c+=d∫d

dx------ ekx( ) kekx=

ddx------ 1

k--- ekx×

1k--- kekx×=

ekx=

ekx dx∫ 1k---ekx c.+=

Antidifferentiate each of the following.

a 3e4x b c (ex − 1)2

THINK WRITE

a Integrate by rule where k = 4. a

Simplify.

b Rewrite the function to be integratedso that the coefficient of the e term isclear.

b =

Integrate by rule where k = −5. =

Simplify the antiderivative. =

=

c Expand the function to be integrated. c =

Integrate each term by the rule. =

e 5x–

4----------

1 3e4x dx∫ 3e4x

4---------- c+=

234---e4x c+=

1e 5x–

4---------dx∫ 1

4---e 5x– dx∫

214---e 5x–

5–------------ c+

314---e 5x– 1

5–------ c+×

120------e 5x–– c+

1 ex 1–( )2 dx∫ e2x 2ex 1+–( ) dx∫2

12---e2x 2ex x c+ +–

9WORKEDExample


Integration of trigonometric functions

Since (sin ax) = a cos ax and (cos ax) = −a sin ax

it follows that sin ax dx = cos ax + c

cos ax dx = sin ax + c

ddx------ d

dx------

∫ 1a---

∫ 1a---

Antidifferentiate the following.a sin 6x b 8 cos 4x c 3 sin

THINK WRITE

a Integrate by rule. a

b Integrate by rule. b

Simplify the result. = 2 sin 4x + c

c Integrate by rule. c =

Simplify the result. =

x2---–

sin 6x dx∫ –16--- cos 6x c+=

1 8 cos 4x dx∫ 84--- sin 4x c+=

2

1 3 sin x2---–

xd∫ 3–12---–

------ cos x2---–

2 6 cos x2---–

c+

10WORKEDExample

Find

THINK WRITE

Integrate each term separately.

Simplify each term where appropriate.

2e4x 5 sin 2x– 4x+( ) d x.∫

1 2e4x 5 sin 2x– 4x+( ) dx∫24---e4x 5–

2------ cos 2x– 4

2---x2 c++=

212---e4x 5

2--- cos 2x 2x2 c+ + +=

11WORKEDExample

remember1. ex dx = ex + c and ekx dx = ekx + c

2. sin ax dx = − cos ax + c and cos ax dx = sin ax + c

∫ ∫ 1k---

∫ 1a--- ∫ 1

a---

remember


Integration of ex, sin x and cos x

1 Antidifferentiate each of the following.

2 Find an antiderivative of (1 + ex)2.

3 Find an antiderivative of (ex − 1)3.

4 Find an antiderivative of x3 − 3x2 + 6e3x.

5

If f ′(x) = e2x + k and f(x) has a stationary point (0, 2), where k is a constant, then:a k is equal to:

b f(1) is equal to:

6 Antidifferentiate the following.

a e2x b e4x c e−x

d e−3x e 5e5x f 7e4x

g h i −3e6x

j −8e−2x k l

m n o ex + e−x

p

A e B e2 C 1 D −1 E −e

A e2 − 1 B e2 + C e2 + D e4 E 1

a sin 3x b sin 4x c cos 7x

d e sin (−2x) f cos (−3x)

g h 8 cos 4x i −6 sin 3x

j −2 cos (−x) k sin l cos

m 3 sin n −2 sin o 4 cos

p −6 cos q 4 sin r 6 cos

s −2 sin t −3 cos u 5 sin πx

v 3 cos w −2 cos x −sin

9B

Mathca

d

Integrator

WORKEDExample

9

e6x

2------- 2e3x

3----------

ex3---

0.1ex4---

3ex2---

3ex–

3------

ex e x––2

------------------


12--- 1

2--- 1

2--- 1

2---

WORKEDExample

10cos 2x

3----------------

4 sin 6x3

-------------------

x3--- x

2---

x4---–

x5--- x

4---

x2---–

2x3

------ 3x4

------

5x2

------ 7x4

------

πx2

------ πx3

------ 4xπ------–

C h a p t e r 9 I n t e g r a t i o n 3557 Find:

8 Find the antiderivative of e4x + sin 2x + x3.

9 Find an antiderivative of 3x2 − 2 cos 2x + 6e3x.


11 In each of the following find f(x) if:

a f ′(x) = cos x and f = 5

b f ′(x) = 4 sin 2x and f(0) = −1

c f ′(x) = 3 cos and f(π) =

d f ′(x) = cos − sin and f(2π) = −2.

12 If , where k is a constant, and y has a stationary point (3, 4), find:

a the value of k

b the equation of the curve

c y when x = 6.

13 A curve has a gradient function f ′(x) = 4 cos 2x + kex, where k is a constant, and astationary point (0, −1). Find:

a the value of k

b the equation of the curve f(x)

c f correct to 2 decimal places.

a b

c d

e f

g h

a b x2 + 4 cos 2x − e−x

c sin d

e 3 sin f

WORKEDExample

11 sin x cos x+( ) dx∫ sin 2x cos x–( ) dx∫cos 4x sin 2x+( ) dx∫ sin x

2--- cos 2x–

dx∫4 cos 4x 1

3--- sin 2x–( ) dx∫ 5x 2 sin x+( ) dx∫

3 sin πx2

------ 2 cos πx3

------+ dx∫ 3e6x 4 sin 8x– 7+( ) dx∫

x3 12x 3+---------------– e2x+

x3--- e

x2---

3x 1–( )4–+ 13x 2–--------------- e4x cos

x5---+ +

x2--- 2 cos x

3--- e

x–5

------–– x 2x 2 sin

πx3

------– 5+ +

Mathcad

Antidifferentiationπ2---

x4--- 9 2

x4--- x

2---

dydx------ sin

πx6

------ k+=

π6---


Integration by recognitionAs we have seen, if

then , where g(x) = f ′(x).

This result can be used to determine integrals of functions, which are too difficult toantidifferentiate, via differentiation of a related function.

Note that the shorter form of the chain rule below can be used to differentiate.

If then

ddx------ [f(x)] g x( )=

g x( ) dx∫ f x( ) c+=

a Find the derivative of the function y = (5x + 1)3.b Use this result to deduce the antiderivative of 3(5x + 1)2.

THINK WRITE

a Write the function and recognise that the chain rule can be used.

a y = (5x + 1)3

Let u equal the function inside the brackets. Let u = 5x + 1

Find .

Express y in terms of u. y = u3

Find .

Write the chain rule.


Replace u with the expression inside the brackets and simplify where applicable.

= 15(5x + 1)2

b Since , express the

relationship in integral notation.

b

Remove a factor from so that it

resembles the integral required.

Divide both sides by the factor in order to obtain the integral required.

, where

Therefore, the antiderivative of

is

1

2

3dudx------ du

dx------ 5=

4

5dydu------ dy

du------ 3u2=

6dydx------ dy

du------ du

dx------×=

7dydx------ dy

dx------ 3u2 5×=

8

1dydx------ dx∫ y c1+= 15 5x 1+( )2 dx∫ 5x 1+( )3 c1+=

2dydx------ 5 3 5x 1+( )2 dx∫ 5x 1+( )3 c1+=

3 3 5x 1+( )2 dx∫ 15--- 5x 1+( )3 c+=

cc1

5-----=

3 5x 1+( )2 15--- 5x 1+( )3 c.+

12WORKEDExample

f x( ) g x( )[ ]n= f′ x( ) ng′ x( ) g x( )[ ]n 1–=


Note that the shorter form of the chain rule below can be used to differentiate.

If then

a Differentiate . b Hence, antidifferentiate 6x2 .

THINK WRITE

a Write the equation and apply the chain rule to differentiate y.

a y =

Let u equal the index of e. Let u = x3

Find .

Express y in terms of u. y = eu

Find .

Find using the chain rule

and replace u.

, applying the chain rule


Multiply both sides by a constant to obtain the integral required.

2

, where c = 2c1.

Therefore, the antiderivative of is .

ex3 ex3

1 ex3

2

3dudx------ du

dx------ 3x2=

4

5dydu------ dy

du------ eu=

6dydx------ dy

dx------ 3x2eu=

3x2ex3=

1dydx------ 3x2ex3 dx∫ ex3 c1+=

2 3x2ex3 dx∫ 2ex3 2c1+=

6x2ex3 dx∫ 2ex3 c+=

6x2ex3 2ex3 c+

13WORKEDExample

y ef x( )= d ydx------ f ′ x( )ef x( )=

a Find the derivative of sin (2x + 1) and use this result to deduce the antiderivative of8 cos (2x + 1).

b Differentiate loge (5x2 − 2) and hence antidifferentiate .

Continued over page

THINK WRITEa Let f(x) equal the rule. a f(x) = sin (2x + 1)

Differentiate using f ′(x) = g′(x) cos [g(x)] where f(x) = sin [g(x)].

f ′(x) = 2 cos (2x + 1)

Express f(x) using integral notation.

Multiply both sides by whatever is necessary for it to resemble the integral required.

Write the integral in the form in which the question is asked.

x5x2 2–------------------

12

3 2 cos 2x 1+( ) dx∫ sin 2x 1+( ) c1+=

4 4 2 cos 2x 1+( ) dx∫ 4 sin 2x 1+( ) c+=

5 8 cos 2x 1+( ) dx∫ 4 sin 2x 1+( ) c+=

14WORKEDExample


THINK WRITE

b Let f(x) equal the rule. b f(x) = loge (5x2 − 2)

Differentiate using

where f(x) = loge [g(x)].

Express f(x) using integral notation.

Take out a factor so that f ′(x) resembles the integral required.

Divide both sides by the factor to obtain the required integral.


is

1

2 f ′ x( ) g′ x( )g x( )------------= f ′ x( ) 10x

5x2 2–-----------------=

310x

5x2 2–----------------- dx∫ loge 5x2 2–( ) c1+=

4 10x

5x2 2–----------------- dx∫ loge 5x2 2–( ) c1+=

5x

5x2 2–----------------- dx∫ 1

10------ loge 5x2 2–( ) c+=

x5x2 2–----------------- 1

10------ loge 5x2 2–( ) c+

Differentiate x cos x and hence find an antiderivative of x sin x.

THINK WRITEWrite the rule. Let y = x cos x

Apply the product rule to differentiate x cos x.

= x (−sin x) + (cos x)(1)

= −x sin x + cos x

= cos x − x sin x

Express the result in integral notation. (Do not add c, as an antiderivative is required.)

∴

Express the integral as two separate integrals.

()

Simplify by integrating. (Do not add c.)

Make the expression to be integrated the subject of the equation.

Simplify.

Therefore, an antiderivative of x sin x is

sin x − x cos x.

1

2dydx------

3 cos x x sin x–( )dx∫ x cos x=

4 cos x dx∫ – x sin x dx∫ x cos x=

5 sin x x sin x dx∫– x cos x=

6 x sin x dx∫– x cos x sin x–=

7 x sin x dx∫ sin x x cos x–=

15WORKEDExample


The Mathcad file shown below may be used to check derivatives.

a Show that . b Hence, find .

THINK WRITE

a Use algebraic long division to divide the numerator into the denominator.

a So

b Write the expression using integral notation.

Express as two separate integrals.

b =

=

Antidifferentiate each part. = 5x − 4 loge (x + 1) + c

5x 1+x 1+--------------- 5 4

x 1+------------–= 5x 1+x 1+--------------- dx∫

x + 1 ) 5x + 15

−(5x + 5)

–4

5x 1+x 1+

--------------- 5 4x 1+------------–=

1

2

5x 1+x 1+

--------------- dx∫ 5 4x 1+------------–

∫5 dx

4x 1+------------ dx∫–∫

3

16WORKEDExample

Mathcad

Differentiator

remember1. To differentiate using the chain rule, use one of the following rules.

(a) If f(x) = [g(x)]n then f ′(x) = ng′(x)[g(x)]n − 1

(b) If y = ef(x),

(c)

(d) f ′(x) = g′(x) cos [g(x)] where f(x) = sin [g(x)]f ′(x) = −g′(x) sin [g(x)] where f(x) = cos [g(x)]

(e) where f (x) = loge[g(x)].

2. To antidifferentiate use where g(x) = f ′(x).

dydx------ f ′ x( )ef x( )=

dydx------ dy

du------= du

dx------×

f ′ x( ) g′ x( )g x( )------------=

g x( ) dx∫ f x( ) c+=

remember


Integration by recognition

1 For each of the following find the derivative of the function in i and use this result todeduce the antiderivative of the function in ii.

2The derivative of (x + 7)4 is 4(x + 7)3.a Therefore, the antiderivative of 4(x + 7)3 is:

b The antiderivative of (x + 7)3 is:

3

If the derivative of (2x − 3)6 is 12(2x − 3)5, then is:

4 For each of the following differentiate i and hence antidifferentiate ii.

5 For each of the following find the derivative of the function in i and use this result todeduce the antiderivative of the function in ii.

a i (3x − 2)8 ii 12(3x − 2)7

b i (x2 + 1)5 ii 5x(x2 + 1)4

c i ii

d i ii

e i (x2 + 3x − 7)4 ii (2x + 3)(x2 + 3x − 7)3

f i ii

A (x + 7)4 + c B (x + 7)4 + c C 4(x + 7)4 + cD 3(x + 7)4 + c E 12(x + 7)4 + c

A (x + 7)4 + c B (x + 7)4 + c C 4(x + 7)4 + cD 3(x + 7)4 + c E 12(x + 7)4 + c

A 2(2x − 3)6 + c B 4(2x − 3)6 + c C (2x − 3)6 + cD 6(2x − 3)6 + c E (2x − 3)6 + c

a i e4x − 5 ii 2e4x − 5

b i e6 − 5x ii 10e6 − 5x

c i iid i ii

a i sin (x − 5) ii cos (x − 5)b i sin (3x + 2) ii 6 cos (3x + 2)c i cos (4x − 7) ii sin (4x − 7)d i cos (6x − 3) ii 3 sin (6x − 3)e i sin (2 − 5x) ii 10 cos (2 − 5x)f i cos (3 − 4x) ii −2 sin (3 − 4x)

g i loge (5x + 2) ii

h i loge (x2 + 3) ii

i i loge (x2 − 4x) ii

9CWORKEDExample

12

Mathca

d

Differentiator

2x 5–1

2x 5–-------------------

4x 3+2

4x 3+-------------------

1x2 1–--------------

4xx2 1–( )2

---------------------


14---

14---


6 2x 3–( )5 dx∫12---

WORKEDExample

13

ex2x ex2

ex x2– 1 2x–( )ex x2–

WORKEDExample

14

205x 2+---------------

12xx2 3+--------------

x 2–x2 4x–-----------------

C h a p t e r 9 I n t e g r a t i o n 3616 Differentiate i and hence find an antiderivative of ii.

7 For each of the following differentiate i and use this result to antidifferentiate ii.

8 a Show that = 3 + . b Hence, find .

9 a Show that = 5 − . b Hence, find .

10 a Show that = 4 + . b Hence, find .

11 a Show that = −3 + . b Hence, find .

12 If y = loge (cos x):

a find b hence find .

13 Differentiate and hence find an antiderivative of .

14 Differentiate loge (3x2 − 4) and hence find an antiderivative of .

15 Differentiate sin (ax + b) and hence find an antiderivative of cos (ax + b). (Here, a andb are constants.)

16 Differentiate cos (ax + b) and hence find an antiderivative of sin (ax + b). (Here, a andb are constants.)

17 Differentiate eax + b and hence find an antiderivative of eax + b. (Here, a and b areconstants.)


a i x cos x + 2 sin x ii x sin x

b i ii

c i ex sin x ii 3ex (sin x + cos x)d i x sin x ii x cos x

e i x ex ii x ex

a i (2x − 3x2)6 ii 6x5(1 − 3x)(2 − 3x)5

b i ii

a sin (3πx + 1) b cos (1 − 4πx) c eπx + 3

d sin e 3 cos f cos x esin x

WORKEDExample

15sin x

x-----------

2 x cos x sin x–( )x2

-------------------------------------------

x3 2x+3x2 2+x3 2x+

----------------------

WORKEDExample

16

3x 2–x 1–

--------------- 1x 1–----------- 3x 2–

x 1–--------------- dx∫

5x 8+x 2+

--------------- 2x 2+------------ 5x 8+

x 2+--------------- dx∫8x 7–2x 3–--------------- 5

2x 3–--------------- 8x 7–

2x 3–--------------- dx∫

6x 5–3 2x–--------------- 4

3 2x–--------------- 6x 5–

3 2x–--------------- dx∫

dydx------ tan x dx∫

cos xsin x------------ 1

sin2x------------

x3x2 4–-----------------

WorkS

HEET 9.1

2πx3

------+ πx

2------ 5+


Approximating areas enclosed by functions

There are several ways of finding an approximation to the area between a graph and thex-axis. We will look at three methods:1. the lower rectangle method2. the upper rectangle method3. the trapezoidal method.

The lower rectangle methodConsider the area between the curve f(x) shown at right, the x-axis and the lines x = 1 and x = 5.

If the area is approximated by ‘lower’ rectangles of width 1 unit then the top of each rectangle is below the graph but touches the curve at one point. (In this case the left-hand corner of the rectangle touches the graph.)So, the height of rectangle R1 is f(1) unitsand the area of R1 = 1 × f(1) square units (area of a rectangle = height × width).Similarly, the area of R2 = 1 × f(2) square units,

the area of R3 = 1 × f(3) square units,the area of R4 = 1 × f(4) square units.

Therefore, the approximate area under the graph between the curve f(x), the x-axisand the lines x = 1 to x = 5 is 1[f(1) + f(2) + f(3) + f(4)] square units, (the sum of thearea of the four rectangles).

If the same area was approximated using rectangle widths of 0.5 there would be 8rectangles and the sum of their areas would be:0.5[f(1) + f(1.5) + f(2) + f(2.5) + f(3) + f(3.5) + f(4) + f(4.5)] square units.

From the diagram it can be seen that the lower rectangle approximation is less thanthe actual area.

x0 1 2 3 4 5

R1 R2 R3 R4

y f(x)

Find an approximation for the area between the curve f(x) shown and the x-axis from x = 1 to x = 3 using lower rectangles of width 0.5 units. f(x) = 0.2x2 + 3

THINK WRITE

Write the number of rectangles and their width.

There are 4 rectangles of width 0.5 units.

Find the height of each rectangle (left) by substituting the appropriate x-value into the f(x) equation.

h1 = f(1) = 0.2(1)2 + 3 = 3.2

h2 = f(1.5) = 0.2(1.5)2 + 3 = 3.45

h3 = f(2) = 0.2(2)2 + 3 = 3.8

h4 = f(2.5) = 0.2(2.5)2 + 3 = 4.25Area equals the width multiplied by the sum of the heights.

Area = width × (sum of heights of 4 rectangles)= 0.5(3.2 + 3.45 + 3.8 + 4.25)= 0.5(14.7)

Calculate this area. = 7.35State the solution. The approximate area is 7.35 square units.

1

2

3

45

17WORKEDExample

x0 10.5 1.5 2 2.5 3

y f(x)

3


The upper rectangle methodConsider the area between the curve f(x) shown at right,the x-axis and the lines x = 1 and x = 5.

If the area is approximated by ‘upper rectangles’ ofwidth 1 unit then the top of each rectangle is above thegraph and touches the curve at one point. (In this case thetop right-hand corner of the rectangle touches the graph.)So, the height of R1 is f(2) unitsand the area of R1 is 1 × f(2) square units.Similarly, the area of R2 = 1 × f(3) square units,

the area of R3 = 1 × f(4) square unitsthe area of R4 = 1 × f(5) square units.

Therefore, the approximate area between the curve f(x), the x-axis and the lines x = 1to x = 5 is (R1 + R2 + R3 + R4) = 1[f(2) + f(3) + f(4) + f(5)] square units.

If the same area was approximated with upper rectangle widths of 0.5 units, the sumof their areas would equal:0.5[f(1.5) + f(2) + f(2.5) + f(3) + f(3.5) + f(4) + f(4.5) + f(5)] square units.

From the diagram it can be seen that the upper rectangle approximation is greaterthan the actual area.

Lower rectangle approximation ≤ actual area ≤ upper rectangle approximation

It can be seen that the lower rectangle approximation (7.35 units) is lessthan the upper rectangle approximation (8.15 units).

If the area is divided into narrower strips, the estimate of the areawould be closer to the true value. Use one of the Mathcad files to inves-tigate the effect of a greater number of strips in a given interval.

x0 1 2 3 4 5

R1 R2 R3 R4

y f(x)

Find an approximation for the area in the diagram in worked example 17 using upper rectangles of width 0.5 units. f(x) = 0.2x2 + 3

THINK WRITE

Find the number of rectangles and the height of each one (from left to right).

There are 4 rectangles:

h1 = f(1.5) = 0.2(1.5)2 + 3 = 3.45

h2 = f(2) = 0.2(2)2 + 3 = 3.8

h3 = f(2.5) = 0.2(2.5)2 + 3 = 4.25

h4 = f(3) = 0.2(3)2 + 3 = 4.8

Area is the width of the interval multiplied by the sum of the heights.

Area = 0.5(3.45 + 3.8 + 4.25 + 4.8)

Calculate the area. = 0.5(16.3)= 8.15

State the solution. The approximate area is 8.15 square units.

1

2

3

4

18WORKEDExample

Mathcad

Rightboxes

Mathcad

Leftboxes


The trapezoidal methodRecall that the area of a trapezium = (a + b).

The trapezoidal method involves a series of straight lineapproximations to the curve which generates strips inthe shape of trapeziums.

Consider the area under the graph of f(x) between the x-axis and the lines x = 1 to x = 5.For each trapezium the width, or height, h = 1 unit.For T1, a = f(1) and b = f(2)For T2, a = f(2) and b = f(3)For T3, a = f(3) and b = f(4)For T4, a = f(4) and b = f(5)

The area of T1 = [f(1) + f(2)]

The area of T2 = [f(2) + f(3)] and so on.The total area of the trapeziums is:

[f(1) + f(2) + f(2) + f(3) + f(3) + f(4) + f(4) + f(5)]= [f(1) + 2f(2) + 2f(3) + 2f(4) + f(5)] square units.The first and last terms are counted only once but all others are counted twice.

Note that the lower rectangle approximation found in worked example 17 was 7.35units and the upper rectangle approximation found in worked example 18 was 8.15

units. The average of these two approximations is or 7.75 units which is

the same as the trapezoidal approximation for the area.

a

h

b

h2---

T1 T2 T3 T4

x

f(x)

0 2 4 5

y

1 312---

12---

12---

12---

Find an approximation for the area enclosed by the graph of f(x) = 0.2x2 + 3, the x-axis and the lines x = 1 to x = 3 using interval widths of 0.5 units and using the trapezoidal method.

THINK WRITE/DRAWSketch the graph of f(x).Draw trapeziums of width 0.5 unit from x = 1 to x = 4.

Evaluate the height of each vertical side of the trapeziums by substituting the appropriate x-value into f(x).

f(1) = 0.2(1)2 + 3 = 3.2f(1.5) = 0.2(1.5)2 + 3 = 3.45

f(2) = 0.2(2)2 + 3 = 3.8f(2.5) = 0.2(2.5)2 + 3 = 4.25

f(3) = 0.2(3)2 + 3 = 4.8Calculate the area by using the formula for the area of a trapezium where h is the width of the interval.

Total area of trapeziums

= (3.2 + 2 × 3.45 + 2 × 3.8 + 2 × 4.25 + 4.8)

= 0.25 × 31= 7.75

Therefore, the area under the curve is approximately 7.75 units.

12

x0 2 4

y

3

–2 1 33

40.52

-------

19WORKEDExample

7.35 8.15+2

---------------------------


Note that this average is between the area of the upper rectangles and the area of thelower rectangles and is closer to the actual area.

With width intervals of 1 unit calculate an approximation for the area between the graph of f(x) = x2 + 2 and the x-axis from x = −2 to x = 3 using:

a lower rectangles b upper rectangles c averaging of the lower and upper rectangle areas.

THINK WRITESketch the graph of f(x) over a domain which exceeds the width of the required area.Draw the lower and upper rectangles.

a Calculate the height of the lower rectangles by substituting the appropriate values of x into the equation for f(x). Note the two rectangles to the right and left of the origin have the same height and are equal in area.

a Lower rectangle heights: f(−1) = (−1)2 + 2 = 3

f(0) = 02 + 2 = 2f(0) = 2f(1) = 12 + 2 = 3f(2) = 22 + 2 = 6

Find the area by multiplying the width by the sum of the heights.

Area = 1(3 + 2 + 2 + 3 + 6)= 16

Using lower rectangles, the approximate area is 16 square units.

b Calculate the height of the upper rectangles by substituting the appropriate values of x into the equation for f(x).

b Upper rectangle heights: f(−2) = (−2)2 + 2 = 6f(−1) = 3 (from above)

f(1) = 3f(2) = 6f(3) = 32 + 2 = 11

Find the area by multiplying the width by the sum of the heights.

Area = 1(6 + 3 + 3 + 6 + 11)= 29

Using upper rectangles, the approximate area is 29 square units.

c Find the average by adding the area of the upper rectangles and lower rectangles and dividing by 2.

c Average of the areas =

= 22.5The approximate area is 22.5 square units when averaging the upper and lower rectangle areas and using widths of 1 unit.

1

2

x0

y

2–2 –1 1 3

2

= Lower rectangles

= Upper rectangles

y = x2 + 2

1

2

1

2

16 29+2

------------------

20WORKEDExample


Approximating areas enclosed by functions

1 Find an approximation for the area between the curve f(x) atright and the x-axis from x = 1 to x = 5 using a lowerrectangle of width 2 units.

2 Find an approximation for the area between the curves belowand the x-axis, from x = 1 to x = 5, by calculating the area ofthe shaded rectangles.

3Consider the graph of y = x2 from x = 0 to x = 4 at right.a The width of each rectangle is:

b The height of the right-hand rectangle is:

c The area between the curve y = x2 and the x-axis from x = 0 to x = 4 can beapproximated by the area of the lower rectangles as:

a b

A 1 unit B 2 units C 3 unitsD 4 units E varying

A 9 units B 4 units C 16 unitsD 12 units E 1 unit

A 20 sq. units B 14 sq. units C 18 sq. units D 15 sq. units E 30 sq. units

remember1. An approximation to the area between a curve and the x-axis can be found by

dividing the area into a series of rectangles or trapeziums which are all the same width. The approximation is found by finding the sum of all the areas of the rectangles or trapeziums.

2. Lower rectangle approximation ≤ actual area ≤ upper rectangle approximation3. Trapezoidal approximation =

lower rectangle approximation + upper rectangle approximation2

---------------------------------------------------------------------------------------------------------------------------------------------------------

remember

9DWORKEDExample

17

x0 1 5

y

2

3(1, 2)

(3, 3)

3

f(x)

x0 1 5

yf(x)

2

4

(1, 2)

(5, 4)

x0 1 5

47

(4, 19)

3

f(x)

42

12

19

3

(3, 12)

(2, 7)

(1, 4)

y


x0

y

1 2

y = x2

3 4

C h a p t e r 9 I n t e g r a t i o n 3674 a Find an approximation for the area in the dia-

gram at right using upper rectangles of width 1unit.

b A better approximation for the area under thiscurve can be found by averaging the upper andlower rectangle areas. State this approximatevalue.

5 Find an approximation for the area enclosed by thegraph of f(x) = x2, the x-axis and the lines x = 1 andx = 3 with interval widths of 1 unit using the trapezoidalmethod.

6 Find the approximate area between the curves below and the x-axis, over the intervalindicated, by calculating the area of the shaded trapeziums.a

7 Find an approximation for the area between the curves below and the x-axis, fromx = 1 to x = 5, by calculating the area of the shaded rectangles.a

Mathcad

Trapezoidalrule

GC program

Left and rightbox estimations

of area

Mathcad

Rightboxes

Mathcad

Leftboxes

WORKEDExample

18

x0

y

1 2

y = x2

3 4

WORKEDExample

19

GC program

Trapezoidalrule

0

y

1

y = x2

2 3x

x0 2 4

y f(x)

3

5

(2, 3)

(4, 5)

x0 2 4

y

6

f(x)

32

5

(2, 2)(4, 3)

(6, 5)

b

x0 1 5

y

6

(1, 8)8

f(x) x0 1 3

y

7

(1, 8)8

f(x)5

(3, 8)

x0 1 2

y

7(4, 7)

(3, 10)

8

f(x)

3

(2, 11)

1011

4 5

b c d

x0

y

3

5

1

(3, 3) f(x)

3 5

(5, 5)

0

y

46

910

1 2

(1, 4)

f(x)

3 4 5

(3, 9)(4, 10)

(2, 6)

x

e f g

x0

y

2

4

1

(1, 4) f(x)

3 5

(3, 2)

x0

y

7

1 2

(1, 7)

(2, 5)

(4, 7)

f(x)

3 4 5

45


8 With width intervals of 1 unit calculate an approximationfor the area between the graph of f(x) = x2 + 4 and thex-axis from x = 1 to x = 4 using:a lower rectanglesb upper rectanglesc averaging of the lower and upper rectangle areas.

9 Find the approximate area between the curves below andthe x-axis, over the interval indicated, by calculating thearea of the shaded rectangles. Give exact answers.

10 Find an approximation for the area between the graphs below, the x-axis, and the linesx = 1 and x = 5, using interval widths as shown in the diagrams below. Give exact answers.

a b c d

a b c

d e

WORKEDExample

20

x0 1 2

y

y = x2 + 4

3 4

x0 1 2

y

x = 1 to x = 4

y = –x2 + 3x + 8

3 4

x0 1 2

y

x = –1 to x = 2

y = ex

–1

0 1 2

y

3 4 5

y = logex

x = 1 to x = 5

x 2 30

y

1 4 5 6

y = (x – 4)2

x = 2 to x = 6

x

y

x0

2 3 4 5 6x = 2 to x = 6

y = x3 – 6x2x0

y

–3 –2.5 –2 –1.5 –1

f(x) = –x2 – 4x

x = –3 to x = –1 x0

y

1 2 3 4 5x = 1 to x = 5

f(x) = x3 – 3x2 + 8x 1 –3

f g

x0

y

1413

15

1 2

f(x)

3 4

10

(2, 14)(3, 15)

(4, 13)

(1, 10)

x0 1 2

y

3 4 5

y = –x2 + 6x – 5

–1 x0

y

1 2 3

y = x3 1 —2

0 1 x

y

–1

y = 10 – x2

– 1 –2

1 –2

0

y

1

y = ex

2 3x

e

C h a p t e r 9 I n t e g r a t i o n 36911 Calculate an approximation for the area between the graph of y = x(4 − x), the x-axis

and the lines x = 1 and x = 4, using interval widths of 1 unit and:a lower rectanglesb upper rectanglesc averaging the lower and upper rectangle areas.

12 Calculate an approximation for the area under the graph of y = x2 − 4x + 5 to thex-axis between x = 0 and x = 3, using interval widths of 0.5 units and:a lower rectanglesb upper rectanglesc averaging the lower and upper rectangle areas.

13 Find an approximation for the area under the graph of y = 2x between x = 0 and x = 3,using interval widths of 1 unit, by averaging the lower and upper rectangle areas.

14 Find an approximation for the area between the graph of f ( x) = ( x − 1)3 and thex-axis, between x = 1 and x = 4, using the trapezoidal rule and:a interval widths of 1 unitb interval widths of 0.5 units.Give answers correct to 1 decimal place.

15 Calculate an approximation for the area under the graph of between x = 0.5and x = 2.5, using the trapezoidal rule and:a interval widths of 1 unitb interval widths of 0.5 unitsc interval widths of 0.25 units.Give answers correct to 2 decimal places.

16 Calculate an approximation for the area under the graph of y = 2 loge (x − 1) betweenx = 2 and x = 6 using the trapezoidal rule and interval widths of 1 unit.

y1x---=

17 At the back of a rectangular block of land, 100metres long, is a park and a river. The distance to theriver from the top of the rectangular block is shownin the table below.

Calculate an approximation for the area of parklandbetween the rectangular block and the river by:a using the area of the ‘upper’ rectangles.b using the trapezoidal rule (use intervals of width

20 metres).

Distance across rectangular block in metres

0 20 40 60 80 100

Distance of river from the block in metres

0 30 20 40 60 50

0 10020 40 60 80

metres

River Park

Block of land


18 Calculate an approximate area under the graph of f(x) = sin x, between x = 0 and

x = π, using the trapezoidal rule and interval widths of units. Give your answer

correct to 2 decimal places.

20 Answer the following statements concerning approximate areas under graphs as Trueor False.a An approximation for the area can be found quickly if very small interval widths

are used.b The smaller the interval width used, the more accurate the approximation for the

area.c The upper rectangle method is always more accurate than the lower rectangle

method.d Averaging the upper rectangle area and the lower rectangle area is more accurate

than using the ‘upper’ or ‘lower’ approximations on their own.

π6---

19 The graph below shows the velocity ofa cyclist (in metres per second) at timet seconds after commencing a race.

a What does the shaded area rep-resent?

b Find the approximate distancetravelled by the cyclist in the first30 seconds using the trapezoidalrule and interval widths of 5 sec-onds.

t

Vel

ocity

(m

/s)

010 20 30 40

V

Time (s)

(5, 5)

(15, 12) (20, 13)

(25, 14) (30, 15)

(10, 10)


The fundamental theorem of integral calculus

Consider the region under the curve f(x) between x = a and x = b, where f(x) ≥ 0 and iscontinuous for all x ∈ [a, b].

Let F(x) be the function that is the measure of the area under the curve between aand x.

F(x + h) is the area under the curve between a and x + hand F(x + h) − F(x) is the area of the strip indicated on the graph.

The area of the strip is between the areas of the upper and lower rectangles;

that is,

or , h ≠ 0 (dividing by h).

As h → 0, f(x + h) → f(x)

or

that is, F ′(x) = f(x) (differentiation from first principles).

Therefore, F(x) =

that is, F(x) is an antiderivative of f(x)

or = F(x) + c

but when x = a,

= F(a) + c

= 0 (as the area defined is zero at x = a)

or c = −F(a).

Therefore, = F(x) − F(a)

and when x = b, = F(b) − F(a).

That is, the area under the graph of f(x) between x = a and x = b is F(b) − F(a).

xb0 a x x + h

y y = f(x)

F(x)F(x + h) – F(x)

f x( )h F x h+( ) F x( ) f x h+( )h<–<

f x( ) F x h+( ) F x( )–h

-------------------------------------- f x h+( )< <

F x h+( ) F x( )–h

----------------------------------------h 0→lim f x( )=

f x( ) dx∫

f x( ) dx∫

f x( ) dx∫

f x( ) dx∫f x( ) dx∫


This is the fundamental theorem of integral calculus and it enables areas undergraphs to be calculated exactly. It can be stated as:

Area =

= [do not add c as F(x) is an antiderivative of f(x)]

= F(b) − F(a)

a and b are called the terminals of this definite integral.

is called the definite integral because it can be expressed in terms of its

terminals a and b, which are usually real numbers. In this case the definite integralevaluates as a real number and not a function.

The function being integrated, f(x), is called the integrand.

Properties of definite integralsDefinite integrals have the following five properties.

1.

2. , a < c < b

3.

4.

5.

Definite integrals

1. Evaluate

2. Evaluate a b

c Compare answers a and b.

3. a b Compare the answer to the answer to 2a.

4. Evaluate a b


5. Evaluate a b


f x( ) dxa

b

∫F x( )[ ]a

b

f x( ) dxa

b∫

2x dx1

1

∫2x dx

1

4

∫ 2x dx 2x dx2

4

∫+1

2

∫

2 x dx1

4

∫

2x x2+[ ] dx0

3

∫ 2x dx x2 dx0

3

∫+0

3

∫

2x dx1

3

∫ 2x dx3

1

∫

f x( ) dxa

a

∫ 0=

f x( ) dxa

b

∫ f x( ) dx f x( ) dxc

b

∫+a

c

∫=

kf x( ) dxa

b

∫ k f x( ) dxa

b

∫=

f x( ) g x( )+[ ] dxa

b

∫ f x( ) dx g x( ) dxa

b

∫+a

b

∫=

f x( ) dxa

b

∫ f x( ) dxb

a

∫–=


Evaluate the following definite integrals.

a

b

THINK WRITE

a Antidifferentiate each term of the integrand and write in the form

.

a

=

Substitute values of a and b into F(b) − F(a).

=

Evaluate the integral. = 42 − 0

= 42

b Express the integrand in simplest index form.

b =


=

Express the integral with a positive index number.

=

Substitute values of a and b intoF(b) − F(a).

=


=

3x2 4x 1–+( ) dx0

3

∫4

2x 1+( )3----------------------- dx

1–

2

∫

1

F x( )[ ]ab

3x2 4x 1–+( ) dx0

3

∫x3 2x2 x–+[ ]

03

2 33 2 3( )2 3–+[ ] 03 2 0( )2 0–+[ ]–

3

14

2x 1+( )3---------------------- dx

1–

2

∫ 4 2x 1+( ) 3– dx1–

2

∫

24 2x 1+( ) 2–[ ] 1–

2

2 2–×--------------------------------------

2x 1+( ) 2––[ ]1–

2

31–

2x 1+( )2----------------------

1–

2

4 152-----–

1–1–( )2

-------------–

51

25------ 1+–

2425------

21WORKEDExample


It is possible to find the numerical value for the definite integral using the graphics

calculator. Using the TI–83, follow these steps to evaluate

1. Ensure radian MODE is selected.2. Enter the function sin(X/6) as Y1 in the Y= menu.3. Press [QUIT].4. Press and select 9:fnInt(5. Complete the expression as shown at right.

(Y1 is found under /Y-VARS/1:Function.

The answer is 2.196 15 which is the answer correct to 5 decimal places. This is thesame as −3 + 3 found in worked example 22.

Find the exact value of each of the following definite integrals.

a b

THINK WRITE

a Antidifferentiate the integrand, writing it in the form .

a =

Substitute values of a and b into F(b) − F(a).

=

=


=

=

b Antidifferentiate the integrand, using . b

Substitute values of a and b into F(a) − F(b).

=

Evaluate. =

= 0

sin x6--- dx

π

2π∫ e3x e 3x––( ) dx1–

1

∫

1F x( )[ ]a

bsin

x6--- dx

π

2π∫ 6 cos

x6---–

π

2π

2 6 cos2π6

------– 6 cosπ6---––

6 cosπ3---– 6 cos

π6---––

3 612---

– 63

2-------

––

3–[ ] 3 3–[ ]–

–3 3 3+

1F x( )[ ]a

b

e3x e 3x––( ) dx1–

1

∫13---e3x 1

3---e 3x–+[ ]

1–

1=

213---e3 1

3---e 3–+[ ] – 1

3---e 3– 1

3---e3+[ ]

313---e3 1

3---e 3– 1

3---e 3–– 1

3---e3–+

22WORKEDExample

Graphics CalculatorGraphics Calculator tip!tip! Finding definite integrals

sinx6---

dx.π

2π∫

2ndMATH

VARS

3


If , find k.

THINK WRITE

Antidifferentiate the integrand, using .

=

So = 36

Substitute the values of a and b into F(a) − F(b). [4k2] − [4(0)2] = 36

Simplify the integral. 4k2 − 0 = 36

4k2 = 36

k2 = 9

k = ±

Solve the equation. k = 3 or −3

8x dx0

k

∫ 36=

1 F x( )[ ]ab

8x dx0

k

∫ 4x2[ ]0k

4x2[ ]0k

2

3

9

4

23WORKEDExample

remember1. The fundamental theorem of calculus is =

= F(b) − F(a).where F(x) is an antiderivative of f(x).

2. The expression is called the definite integral where a and b are the

terminals and represent the upper and lower values of x.

3. Properties of definite integrals

(a)

(b) , a < c < b

(c)

(d)

(e)

fa

b

∫ x( ) dx F x( )[ ]ab

f x( ) dxa

b

∫

f x( ) dxa

a

∫ 0=

f x( ) dxa

b


b

∫+a

c

∫=

kf x( ) dxa

b

∫ k f x( ) dxa

b

∫=

f x( ) g x( )+[ ] dxa

b


b

∫+a

b

∫=

f x( ) dxa

b

∫ f x( ) dxb

a

∫–=

remember


The fundamental theorem of integral calculus

1 Evaluate the following definite integrals.

2 Find the exact value of each of the following definite integrals.

3 If , find the value of .

a b c

d e f

g h i

j k l

m n o

p q r

s t

a b dx c

d e f

g h i

j k l

m n o

p q r

9E

Mathca

d

Integrator

SkillSH

EET 9.2 WORKEDExample

21x2 dx

0

1

∫ x3 dx0

3

∫ (x2 2x) dx–3

4

∫1x2----- dx

2

6

∫ x3 3x2 2x–+( ) dx0

2

∫ (3x12---

2x2) dx+1

4

∫

(6 2x– x2)+ dx1–

1

∫ x3 x 4–+( ) dx4–

2–

∫ 3 x dx4

9

∫

4x 2– 2x 6–+( ) dx1

2

∫ 2 x 4+( )4 dx0

3

∫ 3 5x 2–( )4 dx1–

2

∫

4 2 3x–( )3 dx–2–

0

∫ 52x 7–( )3

---------------------- dx0

3

∫ 2x32---

3x 1––( ) dx1

4

∫2x3 5x2+

x----------------------- dx

1

3

∫ 35x------ dx

1

5

∫ 4–3x 4–( )5

---------------------- dx0

2

∫1

2x 5–------------------- dx

3

7

∫ 6

8 3x–------------------- dx

2–

0

∫

WORKEDExample

22e4x dx

0

2

∫ ex3---

2–

0

∫ 4e 2x– dx–1–

1

∫

3e6x 5x+( ) dx1

2

∫ 5x--- e

x2---+

dx1

4

∫ e2x e 2x––( ) dx3–

1–

∫

sin x dx0

π2---

∫ 3 sin 4x dxπ2---

π

∫ 5 sin x4--- dx

0

π

∫

2 sin x3--- dx–

π

2π

∫ cos 2x dxπ–

0

∫ 8 cos 4x dxπ–2

------

π2---

∫

3 cos x6--- dx

π

2π

∫ 7 cos x2--- dx–

2π–

0

∫ 4 cos 3x dx–π–

π2---–

∫

π 2 sin x4---+

dx0

π

∫ x2 3 6 sin 3x–+( ) dx1

3

∫ 1x--- 3 cos

x2---+

dx1

π

∫

f x( ) dx1

5

∫ 6= 3f x( ) dx1

5

∫


Given that ,

a is equal to:

b is equal to:

5 Evaluate the following.

6 If , find k.

7 If , find k.

8 If , find k.

9 If , find the value of a.

10 If , find the value of k given that 0 < k < .

A 16 B 10 C 11

D 19 E 22

A −6 B 5 C −5

D 6 E 0

a b

c d

e f

g h


f x( ) dx1

5

∫ 6=

f x( ) 1+[ ] dx1

5

∫

f x( ) dx5

1

∫

t2 4t–( ) dt0

3

∫ 2 cos 3t dt0

π2---

∫3

p 3–( )32---

------------------- d p4

7

∫ sin x5---–

dx5π

10π

∫

ex4---

cos 2x–( ) dx0

π

∫ 84m 3–---------------- dm

1

2

∫3–

t t-------- dt

1

4

∫ 3 sin 2x e 3x––( ) dx1–

1

∫

WORKEDExample

23

2x 3+( ) dx0

k

∫ 4=

3x2 dx0

k

∫ 8=

2x--- dx

1

k

∫ loge 9=

ex2--- dx

0

a

∫ 4=

cos 2x dxk

π

∫ 34

-------–= π2---


Signed areasWhen calculating areas between the graph of a function f(x) and the x-axis using the defi-

nite integral , the area is signed; that is, it is positive or negative. If f(x) > 0,

the region is above the x-axis; if f(x) < 0 it is below the axis. We shall now examine thesetwo situations and look at how we calculate the area of regions that include both.

Region above axisIf f(x) > 0, that is, the region is above the x-axis, then

, so the value of the definite integral is positive.

For example, if f(x) > 0, then the area = .

Region below axisIf f(x) < 0, that is, the region is below the x-axis, then

, so the value of the definite integral is negative.

For example, if f(x) < 0, then the area = − , as the

region is below the x-axis or area = (reversing the

terminals changes the sign).Therefore, for areas below the x-axis, the sign of the definite integral must be madenegative, or the terminals reversed, to ensure that the area has a positive value. (Areascannot be negative.)

Combining regionsFor regions which are combinations of areas above andbelow the x-axis, each area has to be calculated by separateintegrals — one for each area above and one for each areabelow the x-axis.

For example, from the diagram,Area = A1 + A2

However, , because the areas are signed.

To overcome this difficulty we find the correct area as:

Area = (= A1 − −A2 = A1 + A2)

or =

f x( ) dxa

b

∫

xb0 a

y y = f(x)

f x( ) dx 0>a

b

∫

f x( ) dxa

b

∫

x

y

b

0

a

y = f(x)f x( ) dx 0<

a

b

∫

f x( ) dxa

b

∫f x( ) dx

b

a

∫

xb0a c

y y = f(x)

A2

A1

f x( ) dxa

b

∫ A1 A2–=

f x( ) dx f x( ) dxa

c

∫–c

b

∫f x( ) dx f x( ) dx

a

c

∫+c

b

∫

C h a p t e r 9 I n t e g r a t i o n 379Note: When calculating the area between a curve and the x-axis it is essential that thex-intercepts are determined and a graph of the curve is sketched over the intervalrequired. The term |x| means make the value of x positive even if it is negative.

a Express the shaded area as a definite integral.b Evaluate the definite integral to find the shaded area, giving

your answer as an exact value.

THINK WRITE

a Express the area in definite integral notation where a = 1 and b = 4.

a Area =

b Antidifferentiate the integrand. b Area =

Evaluate. = [loge 4] – [loge 1]= loge 4 − 0= loge 4

State the solution as an exact answer. The area is loge 4.

0 1 4 x

yy = 1–x

1x--- dx

1

4

∫

1 logex[ ]1

4

2

3

24WORKEDExample

Calculate the shaded area.

THINK WRITE

Express the area in definite integral notation, with a negative sign in front of the integral as the region is below the x-axis.

Area =


Evaluate. =

=

=

=

=

= 7

State the solution. The area is 7 square units.

1 x2 4x–( ) dx

1

3

∫–

213---x3 2x2–[ ]

1

3–

3 13--- 3( )3 2 3( )2–[ ] 1

3--- 1( )3 2 1( )2–[ ]–{ }–

– 9 18–[ ] 13--- 2–[ ]–{ }

–9 123---–( )–[ ]–

–9 123---+[ ]–

713---–( )–

13---

4 13---

25WORKEDExample

x0 1 43

yy = x2 – 4x


To evaluate using

a graphics calculator, use the fnInt command found under 9:fnInt( for each part.

a Express the shaded area as a definite integral.b Calculate the area.

THINK WRITE

a Express the area above the x-axis as an integral and the area below the x-axis as an integral. For the area below the x-axis, take the negative of the integral from 0 to 2.

a Area =

b Antidifferentiate the integrands. b =

Evaluate. = {[ (0)4 − 2(0)2] − [ (−2)4 − 2(−2)2]}

− {[ (2)4 − 2(2)2] − [ (0)4 − 2(0)2]}

Simplify. = [0 − (4 − 8)] − [(4 − 8) − 0]= 4 − (−4)= 8


x3 4x–( ) dx x3 4x–( ) dx0

2

∫–2–

0

∫

114---x4 2x2–[ ]

2–0 1

4---x4 2x2–[ ]

02–

214--- 1

4---

14--- 1

4---

3

4

26WORKEDExample

x0–2 2

yy = x3 – 4x

x3 4x–( ) dx x3 4x–( ) dx0

2

∫–2–

0

∫MATH

a Sketch the graph of y = ex − 2 showing all intercepts and using exact values for all keyfeatures.

b Find the area between the curve and the x-axis from x = 0 to x = 2.

THINK WRITE

a Find the x-intercept by letting y = 0 and solving for x.Take loge of both sides.

a When y = 0, ex – 2 = 0ex = 2

loge ex = loge2x = loge 2 (or approximately 0.69)

so the x-intercept is loge 2.

1

27WORKEDExample


THINK WRITE

Find the y-intercept by letting x = 0.

When x = 0, y = e0 − 2= 1 − 2= −1

so the y-intercept is −1.Note the vertical translation and hence sketch the graph showing the appropriate horizontal asymptote and intercept.Shade the region required.

b Express the area above the x-axis as an integral and the area below the x-axis as an integral. Subtract the area below the x-axis from the area above the x-axis.

b Area = −

Antidifferentiate the integrands. =

Evaluate.(Remember: eloge a = a)

= [e2 − 2(2)] − [elog 2 − 2 loge 2]

− {[eloge 2 − 2(loge 2)] − [e0 − 2(0)]}

= [e2 − 4] − [2 − 2 loge 2] − {[2 − 2 loge 2] − [1 − 0]}

Simplify. = e2 − 4 − 2 + 2 loge 2 − 2 + 2 loge 2 + 1

= e2 − 7 + 4 loge 2State the solution. The area is (e2 − 7 + 4 loge 2) or approximately

3.162 square units.

2

3

x0

–2–2

–1 2loge2

y y = ex – 2

y = – 2 4

1 ex 2–( ) dxe 2log

2

∫ ex 2–( ) dx0

e 2log

∫

2 ex 2x–[ ]loge2

2 ex 2x–[ ]0loge2–

3

4

5

remember1. If f(x) > 0, area = .

2. If f(x) < 0, area = − , as the region is below the x-axis or

= , (reversing the terminals changes the sign).

3. If the required area lies above and below the x-axis:(a) find the intercepts and sketch the graph(b) calculate the integrals and subtract the area below

the x-axis from the area above the x-axis.

4. Area = (= A1 − −A2 = A1 + A2)

or =

5.

f x( ) dxa

b

∫f x( ) dx

a

b

∫f x( ) dx

b

a

∫


c

∫–c

b

∫f x( ) dx f x( ) dx

a

c

∫+c

b

∫elogea a=

remember

xb0a c

y y = f(x)

A2

A1


Signed areas

1 Find the area of the triangle at right.a geometricallyb using integration

2 Find the area of the triangle at right.a geometricallyb using integration

3 Express the following shaded areas as definite integrals.

a

e

i

4 Evaluate each of the definite integrals in question 3 to find the shaded area. Give youranswer as an exact value.

5 For each of the following, sketch a graph to illustrate the region for which the definiteintegral gives the area.

a b c

d e f

g h

9F

x0 4

yy = x

x0 3

3

y

y = 3 – x

WORKEDExample

24a

0 31x

y = 2xy

4

4

y = 4 – x0 x

y

x0 1 2

yy = x2

–3 –1

y = 3x2

x0

yb c d

f g h

x0 31

yy = x3 – 9x2 + 20x

x0–2

y = –x3 – 4x2 – 4x

y

x0–1 1

1

yy = ex

x0 41

y

y = e–2x

j

x0

2y y = 2 sin 2x

–2π 0

1

x

y

3 —2π

y = cos –3x

WORKEDExample

24b

Mathca

d

Definite integrals 4x dx

0

3

∫ 6 x–( ) dx1

2

∫ x2 dx1

3

∫4 x2–( ) dx

1–

1

∫ x1

4

∫ dx 2ex dx3–

0

∫

loge x2--- dx

2

4

∫ 3 sin 2x dx0

π2---

∫

C h a p t e r 9 I n t e g r a t i o n 3836 Calculate each of the shaded areas below.

7

a The area between the graph, the x-axis and the lines x = −2 and x = −1 is equal to:

b The area between the graph, the x-axis and the lines x = −2 and x = 3 is equal to:

8 Express the following shaded areas as definite integrals which give the correct area.

A B C

D E −

A B C

D E

WORKEDExample

25b c da

x0

y

y = x – 2

–2

2

x0

yy = –4 – 2x

–2 –1

x0

y y = x2 – 4

2x0

y

y = 1– x2

–2 –1

f g he

x0

y y = x3

–2x0

y

y = x3 + 2x2 – x –2

–1 1

x0

y

y = –ex

–1 1 x0

y

y = –e–2x

1–2 1

ji

x0

y

y = sin x

2π π—2π3

y = 2 cos –2x

x0

y

π–2 π–


x0

y

y = f(x)

3–2f x( ) dx

1

2

∫ f x( ) dx2–

1–

∫ f x( ) dx1–

0

∫f x( ) dx

2

1

∫ f x( ) dx2–

1–

∫

f x( ) dx f x( ) dx0

2–

∫+0

3

∫ f x( ) dx2–

3

∫ f x( ) dx f x( ) dx1

3

∫+2–

1

∫f x( ) dx

3

2–

∫ f x( ) dx – f x( ) dx0

3

∫2–

0

∫WORKEDExample

26a

x0

y g(x)

31–3x0

y

h(x)2–1–3

x0

y

f(x)

52

b

x0

y

f(x)–5 –2–4

e

x0

y g(x)

–3 –2 2 3

d

a c


9Examine the graph.

a The area between the curve and the x-axis from x = −2 and x = 1 is equal to:

b The area between the curve and the x-axis from x = 1 and x = 3 is equal to:

c The area between the curve and the x-axis from x = −2 and x = 3 is equal to:

10 Sketch the graph of the curve y = x2 − 4, showing all intercepts and using exact valuesfor all key features. Find the area between the curve and the x-axis:a from x = 0 to x = 2b from x = 2 to x = 4c from x = 0 to x = 4.

11 Sketch the graph of the curve y = x3 + x2 − 2x, showing all intercepts. Find the areabetween the curve and the x-axis between the lines:a x = −2 and x = 0b x = 0 and x = 1c x = −2 and x = 1.

12 Sketch the graph of the curve y = 1 + 3 cos 2x over [0, π]. Find the exact area betweenthe curve and the x-axis from:

a x = 0 to x =

b x = to x = π.

13 Sketch the graph of f(x) = and find the area between the curve and the x-axisand the lines x = 2 and x = 3. Give both an exact answer and an approximation to 3decimal places.

14 Find the exact area between the curve , the x-axis and the lines x = and x = 2.

15 Find the exact area bounded by the curve g(x) = ex + 2, the x-axis and the linesx = −1 and x = 3.

A 17 sq. units B 15 sq. units C −17 sq. units

D −15 sq. units E 10 sq. units

A −6 sq. units B 2 sq. units C −5 sq. units

D 5 sq. units E 6 sq. units

A 10 sq. units B 11 sq. units C 22 sq. units



x0

y

y = x3 – 2x2 – 5x + 6

–2 1 3

112------ 3

4--- 1

12------

34---

23--- 1

3---

13--- 2

3---

512------ 3

4--- 5

12------

112------

WORKEDExample

27

π4---

3π4

------

x 1–

y1x---= 1

2---

WorkS

HEET 9.2


Further areasAreas bound by a curve and the x-axisFor graphs with two or more x-intercepts, there is an enclosed region (or regions) between the graph and the x-axis.

The area bound by the graph of f(x) and the x-axis is:

(negative because the area is below the x-axis).

The area bound by the graph of g(x) and the x-axis is:

That is, if the graph has two x-intercepts then one integrand is required.

If the graph has three x-intercepts then two integrands are required, and so on.Note: Wherever possible it is good practice to use sketch graphs to assist in anyproblems involving the calculation of areas under curves.

x0

yy = f(x)

a b

f x( ) dxa

b

∫–

x0

y

y = g(x)

a c b

g x( ) dx g x( ) dxa

c

∫–c

b

∫

a Sketch the graph of the function g(x) = (3 − x)(2 + x).b Find the area bound by the x-axis and the graph of the function.

THINK WRITEa Determine the type of graph by

looking at the number of brackets and the sign of the x-terms.

a g(x) = (3 − x)(2 + x) is an inverted parabola.

Solve g(x) = 0 to find thex-intercepts.

For x-intercepts, g(x) = 0(3 − x)(2 + x) = 0x = 3 and x = −2.The x-intercepts are −2 and 3.

Sketch the graph.

b Shade the region bound by g(x) and the x-axis.

b

Express the area as an integral. Area =

Evaluate. =

= [6(3) + (3)2 − (3)3]

− [6(−2) + (−2)2 − (−2)3]

= (18 + − 9) − (−12 + 2 + )

= 13 − (−7 )

= 13 + 7

= 20State the solution. The area bound by g(x) and the x-axis is 20

square units.

1

2

3

1

2 6 x x2–+( ) dx2–

3

∫3 6x 1

2---x2 1

3---x3–+[ ]

2–3

12--- 1

3---

12--- 1

3---

92--- 8

3---

12--- 1

3---

12--- 1

3---

56---

4 56---

28WORKEDExample

x0

y

y = g(x)

–2 3


The area bound by the graph and the x-axis can bedrawn and evaluated using the graphics calculator.1. Enter the function as Y1 in the Y= menu.2. Press and set Xmin to −4.7 and Xmax

to 4.7. These values ensure the cursor reachesexact integers when tracing.

3. Press 7: f(x)dx.4. Press to the left x-intercept (−2), press , and press to the right

x-intercept (3) and press .Note: This method can be used even if the area required is between non-intercepts.

Finding areas without sketch graphsWhen finding areas under curves which involve functions whose graphs are not easilysketched, the area can be calculated providing the x-intercepts can be determined.Note: The symbol |f(x)| is known as the absolute value of f(x) which means that wemust make the f(x) function or value positive whenever it is negative. For example,

|−5| = 5|3| = 3

|− | =

As areas cannot be negative, taking the absolute value of the integrands involved in aproblem will ensure that all areas are made positive. For example,

The shaded area at right =

or =

Graphics CalculatorGraphics Calculator tip!tip! Finding the area bound by a graphand the x-axis

WINDOW

2nd [CALC] ∫�

ENTER �

ENTER

23--- 2

3---

x0

y

y = f(x)

a c b

f x( ) dxc

b

∫ f x( ) dxa

c

∫+


c

∫+c

b

∫

a Find the x-intercepts of y = sin 2x over the domain [0, 2π].b Calculate the area between the curve, the x-axis and x = 0 and x = π.

THINK WRITEa To find the x-intercepts, let y = 0. a For x-intercepts, y = 0

Solve for x over the given domain. sin 2x = 02x = 0, π, 2π, 3π, 4π, etc.

b Pick the x-intercepts that are between the given end points of the area.

b x = is the only x-intercept between 0 and π.

State the regions for which it is necessary to calculate the area.

Area =

12

x 0π2--- π 3π

2------ 2π, , , ,=

1π2---

2 sin 2x dx0

π2---

∫ sin 2x dxπ2---

π

∫+

29WORKEDExample


THINK WRITE

Evaluate the absolute value of the integral for each region.

=

=

+

=

=

Add the result to give the total area. = 1 + 1= 2


312--- cos 2x–[ ]

0

π2---

12--- cos 2x–[ ]π

2---

π+

12--- cos π 1

2--- cos 0–( )––[ ]

–12--- cos 2π 1

2--- cos π–( )–[ ]

12--- 1

2---–( )–[ ] –1

2--- 1

2---( )–[ ]+

1 1–+4

5

a Differentiate loge (x2 − 1).

b Hence, find an antiderivative of .

c Find the area between the graph of , the x-axis, x = 2 and x = 3, giving your

answer correct to 2 decimal places.

Continued over page

THINK WRITE

a Let y equal the expression to be differentiated.

a Let y = loge (x2 − 1)

Express u as a function of x in order to apply the chain rule for differentiation.(Let u equal the function inside the brackets.)

Let u = x2 − 1

Find .

Write y in terms of u. y = loge u

Find .

Find using the chain rule. So =

=

xx2 1–--------------

xx2 1–--------------

1

2

3dudx------ du

dx------ 2x=

4

5dydu------ dy

du------ 1

u---=

6dydx------ dy

dx------ 1

u--- 2x×

2xx2 1–--------------

30WORKEDExample


THINK WRITE

b Since , express the

relationship in integral notation.

b

Remove a factor from so that it

resembles the integral required.

Divide both sides by the factor in order to obtain the required integral.

An antiderivative of is log (x2 − 1).

c Find the x-intercepts.

(For = 0, the numerator = 0.)

c For x-intercepts, = 0

x = 0

If the x-intercepts are not between the terminals of the area, find the area by evaluating the integrand.

Area =

=

=

=

=

=

State the solution. The area is or approximately 0.49

square units.

1dydx------ dx∫ y c+= 2x

x2 1–-------------- dx∫ loge x2 1–( ) c+=

2dydx------ 2

xx2 1–-------------- dx∫ loge x2 1–( ) c+=

3x

x2 1–-------------- dx∫ 1

2--- log x2 1–( ) c+=

x

x2 1–

-------------- 12---

1

xx2 1–--------------

xx2 1–--------------

2x

x2 1–-------------- dx

2

3

∫12--- loge x2 1–( )[ ]

23

12--- loge 32 1–( )[ ] 1

2--- loge 22 1–( )[ ]–

12--- loge 8 1

2--- loge 3–

12--- loge

83---

12--- loge

83---

312--- loge

83---

remember1. For graphs with two or more intercepts, there is an enclosed region (or regions)

between the graph and the x-axis.2. The number of regions is one less than the number of intercepts.3. Where possible, sketch graphs to make it easier to calculate the areas under

curves, or use a graphics calculator.4. As areas can’t be negative, take the absolute values of the integrals.5. When graphs are not easily drawn, areas can be calculated by finding the

x-intercepts and determining whether they are within the bounds of the required area.

remember


Further areas

In the following exercise give all answers correct to 2 decimal places where appro-priate, unless otherwise stated.

1 i Sketch the graph of each of the following functions.

ii Find the area bound by the x-axis and the graph of each function.Use a graphics calculator to assist.

2 Find the area bound by the x-axis and the graph of each of the following functions.

3

The area bound by the curve with equation y = x2 − 6x + 8 and the x-axis is equal to:

4

The area between the curve at right, the x-axis and the linesx = −3 and x = 4 is equal to:

5

The area between the curve y = x2 − x − 6, the x-axis and the lines x = 2 and x = 4 isequal to:

a f(x) = x2 − 3x b g(x) = (2 − x)(4 + x)

a h(x) = (x + 3)(5 − x) b h(x) = x2 + 5x − 6

c g(x) = 8 − x2 d g(x) = x3 − 4x2

e f(x) = x(x − 2)(x − 3) f f(x) = x3 − 4x2 − 4x + 16

g g(x) = x3 + 3x2 − x − 3 h h(x) = (x − 1)(x + 2)(x + 5)


D 3 sq. units E −1 sq. units

A B

C D

E

A 2 sq. units B sq. units C 5 sq. units


9G

WORKEDExample

28

Mathcad

Definiteintegrals


13--- 2

3---

13---


x0

y

y = f(x)

42–3f x( ) dx3–

4

∫ f x( ) dx f x( ) dx2

4

∫+3–

2

∫

f x( ) dx4

3–

∫ f x( ) dx f x( ) dx3–

2

∫–2

4

∫

f x( ) dx f x( ) dx2

4

∫–3–

2

∫


56--- 2

3---

34--- 1

2---


6 For each of the following:

i sketch the graph of the curve over an appropriate domain, clearly labelling anyx-intercepts in the interval required (use a graphics calculator to assist).

ii find the area between the curve, the x-axis and the lines indicated below.

7 For each of the following functions:

i find the x-intercepts over the given domain

ii calculate the area between the curve, the x-axis and the given lines.Use sketch graphs to assist your workings.

a y = x − 4x−1, x ≠ 0, x = 1 and x = 3

b y = sin x − cos x, x ∈ [0, π], x = 0 and x = π

c y = ex − e, x = 0 and x = 3

d y = x − , x ≠ 0, x = and x = 2

e y = , x = −2 and x = 2

f y = x4 − 3x2 − 4, x = 1 and x = 4

g y = (x − 2)4, x = 1 and x = 3

8 Find the exact area of the region enclosed by the x-axis, y = e3x and the lines x = 1 andx = 2.

9 Find the exact area of the region enclosed by the x-axis, y = −cos x and the lines x = and x = . (Use a sketch graph to assist your calculation.)

10 Find the area bound by y = (x − 1)3, the x-axis and the y-axis.

11 a Sketch the graph of showing all asymptotes and intercepts.

b Find the area under the curve between x = −1 and x = 1.

12 a Give the equation of the asymptotes for the function f(x) = (x + 2)−3.

b Find the area between the curve, the x-axis, x = −1 and x = 1.

a y = 3 − 3x2, x = 0 and x = 2 b y = , x = 1 and x = 3

c y = , x = 1 and x = 2 d y = x3 − 4x, x = −2 and x = 1

e y = e2x, x = −2 and x = 0 f y = e−x, x = 0 and x = 2

g y = 2 sin x, x = and x = h y = cos , x = and x = 2π

i y = sin 3x, x = – and x = – j y = , x ≥ 0, x = 0 and x = 4

2x---

1x2-----–

π6--- π

3--- x

2--- π

2---

π2--- π

6--- x x

WORKEDExample

29

1x2----- 1

2---

ex2---

π3---

5π6

------

y1

x 3–( )2-------------------=

C h a p t e r 9 I n t e g r a t i o n 39113 Find the area bound by the curve y = 3 − e2x, the x-axis, x = −2 and x = 0.

(Find the x-intercepts first.)

14 Find the area bound by the curve y = 4 − sin 2x, the x-axis, x = and x = π.(Check the x-intercepts first.)

15 a Differentiate x loge x. (x > 0)

b Hence, find an antiderivative of loge x.

c Find the area bound by the graph of loge x, the x-axis, x = 1 and x = 4 giving exactanswers.

16 a Differentiate loge (x2 + 2).

b Hence, find an antiderivative of .

c Find the area between , the x-axis, x = −1 and x = 1.

17 a Find the area between the graph of y = x2, the x-axis, x = 0 and x = 2.

b Use this result to calculate the area between the graph, the y-axis and the liney = 4.

18 Find the exact area of the shaded region on the graph y = e2x below.

19 Find the shaded area below.

π2---–

WORKEDExample

30

xx2 2+--------------

xx2 2+--------------

x0

y y = x2

2

(2, 4)

x0

y

y = e2x

2

x0

y

y = 2 sin x

–2π–

2π–


Areas between two curvesWe shall now consider the area between two functions, f(x) and g(x), over an interval[a, b]. Our approach depends on whether the curves intersect or do not intersect overthis interval.

If the two curves f(x) and g(x) do not intersect over the interval [a, b]Here, we may look at three circumstances: when the region is above the x-axis, when itis below the x-axis, and when it crosses the x-axis.

Region above x-axis

The red shaded area =

= .

Note: The lower function is subtracted from the higher function to ensure a positive answer.

Region below x-axisAgain, the lower function is subtracted from the higherfunction to ensure a positive answer.

Red shaded area = , as f(x) is above g(x)

over the interval [a, b].

Region crosses x-axis

Shaded area =

x0

y

g(x)

f(x)

a b

f x( ) dx g x( ) dxa

b

∫–a

b

∫f x( ) g x( )–[ ] dx

a

b

∫

x0

y

g(x)f(x)

a b

f x( ) g x( )–[ ] dxa

b

∫

x

y

0a b

f(x)g(x)

f x( ) g x( )–[ ] dxa

b

∫

a State the definite integral which describes the shaded area on the graph at right.

b Find the area.

THINK WRITEa State the two functions f(x) and g(x). a f(x) = 2x + 1 and g(x) = x

Subtract the equation of the lower function from the equation of the upper function and simplify.

f(x) − g(x) = 2x + 1 − x= x + 1

Write as a definite integral between the given values of x.

Area =

Area =

12

3 f x( ) g x( )–[ ]0

2

∫x 1+( ) dx

0

2

∫

31WORKEDExample

0 x

y

y = x

y = 2x + 1

2


THINK WRITE

b Antidifferentiate. b Area =

Evaluate the integral. Area =

Area = (2 + 2) − (0)Area = 4


112---x2 x+[ ]

02

212--- 2( )2 2+[ ] 1

2--- 0( )2 0+[ ]–

3

a Find the values of x where the functions y = x and y = x2 − 2 intersect.b Sketch the graphs on the same axes. (Check using a graphics

calculator.)c Hence, find the area bound by the curves.

Continued over page

THINK WRITEa State the two functions. a y = x and y = x2 − 2

Find where the curves intersect.

For points of intersection:x = x2 − 2

Solve for x. x2 − x − 2 = 0(x − 2)(x + 1) = 0

x = 2 or x = −1

b Find the key points of each function and sketch.

b For y = x,when x = 0, y = 0when x = 2, y = 2when x = −1, y = −1Line passes through (0, 0), (2, 2) and (−1, −1)For y = x2 − 2,when x = 0, y = −2Hence y-intercept is −2.Parabola also passes through (2, 2) and (−1, −1).

c Define f(x) and g(x). c Let f(x) = x and g(x) = x2 − 2

Write the area as a definite integral between the values of x at the points of intersection.

Area

= dx

=

=

12

3

1

2

f x( ) g x( )–[ ]1–

2

∫x x

2 2–( )–[ ] dx1–

2

∫x x2– 2+( ) dx

1–

2

∫

32WORKEDExample

x0

(2, 2)

(–1, –1)

y

y = x

y = x2 – 2


If the two curves intersect over the interval [a, b]Where c1 and c2 are the values of x where f(x) and g(x) inter-sect over the interval [a, b], the area is found by consideringthe intervals [a, c1], [c1, c2] and [c2, b] separately. For eachinterval care must be taken to make sure the integrand is thehigher function. Subtract the lower function.

So the shaded area equals:

+ +

Therefore, when finding areas between two curves over an interval, it must be deter-mined whether the curves intersect within that interval. If they do, the area is brokeninto sub-intervals as shown above.

As with areas under curves, sketch graphs should be used to assist in finding areasbetween curves.

If sketch graphs are not used, the absolute value of each integral, for each sub-interval, should be taken to ensure the correct value is obtained.

THINK WRITE

Antidifferentiate. =


= (2 − + 4) − ( + − 2)

= (3 ) − (–1 )

= 3 + 1

= 4


312---x2 1

3---x3– 2x+[ ]

1–2

412--- 2( )2 1

3--- 2( )3– 2 2( )+[ ] 1

2--- 1–( )2 1

3--- 1–( )3– 2 1–( )+[ ]–

83--- 1

2--- 1

3---

13--- 1

6---

13--- 1

6---

12---

512---

x0

y

a bc1 c2

f(x)

g(x)

g x( ) f x( )–[ ] dxa

c1∫ f x( ) g x( )–[ ] dxc1

c2

∫ g x( ) f x( )–[ ] dxc2

b

∫

a Find the values of x where the graph of the functions f(x) = and g(x) = x intersect.

b Sketch the graphs on the same axes (check using a graphics calculator) and shade theregion between the two curves and x = 1 and x = 3.

c Find the area between f(x) and g(x) from x = 1 to x = 3.

THINK WRITE

a State the two functions. a f(x) = , g(x) = x

Let f(x) = g(x) to find the values of x where the graphs intersect.

For points of intersection, x =

4x---

14x---

24x---

33WORKEDExample


Note: If worked example 33 was calculated without a graph, the area would be foundby evaluating:

+ or +

as without a graph it would not be known which function was above the other for eitherinterval.To find the area between two curves using a graphics calculator:1. Find the area between the upper curve and the x-axis, and record or store the answer.2. Find the area between the lower curve and the x-axis, and record or store the answer.3. Subtract the two answers, giving a positive value for the area.

THINK WRITE

Solve for x. x2 = 4

x2 − 4 = 0

(x − 2)(x + 2) = 0

x = −2 and x = 2

b Sketch f(x) and g(x) on the same axes and shade the region between the two curves from x = 1 to x = 3.

b

c State the area as the sum of two integrals for the two sub-intervals.

c Area =

Antidifferentiate. =

Evaluate the two integrals. =

+

=

+

=

Add the two values. = 4 loge + 1

State the area. The area is 4 loge + 1 or approximately 2.151

square units.

3

g(x) = x

x0

y

1 2 3

4 –xf(x) =

14x--- x–

dx x4x---–

dx2

3

∫+1

2

∫

2 4 logex 12---x2–[ ]

12 1

2---x2 4 logex–[ ]

23+

3 4 loge2 12--- 2( )2–[ ] 4 loge1 1

2--- 1( )2–[ ]–

12--- 3( )2 4 loge3–[ ] 1

2--- 2( )2 4 loge2–[ ]–{ }

4 loge2 2–[ ] 4 loge1 12---–[ ]–

92--- 4 loge3–[ ] 2 4 loge2–[ ]–{ }

4 loge2 2– 0– 12--- 9

2--- 4loge3– 2– 4loge2+ + +

443---

543---

4x--- x–

dx1

2

∫ 4x--- x–

dx2

3

∫ x4x---–

dx1

2

∫ x4x---–

dx2

3

∫


Areas between two curves

1 State the definite integral which will find the shaded areas on each graph below.

2 Find each of the areas in question 1.

3

Which one of the following does not equal the shaded area?

A B

C D

E

remember1. If two curves f(x) and g(x) do not intersect over the interval [a, b] and

f(x) > g(x) then the area enclosed by the two curves and the lines x = a and x = b is found by using the formula:

2. If two curves f(x) and g(x) intersect over the interval [a, b] it is necessary to find the points of intersection and hence find the area of each section because sometimes f(x) > g(x) and sometimes g(x) > f( x).

3. If sketch graphs are not used to determine which is the upper curve, then it is necessary to take the absolute value or positive value of each integral.

f x( ) dx g x( ) dxa

b

∫–a

b

∫ f x( ) g x( )–[ ] dxa

b

∫=

remember

9HWORKEDExample

31a

x0

yy = x2

y = x

1

a b c d

x0

yy = 2x

1 2

y = x + 1 x0

yy = 3x

y = 4 – x2 1

x0

y

y = 8 – x2

y = x2

2–2

e f g h

0 x

y y = x3 y = 3x

3x0

y y = ex

y = 9 – x2

–1 1

x0

y

y = – ex

y = x

1 2 x0

y

y = –4

y = x2 – 5

1–1

WORKEDExample

31b


x0

y

g(x)

f(x)

51

g x( ) dx f x( ) dx1

5

∫–1

5

∫ g x( ) dx f x( ) dx5

1

∫+1

5

∫

f x( ) dx g x( ) dx1

5

∫–1

5

∫ g x( ) f x( )–[ ] dx1

5

∫

f x( ) g x( )–[ ] dx5

1

∫


The area bound by the curves f(x), g(x) and the lines x = −3 and x = 1 at right is equal to:

5

The shaded area at right is equal to:

A

B

6 In each of the following:i find the values of x where the functions intersectii sketch the graphs on the same axes (check using a graphics calculator)iii hence, find the area bound by the curves.

7 i Find the values of x where the functions intersect.ii Sketch the graphs on the same axes. (Check using a graphics calculator.)iii Find the area between f(x) and g(x) giving an exact answer.

8 Find the area between the pairs of curves below, over the given interval. (Use a graphand graphics calculator to assist if necessary.)a y = x3, y = x2, x ∈ [−1, 1] b y = sin x, y = cos x, x ∈ [0, π]c y = (x − 1)2, y = (x + 1)2, x ∈ [−1, 1] d y = x3 − 5x, y = 6 − 2x2, x ∈ [0, 3]

e y = , y = 4x, x ∈ [ , 1] f y = ex, y = e−x, x ∈ [0, 1]

g y = 2 cos x, y = x − , x ∈ [0, ] h y = ex, y = −ex, x ∈ [−2, 1]

Use sketch graphs and a graphics calculator to assist in solving the following problems.

9 Find the area between the curve y = ex and the lines y = x, x = 1 and x = 3.

10 Find the area between the curve y = x2 and the lines y = + 3, x = 1 and x = 3.

A B

C D

E

C dx D

E

a y = 4x and y = x2 b y = 2x and y = 3 − x2

c y = x2 − 1 and y = 1 − x2 d y = x2 − 4 and y = 4 − x2

e y = (x + 1)2 and y = 1 − x2 f y = and y = x2

a y = x3 and y = x b y = 3x2 and y = x3 + 2x


x0

y

–1–3–4

f(x)

g(x)

f x( ) g x( )–[ ] dx1–

3–

∫ f x( ) g x( )+[ ] dx3–

1–

∫g x( ) f x( )–[ ] dx

3–

1–

∫ f x( ) g x( )–[ ] dx3–

1–

∫f x( ) g x( )+[ ] dx

1–

3–

∫


x0

yg(x)

f(x)

3 4

f x( ) g x( )–[ ] dx0

4

∫g x( ) f x( )–[ ] dx f x( ) g x( )–[ ] dx

3

4

∫+0

3

∫g x( ) f x( )–[ ]

0

4

∫ f x( ) g x( )–[ ] dx0

3

∫f x( ) g x( )–[ ] dx g x( ) f x( )–[ ]

3

4

∫+0

3

∫WORKEDExample

32

x

WORKEDExample

33

Mathcad

Areabetween

twocurves1

x--- 1

4---

π2--- π

2---

x2---


11 Calculate the area between the curves y = sin 2x and y = cos x from x = 0 to x = .

12 Calculate the area between the curves y = − sin 2x and y = sin 2x from x = 0 to x = .

13 Find the exact area bound by the curves y = ex and y = 3 − 2e−x.

14 The graph at right shows the cross-section of a brickedarchway. (All measurements are in metres.)a Find the x-intercepts of f(x).b Find the x-intercepts of g(x).c Find the cross-sectional area of the brickwork.

15 The diagram at right shows theoutline of a window frame. If allmeasurements are in metres,what is the area of glass whichfits into the frame?

16 The diagram at right shows the side view of a concretebridge. (All measurements are in metres.) Find:a the x-intercepts of the curveb the length of the bridgec the area of the side of the bridged the volume of concrete used to build the bridge if the bridge is 9 metres wide.

17 The cross-section of a road tunnel entrance is shown atright. (All measurements are in metres.) The shaded area isto be concreted. Find:a the exact area, above the entrance, which is to be con-

cretedb the exact volume of concrete required to build this

tunnel if it is 200 metres long.

18 A section of a river can be modelled by the equation y = 40 sin , where x ∈ [0, 120]

and x is in metres. On the same model a proposed section of road obeys

the rule y = . The area bound by the road and the river is to have

one tree planted per 12 square metres. How many trees will be planted?

π2---

3π4---

x0

y

g(x) = 3 – x2 1–3

f(x) = 4 – x21–4

0 x

y

1–2

y = 2 – 2x2

y = 2x2

1–2

1–2–

x0

y

y = 4 – x2

—–100

5

x0

y

f(x) = 5 sin ––30πx

πx120---------

x5---


Further applications of integration

Further applications of integration

1 If f ′(x) = (2 − x)2 and the y-intercept of f(x) is , find the rule for f(x).

2 If and the y-intercept is 2, find the exact value of y when x = .

3 The rate of deflection from a hori-zontal position of a 3-metre divingboard when an 80-kg person is xmetres from its fixed end is given by

, where

y is the deflection in metres.a What is the deflection when x = 0?b Determine the equation which measures the deflection.c Hence, find the maximum deflection.

4 On any day the cost per item for a machine producing n items is given by

, where n ∈ [0, 200] and C is the cost in dollars.

a Use the rate to find the cost of producing the 100th item.b Express C as a function of n.c What is the total cost of producing the first 100 items?d Find the average cost of production for: i the first 100 items ii the second 100 items.

5 The rate of change of position (velocity) of a racing car travelling down a straight stretch

of road is given by , where x is measured in metres and t in seconds.a Find the velocity when: i t = 0 ii t = 4.b Determine: i when the maximum velocity occurs ii the maximum velocity.

c Sketch the graph of against t for 0 ≤ t ≤ 16.

d Find the area under the graph between t = 0 and t = 10.e What does this area represent?

9I43---

dydx------ 1 4 cos 2x–= π

12------

0 x

y

DeflectionBoard

(Metres)

(Metres)

dydx------ 0.03 x 1+( )2– 0.03+=

dCdn------- 40 2e0.01n–=

dxdt------ t 16 t–( )=

dxdt------


6 The rate at which water is pumped out of a dam, in L/min, t minutes after the pump is

started is .

a How much water is pumped out in the 40th minute?b Find the volume of water pumped out at any time, t, after the pump is started.c How much water is pumped out after 40 minutes?d Find the average rate at which water is pumped in the first hour.e How long would it take to fill a tank holding 1600 litres?

7 The rate of flow of water into a hot water system during a 12-hour period on a certain day is

thought to be , where V is in litres and t is the number of hours after 8 am.

a Sketch the graph of against t.

b Find the length of time for which the rate is above 10.5 L/h.c Find the volume of water that has flowed into the system between:

i 8 am and 2 pm ii 3 pm and 8 pm.

8 The roof of a stadium has the shape given by the function f: [−25, 25] → R, f(x) = 20 − 0.024x2.

The stadium is 75 metres long and its cross-section isshown at right.a Find the volume of the stadium.b The stadium is to have several airconditioners strategically placed around it. Each

airconditioner can service a volume of 11250 m3. How many airconditioners are required?

9 The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metresdeep. Find the depth of water, to the nearest cm, when the channel is half full.

10 For any point P on the curve y = x3, prove that the area under the curve is one quarterof the area of the rectangle.

11 The arch of a concrete bridge has the shape ofa parabola. It is 6 metres high and 8 metres long.

a Find the rule for the function corre-sponding to the arch of the bridge.

b Find the area of the shaded region.c If the bridge is 10 metres wide, find

the volume of concrete in the bridge.

dVdt------- 5 cos

πt40------+=

dVdt------- 10 cos

πt2-----+=

dVdt-------

x0

y (metres)

(metres)

25–25

5

20

x0

yy = x3

P

x0

y (metres)

(metres)

4 5

6

–4

7

C h a p t e r 9 I n t e g r a t i o n 40112 In the figure at right f(x) and g(x) intersect at O and B.

a Show that the coordinate of B is (loge 2, 1).b Find the exact area of the region bound by f(x)

and g(x).c Show that the sum of the areas under f(x) and g(x),

from x = 0 to x = loge 2, is equal to the area of therectangle OABC.

13 The population of kangaroos on an island is increasing at a rate given byP(t) = 12 loge (t + 1), where t is the number of years since 1 January 1955.a Find the rate of growth when t = 0, t = 5 and t = 40.b Sketch the graph of P(t).c Determine the inverse function P−1(t).d Use the inverse function to assist in finding the area under the graph of P(t)

between t = 0 and t = 40.e What does this area represent?

Concrete chute

Concrete is poured from a mixer down a chute which has a cross-sectional shape as shown.The curved bottom has the shape given by the function f:[−10, 10] → R, f(x) = 0.12x2 − 12.a Find the area of the cross-section.Concrete can flow down the chute at 1.6 m/s.b What volume of concrete can be poured in one

minute?c How long does it take to empty a mixer holding 12 m3 of concrete?

xO

y f(x) = ex – 1

g(x) = –2e–x + 2B

C

A

x0

y (cm)

(cm)

10–10

–12

5


Antidifferentiation rules

• The relationships between f(x) and are:

f(x)

a ax + c

axn

(ax + b)n

loge x + c

log (ax + b) + c

ex ex + c

ekx

sin ax − cos ax + c

cos ax sin ax + c

• •

• , where g(x) = f ′(x) dx • is the indefinite integral

Definite integrals• The fundamental theorem of integral calculus:

= = F(b) − F(a) where F(x) is an antiderivative of f(x).

• is the definite integral

• , a < c < b

•

•

•

summaryf∫ x( ) dx

f∫ x( ) dx

axn 1+

n 1+--------------- c+

ax b+( )n 1+

a n 1+( )----------------------------- c+

1x---

1ax b+--------------- 1

a---

1k---ekx c+

1a---

1a---

f x( ) g x( )±[ ] dx∫ f x( ) dx g∫ x( ) dx±∫= kf x( ) dx∫ k f x( ) dx∫=

g x( )dx∫ f x( ) c+= f x( ) dx∫

f x( ) dxa

b

∫ F x( )[ ]ab

f x( ) dxa

b

∫f x( ) dx

a

b


b

∫+a

c

∫=

kf x( ) dxa

b

∫ k f x( ) dxa

b

∫=

f x( ) g x( )±[ ] dxa

b


b

∫±a

b

∫=

f x( ) dxa

b

∫ f x( ) dxb

a

∫–=

C h a p t e r 9 I n t e g r a t i o n 403Approximating areas under curves• An approximation to the area between a curve and the x-axis can be found by

dividing the area into a series of rectangles or trapeziums which are all the same width. The approximation is found by finding the sum of all the areas of the rectangles or trapeziums.

• Lower rectangle approximation ≤ actual area ≤ upper rectangle approximation• Trapezoidal approximation =

• Trapezoidal rule is

where h is the interval width.

Area under curves

• Area = , if f(x) > 0 for x ∈ [a, b]

• Area = − , if f(x) < 0 for x ∈ [a, b]

• Area =

= , if f(x) > 0 for x ∈ [c, b]

and f(x) < 0 for x ∈ [a, c]

Area between curves

• Area = , if f(x) > g(x) for x ∈ [a, b]

• Area = ,

if g(x) > f (x) for x ∈ [a, c]and f (x) > g(x) for x ∈ [c, b]

lower rectangle approximation + upper rectangle approximation2

---------------------------------------------------------------------------------------------------------------------------------------------------------

f x( ) dxh2--- f a( ) 2f a h+( ) 2f a 2h+( ) … 2f b h–( ) f b( )+ + + + +[ ]≅

a

b

∫

f x( ) dxa

b

∫

xb0 a

y y = f(x)

f x( ) dxa

b

∫x

y

b

0

a

y = f(x)


c

∫–c

b

∫

xb0a c

y y = f(x)

A2

A1f x( ) dx f x( ) dx

a

c

∫+c

b

∫

f x( ) g x( )–[ ] dxa

b

∫x0

y

g(x)

f(x)

a b

g x( ) f x( )–[ ] dx f x( ) g x( )–[ ] dxc

b

∫+a

c

∫

x0

y g(x)

f(x)

bca


Multiple choice

1 The antiderivative of is:

2 The indefinite integral is equal to:

3 An antiderivative of 2(3x + 4)−4 is:

4 The antiderivative of 6e−3x is:

5 The indefinite integral is equal to:

6 An antiderivative of x3 + sin 4x + e4x is:

7 If f(x) has a stationary point at (0, 3) and f ′(x) = ex + k, where k is a constant, then f(x) is:

8 If the derivative of (x − x2)8 is 8(1 − 2x)(x − x2)7 then an antiderivative of 24(1 − 2x)(x − x2)7 is:

A x4 − loge (1 − x) + c B x4 + loge (1 − x) + c C 16x4 − loge (1 − x) + c

D 16x4 + loge (1 − x) + c E

A 25(5x − 4)5 + c B 5(5x − 4)5 + c C (5x − 4)5 + c

D (5x − 4)5 + c E (5x − 4)5 + c

A − (3x + 4)−3 B − (3x + 4)−3 + 5 C − (3x + 4)−3

D − (3x + 4)−5 E −8(3x + 4)−3

A −2e−3x + c B −3e−3x + c C −18e−3x + c

D −2e−3x + 1 + c E − e−3x + c

A sin + cos 3x + c B sin + 3 cos 3x + c C −3 sin + cos 3x + c

D 3 sin + cos 3x + c E 3 sin − cos 3x + c

A 4(x4 − cos 4x + e4x) B x4 + cos 4x + e4x C x4 − 4 cos 4x + e4x

D (x4 − cos x + e4x) E (x4 − cos 4x + e4x)

A ex − 2x + 2 B −ex − x + 2 C ex − x + 2D e–x + x + 2 E ex + 2x + 1

A 2(x − x2)8 B 3(x − x2)8 C (x − x2)8

D (x − x2)8 E 8(x − x2)8

CHAPTERreview

9A 4x3 11 x–-----------–

x4 11 x–( )2

-------------------– c+

9A 5x 4–( )4 dx∫15--- 1

25------

9A 23--- 2

3--- 2

9---

29---

9B12---

9B cos x3--- 3 sin 3x–

dx∫x3--- 1

3---

x3--- x

3---

x3--- x

3---

9B 14--- 1

4--- 1

4--- 1

4---

14--- 1

4---

9B

9C 12---

13---


9 If the derivative of ex3 + 3x is 3(x2 + 1)ex3 + 3x, then the antiderivative of (x2 + 1)ex3 + 3x is:

10 If the derivative of loge (5 − x2) is then the antiderivative of is:

11 The approximation for the area under the graph at right from x = 2 to x = 4, using the ‘lower rectangles’ is:

12 The area under the graph at right from x = −5 to x = −1 can be approximated by the area of the ‘upper rectangles’ and is equal to:

13 Using the trapezoidal rule the area under the curve at right from x = 1 to x = 4 is approximately equal to:


15 The exact value of the definite integral is:

16 The exact value of is:

17 The shaded area on the graph at right is equal to:

A 3ex3 + 3x + c B C e3x2 + 3 + c

D ex3 + 3x + c E undefined

A − loge (5 − x2) + c B −2 loge (5 − x2) + c C loge (5 − x2) + cD 2 loge (5 − x2) + c E undefined

A 22 sq. units B 14 sq. units C 11 sq. unitsD 10 sq. units E 20 sq. units



A 2 B 8 C −2 D 20 E 16

A 3e4 − e−4 B 2e4 − e−4 C e4 − 2e−4 D e4 + e−4 E e4 − e−4

A −3 B 3 C D E


9C13 x2 1+( )---------------------- c+

13---

9C2x–

5 x2–-------------- x

5 x2–--------------

12--- 1

2---

9D

0 x

y

(2, 3)(2.5, 4)

(3, 6)(3.5, 9)

2 43

9D

x0

y

(–2, 4)(–3, 5)

(–4, 6)

(–5, 8)

–5 –1

12---

9D

0 x

y

(1, 4)(2, 5)

(3, 7)(4, 10)

4321

9E3 x x–( ) dx0

4

∫12---

9E4e2x 2e 2x––( ) dx2–

2

∫

9E2 cos x3---–

dx0

π∫

3 3– 3 3 3 2

9F

x

y = (x – 2)3

0

y

2 4


18 The shaded area on the graph at right is:

19 The area bound by the curve on the graph at right and the x-axis is equal to:

Questions 20 to 22 apply to the curve with equation f(x) = ex − 1.

20 The graph of f(x) is best represented by:

21 The area bound by the graph of f(x), the x-axis and the line x = 2 is equal to:

22 The area bound by the graph of f (x), the y-axis and the line y = e2 − 1 is equal to:

Use the graph at right to answer questions 23 and 24.

23 The two graphs intersect where x is equal to:

24 The area bound by the two graphs is equal to:

Short answer1 Find the equation of the curve f(x) if it passes through (1, −3) and f ′(x) = .

2 A particular curve has , where k is a constant, and it has a stationary point (2, 1). Find:a the value of kb the equation of the curvec the value of y when x = 6.

A 20 sq. units B −20 sq. units C −16 sq. unitsD 16 sq. units E −18 sq. units


D −10 sq. units E 20 sq. units

A B C D E

A e2 − 1 B e2 − 2 C e2 + 1 D e2 + 2 E e2 − 3

A e2 − 5 B e2 − 3 C e2 + 2 D e2 + 1 E 5 − e2

A 1 and −3 B −1 and 3 C 1 and 2 D −1 and −2 E 1 and 3

A 10 sq. units B 7 sq. units C −7 sq. units


9Fx

y = –1 – 3x2

0

y

2–2

–1

9G

x

y = x(x + 2)(x – 3)

0

y

512------ 1

12------ 5

12------

512------ 7

12------

x

f(x)

0

y

2x

f(x)

0

y f(x)

0

–1

x

y

x

f(x)

0–1

y

x

f(x)

0–1–1

y

9G

9G

x

y = x2

y = 2x + 3 0

y

9H

9H23--- 1

3--- 1

3---

13--- 2

3---

9A3x3 2x2–

x-----------------------

9Bdydx------ cos

πx4

------ k+=


3 If y = sin (x2 + 2x), find and hence antidifferentiate (x + 1) cos (x2 + 2x).

4 Apply the trapezoidal method to find the area between y = e2x − 1 and the x-axis, from x = 0 to x = 4, using intervals of width 1 unit. (Give an exact answer.)

5 Calculate an approximation for the area under the curve y = loge x from x = 2 to x = 4, using interval widths of 0.5 units.

6 Evaluate each of the following definite integrals.

7 Given that , find two possible values for k.

8 a Sketch the graph of the function f(x) = .

b Find the exact area between the graph of f(x), the x-axis and the lines x = 3 and x = 6.

9 Find the area bound by the curve g(x) = (4 − x)(6 + x) and the x-axis.

10 Calculate the area between the curve y = 2 cos x and the lines y = −x, x = 0 and x = .

Analysis1 From past records it has been found that the cost rate of maintaining a certain car is

, where C is the accumulated cost in dollars and t is the time in years

since the car was first used. Find:a the initial maintenance cost,b C as a function of tc the total maintenance cost during the first 5 years of use of the card the total maintenance cost from 3 to 5 yearse the maintenance cost for the second year.

2 Over a 24-hour period on a particular autumn day, starting at 12 am, the rate of change of the

temperature for Melbourne was approximately , where T is the

temperature in °C and t is the number of hours since midnight when the temperature was 10°C. Find:a the temperature at any time, tb whether the temperature reaches 17°C at any time during the dayc the maximum temperature and the time at which it occursd the minimum temperature and the time at which it occurse the temperature at

i 2 amii 3 pm

f the time when the temperature first reaches 14.33°C.

a b

9Cdydx------

9D

9D

9E9

2x 3+( )4---------------------- dx–

1–

0

∫ cos 2x dxπ3---

2π3

------

∫

9E4x 5–( ) dx0

k

∫ 2–=

9F1

x 2–-----------

9G9H

π2---

dCdt------- 75t2 50t 800+ +=

dTdt------- 5π

12------ cos

πt12------–=


3 The diagram at right shows part of the curve with equation .Find:a the coordinate of point Ab the equation of the normal to the curve at point Ac the coordinate of point Bd the coordinate of point Ce the area bound by the curve and the lines AB and BC.

4 a Find the derivative of x loge x.b Hence, find an antiderivative of loge x.The cross-section of a platform is shown at right. (All measurements are in metres.)c Find the height of the platform.d Find the cross-sectional area of the platform.e Find the volume of concrete required to build this platform if it is 20 metres long.

y ex2---=

x2

AC

B

0

y

y = e–2xnormal

xe1

1

e–2

f(x) = logex

0

y

testtest



9

calculus made ridiculusly easy

Documents