applications ofintegration world mathematical methods units...applications of integration 301...

88

Applications ofintegration

The

tangent-line problem

(finding the slope of a curve at a point) and the

area problem

(finding areas bound by curves) are the two main geometric problems of calculus. Asyou have seen, the derivative, and associated methods for calculating it, solve thetangent-line problem. We now turn our attention to using integral calculus to find theareas of non-trivial shapes, and applications of the techniques learnt in chapter 6.

DowseyMathsMethodsCh08.fm Page 295 Saturday, January 14, 2006 3:40 PM

296

MathsWorld

Mathematical Methods Units 3 & 4

Graph of an antiderivative function

If

f

is a function with derivative

f

′

, then the process of antidifferentiation can be used to find

f

(

x

) from

f

′

(

x

), i.e.

f

(

x

)

=

f

′

(

x

)

dx

. It is possible to identify a family of curves with a given gradient function.

Consider a curve with gradient function

f

′

(

x

)

=

2

x

, then

f

(

x

)

=

x

2

+

c

, where

c

is a real constant. This describes a family of parabolas, some of which are sketched below.

Depending on the value of

c

, different parabolas are obtained. Each one is a translation of the curve

y

=

x

2

parallel to the

y

-axis.

Similarly, if

f

′

(

x

)

=

1 then

f

(

x

)

=

x

+

c

where

c

is a real constant. This describes a family of straight lines, some of which are sketched below.

Depending on the value of

c

, different straight lines are obtained. Each one is a translation of the line

y

=

x

parallel to the

y

-axis.

⌠⌡

20 4 6 8 10–10 –8 –6 –4 –2 x

y

5

–5

y = x2 +5y = x2 – 1

y = x2 – 5y = x2 – 3

y = x2 – 8

y = x2 +2y = x2 +1y = x2

All of these curveshave the samegradient functiondy

= 2xdx

y = x + 5y = x + 2

y = x – 3

y = x – 6

y = x

20 4 6 8 10–10 –8 –6 –4 –2 x

y

5

–5All of these lineshave the samegradient functiondy

= 1dx

8.18.1

DowseyMathsMethodsCh08.fm 96 Saturday, January 14, 2006 3:40 PM

Applications of integration

297

chap

ter

8

Example

1

Sketch two graphs, each of which could be the graph of an antiderivative of

f

′

(

x

)

=

−

2.

Solution

f

(

x

)

=

f

′

(

x

)

dx

=

−

2

dx

=

−

2

x

+

c

Choosing two values of

c

, say

c

=

0 and

c

=

1:

c

=

0,

f

(

x

)

=

−

2

xc

=

1,

f

(

x

)

=

−

2

x

+

1

It is possible to determine a unique antiderivative if more information is provided.

Example 2The graphs of y = f ′(x ) and y = f (x ) are as shown.

Find the rule for:a f ′(x ) b f (x )

Solutiona f ′(x ) = mx

(1, −1) is on this line).So −1 = m and hence f ′(x ) = −x.

0 1 2–2 –1 x

y

1

2

–1

–2

y = −2x+1

y = −2x

⌠⌡⌠⌡

2(1, –1)

0 4 6–6 –4 –2

y

2

4

–2

–4y=f ′ (x)

(0, 4)

y=f(x)

20 4 6–6 –4 –2 x

y

2

4

–2

–4

b f (x ) = f ′(x ) dx

=

=

(0, 4) is on this curveSo 4 = c and hence f (x ) = .

⌠⌡⌠⌡ x– xd

x2

2-----– c+

x2

2-----– 4+

8.1


298

MathsWorld Mathematical Methods Units 3 & 4

exercise 8.11 For each of the following, sketch two graphs, each of which could be the graph of an

antiderivative of f ′(x).

a f ′(x) = −3 b f ′(x) = 4x c f ′(x) = x2

2 In each of the following, the graphs of y = f ′(x) and y = f(x) are as shown. Find a possible rule for:

i f ′(x) ii f(x)

a

b

c

d

10 2–4 –2 x

y

y = f ′(x)

2

–2

–4y=f(x)

20 3 4–2 –1 1 x

y

–2

–4

–6

20 4–2 31–1 x

y

–2

–4

–6

y = f ′(x)

20 4–2 31–1 x

y

4

6

2

y=f(x)

20 4–2 3

(1, –3)

1–1 x

y

4

6

2

–2

y = f ′(x)

20 4–2 3

(0, 3)

1–1 x

y

4

6

2

–2

y = f(x)

0 1 2–1–2 x

f ′(x)

10

15

5(1, e 2)

y = f ′(x)

f(x)

0 1 2–1–2x

10

15

5

y = –1

y = f(x)



299

chap

ter

8Sketching antiderivativesGraphs of antiderivatives can be sketched from graphical information alone.

Recall that when interpreting the graph of a derivative:

. x-axis intercepts correspond to stationary points on the original function.

. If the graph is above the x-axis, then the original function has positive gradient and therefore it is increasing as x increases.

. If the graph is below the x-axis, then the original function has negative gradient and therefore it is decreasing as x increases.

If the graph of a function is given and the graph of the antiderivative is required, the graph that is given should be interpreted as the graph of the gradient function of the required curve.

Example 3Sketch the graph of the antiderivative of the function shown, given that it passes through the origin.

Solution

Interpreting the graph above as the gradient function of the required curve, it can be seen that the graph of the antiderivative has a minimum turning point at x = −2 and a maximum turning point at x = 1.The gradient of the antiderivative function is negative for x < −2 and x > 1 and positive for−2 < x < 1. The antiderivative function has maximum positive gradient at x = −0.5.The graph of the antiderivative is required to passthrough (0, 0).A suitable graph of the antiderivative is sketchedat right.

10 2–2 –1 3 4–3–4 x

y

10

For the required curve, at x = –2 the gradient is zero and therefore there is a stationary point. The gradient changes from negative to positive at this point, so the stationary point is a minimum.

For the required curve, at x = 1 the gradient is zero and therefore there is a stationary point. The gradient changes from positive to negative at this point, so the stationary point is a maximum.

2–2 –1 3 4–3–4 x

y

10–2 x

y

8.1


300


Example 4Sketch the graph of the antiderivative of the function shown, given that it passes through the origin.

SolutionInterpreting the graph above as a gradient function, it can be seen that the graph of the antiderivative has a stationary point of inflectionat x = −2 and a minimum turningpoint at x = 0.The gradient of the antiderivative function is negative for x < −2 and−2 < x < 0 and positive for x > 0.The graph of the antiderivativeis required to pass through (0, 0), so the minimum turning point, in this case, is at the origin.

A suitable graph of the antiderivativeis sketched at right.

Example 5The graph of y = f ′(x ) is shown. Sketch a possible graph of y = f (x ).

SolutionThe graph of y = f (x ) will have a maximum turning point at x = −3 and a minimum turning point at x = 2. The gradient of this curve is positive for x < −3 and x > 2 and negative for −3 < x < 2.

10–2 –1 x

y

1

At x = –2 the gradient is zero, so there is a stationary point. The gradient does not change sign at this point,so the stationary point is a stationary point ofinflection.

At x = 0 the gradient is zero, so there is a stationary point. Thegradient changes from negative to positive at thispoint, so the stationarypoint is a minimum.

0–2 –1 x

y

10–2 –1 x

y

10 2–2 –1 3 4 5 6–3–4–5–6 x

y

y = f ′(x)



301

chap

ter

8A possible graph of y = f (x ) is shown below.

exercise 8.13 For each of the graphs of y = f ′(x) shown below, sketch the graph of the antiderivative,

y = f(x), given that it passes through the origin.

a b

c d

tip. When we find the indefinite integral of a

function, it always includes an arbitrary constant, the constant of integration. As a result, there is not a unique antiderivative graph for the graph of a given function. It may be translated any distance parallel to the y-axis.

. When sketching graphs of antiderivatives, the basic shape can be accurately represented by key features with associated x-coordinates, but its vertical translation is indeterminate unless extra information is provided.

0 2–3 x

y

y = f(x)

continued

10 5–2 –1

1

–1

–2

–3

–4

–5

2

4

5

3

2 3 4–3–4–5 x

y

y = f ′(x)

10 5–2 –1

1

–1

–2

–3

–4

–5

2

4

5

3

2 3 4–3–4–5 x

y

y = f ′(x)

10 5–2 –1

1

–1

–2

–3

–4

–5

2

4

5

3

2 3 4–3–4–5 x

y

y = f ′(x)

10 5–2 –1

1

–1

–2

–3

–4

–5

2

4

5

3

2 3 4–3–4–5 x

y

y = f ′(x)

8.1


302


e f

g h

4 The graph of y = f ′(x) is shown at right. Sketch the graph of y = f(x) if f(0) = 0.

10 5–2 –1

1

–1

–2

–3

–4

–5

2

4

5

3

2 3 4–3–4–5 x

y

y = f ′(x)

10 5–2 –1

1

–1

–2

–3

–4

–5

2

4

5

3

2 3 4–3–4–5 x

y

y = f ′(x)

10 65–2 –1

1

–1

–2

–3

–4

–5

2

4

5

3

2 3 4 x

y

y = f ′(x)

10 5–2 –1

2

–2

–4

–6

–8

–10

4

8

10

6

2 3 4–3–4–5 x

y

y = f ′(x)

10 65–2 –1

2

–2

–4

–6

–8

–10

4

8

10

6

2 3 4–3–4–5–6 x

y

y = f ′(x)



303

chap

ter

85 The graph of y = f ′(x) is shown. Sketch the graph of

y = f(x) if f(0) = 3.

6 Sketch a graph showing the general shape of y = f(x) for each of the graphs of y = f ′(x) sketched below.

a b

c d

10–2 –1

–10

–5

10

5

2 3 4–3–4 x

y

y = f ′(x)

10–2 –1 2 3 4–3–4 x

y

y = f ′(x)10–2 –1 2 3 4 5 x

y

y = f ′(x)

x0

2ππ

y

y = f ′(x)

10–2 –1 2 3–3 x

y

y = f ′(x)

8.1


304


Areas under and between curves

The fundamental theorem of calculusLet A(x) be the area bounded by the curve with equation y = f(x), the x-axis and the lines x = a and x = b.

Let P be the point (x, f(x)),

Q the point (x + δx, f(x + δx)) and

A the point (a, f(a)).

A(x) is the red shaded area.

A(x + δx) is the sum of both shaded areas.

A(x + δx) − A(x) is the blue shaded area.

It is clear from the diagram that f(x)δx ≤ A(x + δx) − A(x) ≤ f(x + δx)δx.

Therefore, (dividing each expression by δx)

If we let δx → 0,

So f(x) ≤ ≤ f(x)

Therefore, since it is sandwiched between two equal quantities.

That is, A(x) = F(x) + c, where F ′(x) = f(x).

When x = a, A = 0, so 0 = F(a) + c, i.e. c = −F(a).

When x = b, A(b) is the area bounded by the curve with equation y = f(x), the x-axis and the lines x = a and x = b. Call this area A. Then A = F(b) + c = F(b) − F(a), where F ′(x) = f(x).

x

y y = f(x)(b, f(b))

(a, f(a))

(x, f(x))

A

Q

P

(x + x, f(x + x))

f x( ) A x δx+( ) A x( )–δx

------------------------------------------- f x δx+( )≤ ≤

f x( )δx 0→lim

A x δx+( ) A x( )–δx

-------------------------------------------δx 0→lim f x δx+( )

δx 0→lim≤ ≤

dAdx-------

dAdx------- f x( )=

Fundamental theorem of calculusLet f be a continuous function on the interval [a, b].

Let F be an antiderivative of f on [a, b] so that F ′(x ) = f (x ).

Then f (x ) dx = F ′(x ) dx = F (b) − F (a).

If the graph of y = f (x ) lies above the x-axis, this represents the area between the graph and the x-axis.

⌠⌡a

b⌠⌡a

b

8.28.2



305

chap

ter

8Example 1Consider the graph of y = 3x2 − 12. Find the area, in square units, enclosed by the curve, the x-axis and the lines:a x = 2 and x = 4 b x = 0 and x = 2 c x = 0 and x = 4

Solution

a (3x2 − 12) dx

= (antidifferentiating)= (substituting for x )= 32The area is 32 square units.

b If (3x2 − 12) (dx) is evaluated, the result isnegative:

(3x2 − 12) dx

= = = −16

The presence of the minus sign indicates thatthe area is located below the x-axis. The area is 16 square units.

c The area of the region bounded by the graphof y = 3x2 − 12, the x-axis and the lines x = 0 and x = 4 is 48 square units.As y is positive for some values of x and negative for other values of x on the interval[0, 4], the area below the x-axis and the area above the x-axis must be calculated separately and then added together in order to determine the required area.

Signed areaThe calculation in the warning box above gives what is known as the signed area. Regions below the x-axis have a negative signed area. In some applications, the signed area rather than the actual area, is required.

40–4 62–6 –2 x

yy = 3x

2 − 12

20

30

10

–10

A=32

⌠⌡2

4

x3 12x–[ ]24

43 12 4( )–( ) 23 12 2( )–( )–

40–4 62–6 –2 x

y

20

30

10

–10 A=16

y = 3x2 − 12

40–4 6

A = 16 + 32 = 48

2–6 –2 x

y

20

30

10

–10

y = 3x2 − 12

⌠⌡0

2

⌠⌡0

2

x3 12x–[ ]02

23 12 2( )–( ) 0–

Warning Look out for x-interceptsThe value in example 1 part c is very different from the evaluation of (3x2 − 12) dx:

(3x2 − 12) dx =

= = 16

⌠⌡0

4

⌠⌡0

4x3 12x–[ ]

04

43 12 4( )–( ) 03 12 0( )–( )–

8.2


306


In general, when finding the area enclosed between a curve with equation y = f(x), thex-axis and the lines x = a and x = b, areas above and below the x-axis need to be calculated using separate integrals.

If the required area lies entirely above the x-axis, it can be calculated using a single integral, as shown above right.

If the required area lies below the x-axis,it is calculated as shown below. In this case

f(x) dx is negative so the area is given by

− f(x) dx, which will be positive.

If the required area lies partly above and partly below the x-axis, separate calculations are required as shown in example 1.

Example 2a Find the exact area of the region enclosed by the curve with equation y = 2e−3x + 1, the

coordinate axes and the line x = 1.b Give the answer in part a correct to 3 decimal places, and check using technology.

Solutiona The required region lies entirely above the x-axis and

can therefore be calculated using a single integral.

Area = (2e−3x + 1) dx

=

=

= (square units)

b Correct to 3 decimal places, the answer is 1.633. The screenshots below provide a check on the numerical answer. The one on the left uses fnInt at the home screen; the one on the right finds the area interactively on the graph. Note that on the TI-89, the exact as well as the approximate answer can be found using the integral command at the homescreen.

x

y

x=a x=b

y = f(x)

f(x) dx⌠⌡a

b

x

y

x=a x=b

y = f(x)

− f(x) dx⌠⌡a

b

⌠⌡a

b

⌠⌡a

b

20 1–1 x

y

y = 2e−3x

+ 1

10

15

5

⌠⌡0

1

23---– e 3x– x+

0

1

23---– e 3– 1+⎝ ⎠

⎛ ⎞ 23---e0–⎝ ⎠

⎛ ⎞–

53--- 2

3e3---------–

GC 5.5CAS 5.5

[1, 2] by [−1, 5]



307

chap

ter

8Example 3Calculate the exact area of the region bounded by the curve y = x3 − 4x2 and the x-axis.

SolutionSketch the graph and identify the required region.The required region lies entirely below the x-axis and can therefore be calculated using a single integral.

Area = − (x3 − 4x2) dx

=

=

=

The area is square units.

Example 4Find the exact area of the region enclosed by the curve with equation y = cos 2x, the x-axis

and the lines x = 0 and .

SolutionSketch the graph and identify the required region.The area is made up of one region above the x-axis and one below the x-axis. The shaded area can be calculated using two integrals.

Area = cos 2x dx − cos 2x dx

=

=

=

=

=

The area is square units.

–10

4–4 62–6 –2 x

y

y = x3 − 4x

2

–2

–4

2

–6–8

⌠⌡0

4

x4

4----- 4x3

3--------–

0

4

–

4( )4

4---------- 4 4( )3

3--------------–⎝ ⎠

⎛ ⎞–

643

------tipThe region in question is enclosed byonly the x-axis and the curve; there areno vertical boundaries.64

3------

x 2π3

------=

x

y

y = cos 2x

4

1

–1

0

43

32

π

π

π π

tipThis example shows that it is critical to initially identify the x-intercepts of the graph as a first step so that the correct integral bounds are used.

⌠⌡0

π/4⌠⌡π/4

2π/3

12--- sin 2x

0

π 4⁄ 12--- sin 2x

π 4⁄

2π 3⁄–

12--- sin π

2--- sin 0–⎝ ⎠

⎛ ⎞ sin 4π3

------ sin π2---–⎝ ⎠

⎛ ⎞–⎝ ⎠⎛ ⎞

12--- 1 3

2-------– 1–⎝ ⎠

⎛ ⎞–⎝ ⎠⎛ ⎞

12--- 2 3

2--------+⎝ ⎠

⎛ ⎞

1 34

--------+

1 34

--------+

8.2


308


Example 5

A section of the graph with equation is shown below. The shaded area is equal to 16 square units. Find the value of k.

Solution = 16

= 16

−2k (cos π − cos 0) = 16−2k(−2) = 16

4k = 16k = 4

exercise 8.21 Find the exact area of the region bounded by the each of the following curves and the x-axis.

a y = 3 − x2 b y = x3 + 5x2 − x − 5

c y = x2(2 − x) d y = x4 − 4x2

2 Find the area of the shaded region in each of the following.

a b

y k sin x2---=

x

y

k

π0π2

tipThe value of k can be found using the Solver on the TI-83/84 or the solve command on the TI-89.GC 5.5, 10.2

CAS 5.5, 7.1

⌠⌡0

2πk sin x

2--- xd

2k cos x2---–

0

2π

y = 2x

x

y

8

6

4

2

00.5 1 1.5 2 2.5 3 3.5

20–2 31–1–3 x

y

34

21

–1–2–3–4

y = 4 – x2



309

chap

ter

8c d

3 Find the exact area of the region enclosed by the curve y = 2ex − 1 and the coordinate axes.

4 Find the exact area of the region enclosed by the curve with equation and the lines x = 1 and x = 4.

5 a Sketch the graph of the curve with equation .

b Find the exact area of the region bounded by the curve, the x-axis and the lines x = 3 and x = 5.

6 Calculate the exact area of the region bounded by the graph of y = (x − 1)3 and the linesx = 0 and x = 2.

7 Find the exact area of the region enclosed by the curve y = 3 sin 2x, the x-axis and the

lines and .

8 Find the area of the region bounded by , the x-axis and the line x = 1.

9 Find the exact area of the shaded region.

y = x2 +2x – 3

20–2 31–1–3 x

y

34

21

–1–2–3–4

x

y

8

6

4

2

00.5 1 1.5 2 2.5 3 3.5

y = 2x + 1

9

y = x

x

y

2.5

2

1.5

0.5

1

00.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5

e

x

y

2.5

2

1.5

0.5

1

00.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5

y = x

f

y 12x------=

y 3 1x 2–------------–=

x π4---= x 2π

3------=

y 4 sinπx2

------=

y =1 – ex

0 1–1 –0.5 x

y

1

–1

0.5

8.2


310



11 a Find .

b Hence, find the exact area of the regionshaded in the diagram below.

12 The graph of y = f(x), wheref: [0, π] → R, f(x) = k sin(x), is shown. Find the value of k,if the area of the shaded regionis 2 square units.

13 The graph with equation y = ax(4 − x)is shown. The area of the shaded region is 40 square units. Find the value of a.

14 The graph of is shown.

a Find the values of a and b.

b Hence find the exact area of the shaded region.

y = x3 – x

20–2 x

y

6

4

2

y =(2x+1)

3

0.40.2 0.60 0.8 1 x

y

1.5

1

0.5 1

⌠⎮⌡

1

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

x

y

y = f(x)

π π2

0

x

y

0 0.5 1 1.5 2 2.5 3 3.5 4

y = ax(4 − x)

x

y

0

(0, 1)

(2, 7)

0.5 1 1.5 2 2.5 3 3.5 4

y = a − b cos x2

π

y a b cosπx2

------–=



311

chap

ter

815 Differentiate y = loge (x2 + 1).

Hence find the area of theshaded region.

Area between curvesConsider two curves with equations y = f(x) and y = g(x), where f(x) > g(x) for all x ∈ [a, b].

The area enclosed between the two curves can be determined by finding the area of the region enclosed by the curve with equation y = f(x), the x-axis and the lines x = a and x = b and subtracting the area of the region enclosed by the curve with equation y = g(x), the x-axis and the lines x = a and x = b.

Area = f(x) dx − g(x) dx

= (f(x) − g(x)) dx

Example 6Find the exact area of the region enclosed between the line with equation y = 2x and the

curve with equation .

20–2 2.51.51–1 –0.5–1.5–2.5 x

y

0.5

x

x2 + 1

y =

xba

y

f(x)

g(x)

y =

y =

⌠⌡a

b⌠⌡a

b

⌠⌡a

b

Visualising the area between curvesArea between curves = (top curve − bottom curve) dx = (f (x ) − g(x)) dx⌠

⌡a

b⌠⌡a

b

ba

y

x x x

y y

ba

f(x)

= −g(x)

ba

f(x)

g(x)

y =

y = y =

y = f(x)

g(x)y =

y =

This rule applies even if the curves are partly above and partly below the x-axis. When calculating the area between two curves it is only necessary to use more than one definite integral if the top curve and the bottom curve interchange.

tip

y 12---x2=

8.2


312


SolutionSketch the graph and identify the required region.The required area is enclosed between the upper graph, y = 2x, and the lower graph,

, between their points of intersection.

In this case the x-coordinates of the points of intersection can be easily read from the graph, but it is also possible to determine these algebraically as follows.

When y = 2x and intersect:

2x =

0 =

0 =

x = 0, 4

The required area is given by:

Area =

=

=

=

The area of the region enclosed between the two curves is square units.

Example 7Find the exact area of the shaded region.

0 64–4 –2–6 x

y

6

8

4

2

–2

–4

2

y=2xy= x

212

y 12---x2=

tipThe required area can be shown using the shade command. However, to find the area required enter the rule Y3=Y1–Y2, adjust the window dimensions and turn off Y1 and Y2. The area can be found interactively as shown below right. Alternatively, the fnInt command on the TI-83/84 or the integration command on the TI-89 can be used on the home screen.

[0, 4] by [−2, 8] [0, 4] by [−1, 3]

y 12---x2=

12---x2

12---x2 2x–

12---x x 4–( )

⌠⎮⌡0

4

2x 12---x2–⎝ ⎠

⎛ ⎞ xd

x2 16---x3–

0

4

4( )2 4( )3

6----------–

163

------

163

------

x

y

π2

–1

0ππ 2π

23

1

y=sin x

y=cos x



313

chap

ter

8SolutionThe x-coordinates of the points of intersection must be determined.

sin x = cos xtan x = 1 (dividing both sides by cos x)

There is one solution in the first quadrant and one in the third quadrant.

Solution in first quadrant:

Solution in third quadrant: x =

=

Hence the curves intersect at and .

On the interval , y = cos x is the top curve and y = sin x is the bottom curve.

On the interval , y = sin x is the top curve and y = cos x is the bottom curve.

On the interval , y = cos x is the top curve and y = sin x is the bottom curve.

Three integrals are required to calculate the area of the shaded region.

Area = (cos x − sin x ) dx + (sin x − cos x ) dx + (cos x − sin x ) dx

= sin x + cos x + −cos x − sin x + sin x + cos x

=

=

The area of the shaded region is square units.(Note that each of these definite integrals is positive.)

exercise 8.216 Find the total area of the region or regions bounded by the graphs with the equations given.

a y = x2, y = 16 b y = 1 − x2, y = x + 1 c y = x, y = 2x2

d y = x − 1, y = x2 − 4x + 3 e y = x2, y = x f y = x3, y = x

g h y = 4 − x2, y = 2x2 − 5x + 2

17 Find the area of the region enclosed by the curves with equations and and the lines x = 2 and x = 4.

18 Find the area of the region enclosed by y = x2 + x − 2 and y = 3x + 1.

19 Find the area of the region enclosed between the graphs of and .

x π4---=

π π4---+

5π4

------

x π4---= x 5π

4------=

0 π4---,

π4--- 5π

4------,

5π4

------ 2π,

⌠⌡0

π/4⌠⌡π/4

5π/4⌠⌡5π/4

2π

⎡⎢⎣

⎤⎥⎦ 0

π/4 ⎡⎢⎣

⎤⎥⎦ π/4

5π/4 ⎡⎢⎣

⎤⎥⎦ 5π/4

2π

22

------- 22

-------+⎝ ⎠⎛ ⎞ 0 1+( )–⎝ ⎠

⎛ ⎞ 22

------- 22

-------+⎝ ⎠⎛ ⎞ 2

2-------– 2

2-------–⎝ ⎠

⎛ ⎞–⎝ ⎠⎛ ⎞ 0 1+( ) 2

2-------– 2

2-------–⎝ ⎠

⎛ ⎞–⎝ ⎠⎛ ⎞+ +

4 2

4 2

continued

y1x--- y, 5

2--- x–= =

y ex 2⁄= y 2x---=

y x2= y x=

8.2


314


20 Find the area of the region enclosed by the curves with equations y = x3 and y = 16 − x3 and the straight lines x = −1 and x = 3.

21 The tangent to the curve y = x3 − 5x2 + 6x at x = 1 cuts the curve again at x = b. Find the value of b and hence find the area of the region enclosed between this tangent and the curve.

22 a Find the derivative of x sin x and hence find an antiderivative of x cos x.

b Use the answer to part a to find the area of the region enclosed by the curves with equations y = x cos x and y = −cos x and the lines x = 0 and x = π.

23 Find the area of the region enclosed by the graphs of y = ex2 and y = 4, correct to 2 decimal

places.


25 Find the area of the region shaded.

26 Find the value of k such that the area bounded by the line with equation y = 2x,the x-axis and the line x = 4 is divided into two regions of equal area by the linewith equation y = k.

27 Find the exact area of the region in the first quadrant bounded by the x-axis and the curves with equations and .

x

y

0.5 1 1.50

1.5

1

0.5

y= x

y= 1

y=sin

x

x

y

π2

–1

0ππ 2

2

π23

1

y=cos x2

y x= y 3 x–=



315

chap

ter

828 a Find the area of the region enclosed by the graphs of y = x3 and y = x1/3 in the first

quadrant.

b Repeat part a for the graphs of

i y = x4 and y = x1/4 ii y = x5 and y = x1/5

c From the results above, make a conjecture about the area of the region enclosed by the graphs of y = xn and y = x1/n where n is a positive integer. Verify your conjecture using integration.

A parabolic arch is defined as having a base and height as shown.

a Find the area under the parabolic arch shown on the right with equation y = 8 − 2x − x2, −4 ≤ x ≤ 2.

b What is the base length of this arch?c Find the height of this arch.

Archimedes was a very great mathematician who discovered that the area under a parabolic arch isalways equal to two-thirds of the base length multiplied by the height.

d Test Archimedes’ law for the arch defined in question a.

e Sketch the graph of the parabola

defined by the equation ,

showing the coordinates of the x-intercepts and the turning point. Assume thatb and h are positive constants.

f Use calculus to find the area enclosedby the curve and the x-axis and henceverify Archimedes’ law.

Height

Base

40 2–4 –2 x

y

6

8

10

42

–2

y h4h

b2------⎝ ⎠

⎛ ⎞ x2–=

analysis task 1—the parabolic arch

SAC

8.1The parabolic arch

CD

SAC

analysis task

8.2


316


Hoa plans to construct a new fish pond in her back garden. Her proposed design is made up of a shallow section bounded by a parabola of constant depth, in which her fish can be easily visible, attached to a rectangular pool of changing depth in which the temperature of the water is cooler on hot days.

The proposed plan for the pond, as viewed from above, is shown. The rectangular section PQRS is 3 metres long and 1 metre wide.

The equation of the parabolic boundary ofthe shallow section is of the form y = ax2 + c.

a Find the values of a and c.

b Find the exact area of the total pond surface.

A view of the vertical cross-section of the pond along TU is displayed on a different set of axes in the diagram. All measurements are in metres.

c What is the depth of water in the shallow section?

d Determine the volume of water required to fill the shallow section.

e Write down the coordinates of the points C and D.

f The curve DC has equation y = −ekx + c. Find the exact values of k and c.

g Find the area of the region OUCD, correct to 4 decimal places.

h Find the volume of water required to fill the rectangular section, correct to 3 decimal places.

i Find the total volume of water required to fill the pond, correct to 3 decimal places.

xQ

O

P

S RU

Shallow section

T

y

–1.0

0.5–0.5

PLAN

x

DE

ELEVATION

Shallowsection

C

T U

y

1

1.5

0.5

3

O U

analysis task 2—Hoa’s fish pond

SAC

8.2Hoa’s fish pond

CD

SAC

analysis task



317

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8Straight-line motion

The study of the motion of a particle is called kinematics. When the motion of a body, usually referred to as a particle, is in a straight line only, it is referred to as rectilinear motion.

Position and displacementThe position, x, of a particle on a straight line is defined as where a particle is relative to some fixed point. This point is usually referred to as the origin, O. The position of the particle is usually described as being positive when it is to the right of the origin and negative when it is to the left of the origin.

Displacement is defined as the change in the position of a particle, while distance is defined as how far a particle has travelled. The distinction between distance and displacement is most easily seen through an example.

Consider a particle which starts at the origin. It moves 15 metres to the left and then 10 metres to the right.

The distance travelled by the particle is 25 metres but the displacement of the particle is −5 metres.

VelocityIn everyday language the terms velocity and speed are used interchangeably, but in the study of motion there is a very important distinction between the two quantities.

The speed of a particle refers to how fast the particle is travelling, while velocity also takes the direction of motion into account. A particle travelling with a constant velocity of 5 metres per second is conventionally considered to be travelling to the right, but if it travels with a velocity of −5 metres per second it is considered to be travelling to the left. In both cases the particle has a speed of 5 metres per second.

x–15 –14 –13 –12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 O 1 2 3 4 5

Average speed and velocityAverage speed and average velocity are both examples of average rates of change:

average speed =

average velocity = = .

distance travelledtime taken

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

change in positionchange in time

------------------------------------------------- displacementtime taken

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

8.38.38.3


318


Example 1A particle starts at the origin and travels 5 metres to the right and then 5 metres to the left in 5 seconds. Find:a the distance travelled by the particle. b the displacement of the particle.c the average speed of the particle. d the average velocity of the particle.

SolutionIt is often useful to consider the motion of the particle through a diagram.a The particle travels a distance of 10 m.b The displacement of the particle is 0 m.

c Average speed =

=

= 2The average speed of the particle is 2 m/s.

d Average velocity =

=

= 0The average velocity of the particle is 0 m/s.

Example 2The position of a particle, x metres from the point O at time t seconds, is given byx(t ) = t2 − 4t − 5, t ≥ 0. Find:a the initial position of the particle.b the position of the particle after 6 seconds.c the distance travelled by the particle in 6 seconds.d the average speed over the first 6 seconds.e the average velocity over the first 6 seconds.f the particle’s initial velocity.g when and where the particle is instantaneously at rest.h the velocity of the particle as it passes through O.

x54321O


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

105

------


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

05---

Instantaneous velocityVelocity measures the rate of change of position of a particle. Instantaneous velocity, v, is

the rate of change of position with time, i.e. .v dxdt------=

When the velocity of a particleis 0, we say that the particle isat rest, even when this is onlyfor an instant.

tip



319

chap

ter

8SolutionIt is useful to consider the position–time graph when answering the questions.a x(0) = −5

The particle starts 5 metres to the left of O.

b x(6) = 62 − 4(6) − 5= 7

The particle is 7 metres to the right of O after 6 seconds.

c Using completing the square gives:x(t ) = (t − 2)2 − 9

Hence the turning point is at (2, −9).The particle travels from x = −5 to x = −9,4 metres to the left, then from x = −9 tox = 7, 16 metres to the right.The distance travelled is 20 metres.

d Average speed =

=

=

The average speed over the first

6 seconds is m/s.

f v =

= 2t − 4The initial velocity is the velocity when t = 0. Hence the initial velocityis −4 m/s.

h The particle passes through O when x = 0.0 = t2 − 4t − 50 = (t − 5)(t + 1)t = 5, −1

As t ≥ 0, the particle passes through the origin after 5 seconds.Substituting t = 5 gives v = 6.The velocity of the particle as it passes through the origin is 6 m/s.

t

x

1 2 3 4 5 6 7 8O

5

10

–5

–10 (2, –9)

x = t2 – 4t –5

e Average velocity =

=

= 2The average velocity over the first 6 seconds is 2 m/s.

g The particle is at rest when v = 0.0 = 2t − 4t = 2When t = 2,

x = (2)2 − 4(2) − 5= −9

The particle is at rest after 2 seconds when it is 9 metres to the left of the origin. (Note that (2, −9) is the minimum point on the graph above.)

displacementtime taken

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

7 5–( )–6

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


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

206

------

103

------

103

------

dxdt------

8.3


320


Calculating distance and displacement from velocity–time graphsVelocity is the rate of change of position with respect to time, i.e. . Hence x = v dt.

Example 3Consider a particle travelling with velocity v m/s at time t seconds given by , where 0 ≤ t ≤ 8. Find:a the distance travelled in the first 8 seconds.b the displacement of the object after 8 seconds.

SolutionThe distance travelled by the particle in 8 seconds is given by the total area because, although the section below the t-axis represents velocity in the opposite direction to that in the previous section, it nevertheless represents movement. So, although the displacement is decreasing, distance travelled is increasing.

a Distance = −

= 20The distance travelled is 20 metres.Note: In this case it is not necessary to use calculus to find the required area. It could be easily found from the areas of two triangles each with base 4 and height 5.

b The displacement of the particle is given by the signed area i.e. the definite integral calculated over the full 8 seconds.

The displacement of the particle is 0 metres. Since the two triangles are congruent, the sum of their signed areas is zero.

Example 4The velocity of an object, v m/s, at time t seconds is given by . Find:a the distance travelled in the first 10 seconds.b the displacement of the object after 10 seconds.

Solutiona This problem involves finding area,

so a graph is required.The graph cuts the t-axis where v(t) = 0, i.e. t = 8 (since t > 0).

v dxdt------= ⌠

⌡

v t( ) 5 54---t–=

t

v

2 4 6 80

2

4

–2

–4

v(t) = 5 − 5t

4

⌠⎮⌡0

4

5 54---t–⎝ ⎠

⎛ ⎞ td⌠⎮⌡4

8

5 54---t–⎝ ⎠

⎛ ⎞ td

⌠⎮⌡0

8

5 54---t–⎝ ⎠

⎛ ⎞ td 0=

v t( ) 16 14---t2 t 0≥,–=

t

v

2 4 6 8 100

10

5

15

–5

–10



321

chap

ter

8Area = −

=

=

= 94The distance travelled by the object in the first 10 seconds is 94 metres.

b The displacement of the object is given by .

=

=

=

The displacement of the particle is metres.

exercise 8.31 The position of an object on a straight line is given by x = t2 − 2t, t ≥ 0. Displacement is

measured in metres and time in seconds. Find:

a the initial position of the object.

b the position of the object after 2 seconds.

c the distance travelled by the object in the first 2 seconds.

d the average speed of the object over the first 2 seconds.

e the velocity of the object after 2 seconds.

f when and where the object is instantaneously at rest.

2 The position of an accelerating object is given by . Find the velocity of the body when:

a t = 0 b t = 1 c t = 4 d t = 9

3 A particle moves in a straight line so that its position x centimetres relative to O at time t seconds is given by x = t2 − 7t + 6, t ≥ 0. Find:

a the initial position of the particle.

b the position of the particle after 6 seconds.

c the distance travelled by the particle in 6 seconds.

d the average speed over the first 6 seconds.

e the average velocity over the first 6 seconds.

⌠⎮⌡0

8

16 t2

4----–⎝ ⎠

⎛ ⎞ td⌠⎮⌡8

10

16 t2

4----–⎝ ⎠

⎛ ⎞ td

16t t3

12------–

0

8

16t t3

12------–

8

10

–

16 8( ) 8( )3

12----------– 0( )–⎝ ⎠

⎛ ⎞ 16 10( ) 10( )3

12-------------– 16 8( ) 8( )3

12----------–⎝ ⎠

⎛ ⎞–⎝ ⎠⎛ ⎞–

⌠⎮⌡0

10

16 t2

4----–⎝ ⎠

⎛ ⎞ td

⌠⎮⌡0

10

16 t2

4----–⎝ ⎠

⎛ ⎞ td 16t t3

12------–

0

10

16 10( ) 10( )3

12-------------–

2303

----------

7623---

x 10t3 2⁄ 0 t 10≤ ≤,=

8.3


322


f the particle’s initial velocity.

g when and where the particle is at rest.

h the velocity of the particle as it passes through O.

4 The velocity of a particle moving in a straight line, in metres per second, is given byv = 4t2 + 2t, t ≥ 0. Find the distance travelled by the particle in the first 5 seconds.

5 A body starts at the origin and moves in a straight line. Its velocity, in kilometresper hour, is given by v = 3t2 + t, t ≥ 0. How far does it travel in 3 hours?

6 A particle starts at the origin and travels in a straight line with velocity v m/s given by v = 3t2 − 4t − 4, t ≥ 0. Find:

a the displacement of the particle after 3 seconds.

b the distance travelled by the particle in the first 3 seconds.

c when and where the particle is at rest.

7 A particle attached to a spring moves in a straight line so that at time t seconds its

velocity, v cm/s, is given by . Find:

a the displacement of the particle after 18 seconds.

b the distance travelled by the particle in the first 18 seconds.

8 The velocity of an object is given by v = 3t2 − 2t, t ≥ 0. If the object starts at the origin, at what time is it 100 units to the right of the origin?

9 At time t seconds, the velocity v metres per second of an object is given by v = 5 sin t. What is the total distance travelled by the object for:

a b

v 4π sin πt6----- t 0≥,=

0 t2π3

------≤ ≤ 2π3

------ t 2π≤ ≤

Numbersense with the spence

8888% of an iceberg will always be under water. And there are 88 keys on a standard piano. There was a piano on the Titanic. Spooky.



323

chap

ter

8Finding a function given its derivative

In section 8.1 we looked at how to sketch the graph of an antiderivative. This is not a unique curve, as f ′(x)dx = f(x) + c, where c is any real number.

The value of c can be determined if more information is provided.

Example 1Find the rule for the function f for which f ′(x ) = 4x − 5 and f (2) = 1.

Solutionf ′(x ) = 4x − 5Antidifferentiate to find f (x ).

f (x ) = 2x2 − 5x + cDetermine the value of c, using the information f (2) = 1.

1 = 2(2)2 − 5(2) + c1 = −2 + cc = 3

So, f (x ) = 2x2 − 5x + 3.

Example 2

Find the equation of the curve defined by , given that the curve passes through (0, 5).

Solution

Antidifferentiate to find y.

Find c, using the information that when x = 0, y = 5.

5 =

5 =

c =

The equation of the curve is .

⌠⌡

tipThese types of problems can

be solved using a TI-89. First findthe antiderivative, then solve for cto find the unique antiderivative.

CAS 10.2, 10.7

dydx------ e2x=

dydx------ e2x= tip

The graphs of , for

c ∈{−10, −5, 0, 5, 10} help to show

that only the graph of

(shown in thick style) passes through the point (0, 5).

y12---e2x c+=

y12---e2x 9

2---+=

[−1, 2] by [−10, 15]

y 12---e2x c+=

12---e0 c+

12--- c+

92---

y 12---e2x 9

2---+=

8.48.48.4


324


exercise 8.41 Find f(x) for each of the following.

a f ′(x) = 6 − 2x and f(2) = 7

b f ′(x) = 3(x + 2)2 and f(1) = 23

c f ′(x) = 9e3x and f(0) = 5

d f ′(x) = x(2 − x) and the graph of y = f(x) passes through the point (3, 0)

e and the graph of y = f(x) passes through the point (1, −2)

f and

g and f(1) = −1

2 Find y in terms of x if and y = 0 when x = 2.

3 Find the function f if f ′(x) = sin 2x − sin x and f(0) = 2.

Practical applicationsExample 3The costs associated with maintaining a car increase as the car gets older. The rate of increase in maintenance costs, C, in dollars per year, for a particular car can be approximated by:

C ′(t) = 30t2 + 1000, 0 ≤ t ≤ 6where t is the age of the car in years.Find the total maintenance costs from the third to fifth year.

SolutionC′(t ) = 30t2 + 1000, 0 ≤ t ≤ 6

So, C(t ) = (30t2 + 1000)dtThe total maintenance costs from the third to fifth year is given by:

This can be calculated using a definite integral.

C ′(t ) dt = (30t2 + 1000) dt

=

= (10(5)3 + 1000(5)) − (10(3)3 + 1000(3))= 2980

The total maintenance cost is $2980.

f ′ x( ) x 3x2–=

f ′ x( ) cos x2---= f

π2---⎝ ⎠

⎛ ⎞ 1 2+=

f ′ x( ) 1 x2+

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

dydx------ 1

3 x–------------=

⌠⌡

C 5( ) C 3( )– C t( )[ ]35=

⌠⌡3

5⌠⌡3

5

10t3 1000t+[ ]3

5



325

chap

ter

8exercise 8.44 A model for the rate at which a typist types is given by words per

minute. Estimate the number of words typed during:

a the first 3 minutes.

b the 3rd minute.

5 A population of organisms grows according to the rule where P million is the number of organisms present after t days.

a If there are initially 500 million organisms, find a rule for the number present after t days.

b Find the number of days taken, correct to 2 decimal places, for the population to double.

6 A ball is thrown vertically upwards from a height of 20 metres above the ground so that t seconds after its release it is h metres above the ground. The rate of change of height of

the ball, in metres per second, is given by . Find:

a a rule for h in terms of t.

b after how many seconds, correct to the nearest second, the ball reaches its maximum height above the ground.

c the maximum height above the ground reached by the ball, correct to the nearest centimetre.

7 The rate of change of depth of water in a tidal river is modelled by , where

h is the depth of water in metres and t is the number of hours after midnight on a given day.

a Find an expression for h in terms of t, if the water is 5 metres deep at midnight.

b What are the maximum and minimum depths of the water?

c For how many hours a day is the depth of water 6 metres or more?

continueddWdt

--------- –6t2 14t 90+ +=

dPdt------ 300e0.2t=

dhdt------- 10 9.8t–=

dhdt-------

π3--- cos

πt6-----=

8.4


326


8 The rate of decay of a radioactive substance is such that N ′(t) = −e−0.05t, where N is the number of grams present after t days.

a If there are 20 grams of substance initially, find an expression for N(t).

b Find the time taken for the initial amount of substance to halve. Give your answer correct to 2 decimal places.

9 A body moves with velocity v metres per second at time t seconds, wherev = 6 − 2t, t ≥ 0. If the body starts 3 metres to the left of O, find:

a the displacement of the body after 5 seconds.

b the distance travelled by the body in the first 5 seconds.

c when and where the body is instantaneously at rest.

10 The sales of a particular product are predicted to grow at a rateS′(t) dollars per day, where S′(t) = 100et and t is the time in days since sales of the product commenced.

a How much money is expected to be made from sales over the first 8 days, correct to the nearest thousand dollars.

b On what day are the accumulated sales predicted to first exceed $150 000?

c How much money is expected to be made from sales during the 6th, 7th and 8th days, correct to the nearest thousand dollars.

11 The flow of blood in a blood vessel is slower towards the outside of the vessel and faster toward the middle. The speed of the blood is given by v(r) = k(R2 − r2), where k is a constant related to the pressure and the viscosity of the blood and the length of the blood vessel, R is the radius of the blood vessel and r is the distance of the blood from the middle of the vessel. Find the total blood flow in a vessel in terms of k if this is given by

2π (v(r) × r) dr.

12 The rate, in kilowatt hours per hour, at which electrical energy is used on a particular day

in a particular household, is given by the rule , where K is the

number of kilowatt hours used and t is the time in hours after midnight.

a Differentiate te−t and hence find an antiderivative of 15te−t.

b How many kilowatt hours, correct to 2 decimal places, are used in the household over the first 4 hours of the day?

c Find an expression that could be used to determine the number of kilowatt hours used during the first T hours of the day.

13 A concrete bridge of width 15 metres spans a river. The cross section of the bridge is the region between the graphs of y = 0.4x(10 − x) and y = 12 as shown inthe screenshot on the right. Linear measurements arein metres. Find the volume of concrete in the bridge.

⌠⌡0

R

dKdt------- 15te t– 0 t 24≤ ≤,=

[0, 10] by [−2, 14]

DowseyMathsMethodsCh08.fm 6 Saturday, January 14, 2006 3:40 PM


327

chap

ter

8Average value of a function

One day, when cleaning the glass of her new fish tank, Natalie notices that waves form at the surface of the water. The water now has peaks and troughs. Eventually the waves subside andthe water once again reaches a mean or average level.

If a function could be found to model the depth of the water in the tank when the waves are present at the surface, then the average value of this function would be the mean level to which the water returns.

The area under the curve shown with equation y = f(x) on the interval [a, b] can be

calculated using f(x) dx.

The same area could be calculated using a rectangle with width = b − a and height = average value of f over the interval [a, b].

It follows that:

f(x) dx = (b − a) × average value of f

Average value of f = f(x) dx

The average value of f is often denoted by fave.

Not level

Mean level

⌠⌡a

b

xx=a x=b

y

A1

A2

Rectangle withaverage heightof f(x) on theinterval (a, b)

Note: the averageheight of f(x) is theheight that makesA1 = A2.f(x)y =

⌠⌡a

b

1b a–------------

⌠⎮⌡a

b

Average (or mean) value of a functionIf f is a continuous function on the interval [a, b], then the average (or mean) value of f on [a, b] is given by:

f (x ) dxfave 1b a–------------=

⌠⎮⌡a

b

8.58.58.5CAS

only


328


Example 1Find the average value of the function on the given interval.a f (x ) = x2; [1, 5] b f (x ) = sin 2x; [0, π]

Solution

a fave = f (x ) dx

= x2 dx

=

=

=

b fave = f (x ) dx

= sin 2x dx

=

=

= 0

Example 2Find the value of c such that fave = f (c) for f (x ) = x2 − 4 on the interval [0, 3].

SolutionIt is first necessary to determine the average value of f on the interval [0, 3].

fave =

=

=

= −1Now, fave = f (c), so:

−1 = c2 − 43 = c2

c = But as c must be in the interval [0, 3], .

tipThis answer to part a can also be

obtained using a TI-89 by defining a function as shown below.

The graph of y = f(x) and its average value on [1, 5] is shown below.

Note that A1 = A2.

A1

A2

CAS 5.5, 10.1

1b a–------------

⌠⎮⌡a

b

15 1–------------

⌠⎮⌡1

5

14--- x3

3-----

1

5

112------ 125 1–( )

313

------

1b a–------------

⌠⎮⌡a

b

1π 0–------------

⌠⎮⌡0

π

1π--- 1

2--- cos 2x–

0

π

12π------ cos 2π cos 0–( )–

CAS 10.1, 10.2

tipThis solution can also be obtained

using the aveval function defined above.. First find the average value of f on

the interval [0, 3]. Now solve f(c) = −1 for c on the

interval [0, 3].

13---

⌠⎮⌡0

3

x2 4–( ) xd

13--- x3

3----- 4x–

0

3

13--- 27

3------ 12–⎝ ⎠

⎛ ⎞

± 3c 3=



329

chap

ter

8Example 3The number of people living in a country is given by N (t ) = 200e0.008t, where N is in millions and t is the number of years since 1985. Find the average number of people living in this country between 1985 and 1990, correct to the nearest million.

SolutionAverage number of people =

=

=

= 5000(e0.04 − 1)= 204.054

The average number of people living in the country between 1985 and 1990 was 204 million.This result can be confirmed using a TI-89 as shown above.

exercise 8.51 Find the average value of the function over the interval indicated by hand. Confirm your

answer using CAS.

a x3; [−2, 1] b 5 − 3x2; [0, 4]

c ; 0 ≤ x ≤ 16 d ; [2, 4]

e 3x + 2; [−5, 2] f cos x;

g e−2x; [−1, 1] h ; −3 ≤ x ≤ −2

i x2 − 5x + 4; [0, 5] j ; 1 ≤ x ≤ 4

2 For each of the following:

i Find the value of fave over the interval given by the domain of the function.

ii Find the value(s) of c in the given interval such that f(c) = fave.

iii Sketch the graph of f and construct a rectangle whose area is the same as the area under the graph of f over the interval.

a f : [−2, 4] → R, f(x) = 4x b f : [−1, 4] → R, f(x) = x2

c f : [−3, −1] → R, d f : [0, 1] → R, f(x) = e−x

3 The amount of drug, A milligrams, in the bloodstream at time t hours after it is administered is given by A = 2e−0.1t.

a What is the initial dose of the drug?

b What is the average amount of drug present in the bloodstream over the first 2 hours after it is administered? Give your answer correct to 2 decimal places.

15 0–------------

⌠⎮⌡0

5

N t( ) td

15---

⌠⎮⌡0

5

200e0.008t td

15--- 200

0.008--------------- e0.008t[ ]

05×

x1x---

π2--- x π≤ ≤

11 x–------------

1

x2-----

f x( ) 1

x2-----=

8.5


330


4 The air temperature in a particular suburb during a period of 12 hours is modelled byT = 12 + 3t − 0.17t2, 0 ≤ t ≤ 12, where t is measured in hours and T in degrees Celsius. Find, correct to the nearest tenth of a degree, the average temperature during:

a the first 8 hour period b the entire 12 hour period.

5 The average value of f(x) = −3x2 + 6x + 4 is equal to 3 on the interval [0, k] and k is a positive real number. Find the exact value of k.

6 The velocity of a falling object is given by v(t) = 10 + 9.8t metres per second at time t seconds after its release. Find the average velocity for 0 ≤ t ≤ 4.

7 The depth in metres of water in a tidal pool at time t hours is . What is the average depth for 0 ≤ t ≤ 10?

8 The speed at which a typist can type over a 4 minute interval is given byS(t) = 70 + 14t − 6t2, 0 ≤ t ≤ 4, where S(t) is the number of words per minute attime t minutes. Find:

a the speed at the beginning of the interval

b the maximum speed, correct to 2 decimal places

c the average speed over the interval.

9 A function y = f(x) is positive for all values of x. If the average value of the function on the interval [1, 5] is 3, what is the area under the curve on the same interval?

d 3.5 1.2 sinπt5-----+=

In an alternating current electric circuit, the current, I(t) amperes is given by the expression I(t) = A cos ωt, where A is the amplitude and ω is the frequency measured in cycles per second.a Write down an expression for the current in an average household circuit for which the

frequency is 60 cycles per second.The power in an electric circuit of this type is given by P(t) = (I(t))2.

b Write down an expression for the power in the circuit defined in question a.c Sketch a graph of the function defined in question b, showing one complete cycle.d Write down, but do not attempt to evaluate, a suitable integral to find the average

power in this circuit over a complete cycle.

e i It can be shown that .

Use this result to express cos260t in the form

, where a is a positive constant.

ii Hence evaluate the integral found in question d.

f It has been found that average power in an electric circuit is given by kA2, where k is a positive real constant and A is the amplitude. Find the value of k.

cos2 x12--- 1 cos 2x–( )=

12--- 1 cos at–( )

analysis task 3—average power in an electric circuit

SAC

8.3Average pow

er in an electric circuit

CD

SAC

analysis task



331

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8

In a dry country like Australia, farm dams are part of the lifeblood of rural life. Knowing the volume of water at any time is important for such things as planning the number and distribution of livestock, and estimating when the supply is likely to run out if there are drought conditions.

Methods of estimating the volume of a partly empty dam depend on interpreting physically observable signs. If depth markers are embedded when the dam is excavated, the volume can be estimated from the water level measured on the marker. In general, however, dams do not have markers and some other method is needed. One readily obtainable piece of data is the distance the water surface has receded from its position when the dam was full—a distance obtained by stepping out the distance taken to walk from the edge of the dam directly to the current water level.

The mathematics required in this problem involves Pythagoras’ theorem, simple trigonometry and integral calculus, with some ability to handle formulas carefully and systematically. The problem is ideally suited to the use of technology, as in a spreadsheet and graphics calculator.

modelling task—dam it!

SAC

8.4Dam

it!

CD

SAC

modelling task

8.5


Chapter review

332


SummaryGraph of an antiderivative function. f(x) = f ′(x)dx

. Graphs of antiderivatives can be sketched from graphical information alone. When interpreting the graph of a derivative:

a x-axis intercepts correspond to stationary points on the original function.

a If the graph is above the x-axis then the original function has positive gradient and therefore it is increasing as x increases.

a If the graph is below the x-axis then the original function has negative gradient and therefore it is decreasing as x increases.

Areas under and between curvesArea under a curve. When finding the area enclosed between a curve with equation y = f(x), the x-axis and the

lines x = a and x = b, the areas above and below the x-axis need to be calculated using separate integrals.

. If the required area lies entirely above the x-axis it can be calculated using f(x) dx.

. If the required area lies below the x-axis then it is calculated using − f(x) dx.

Area between curves. The area enclosed between the two curves can be determined by finding the area of the

region enclosed by the curve with equation y = f(x), the x-axis and the lines x = a andx = b, then subtracting the area of the region enclosed by the curve with equation y = g(x), the x-axis and the lines x = a and x = b.

Area = f(x) dx − g(x) dx

= (f(x) − g(x)) dx

This can be easily remembered by

Area between curves = (top curve − bottom curve) dx

Straight-line motion. Position, x, of a particle is defined as its location relative to some fixed point.

. Displacement is defined as being the change in position of a particle. Distance refers to how far a particle has travelled.

. The speed of a particle refers to how fast the particle is travelling. Velocity takes the direction of motion into account.

Average speed =

Average velocity = =

. Velocity is the rate of change of position with respect to time, i.e. . Hence x = vdt.

⌠⌡

⌠⌡a

b

⌠⌡a

b

⌠⌡a

b⌠⌡a

b

⌠⌡a

b

⌠⌡a

b


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


------------------------------------------------ displacementtime taken

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

v dxdt------= ⌠

⌡


Applications of integration 8chap

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333

Finding a function given its derivative. Antidifferentiation gives f ′(x)dx = f(x) + c. The constant c is determined from additional

information.

Average value of a function. If f is a continuous function on the interval [a, b], then the average value of f from x = a to

x = b is given by f(x) dx. The average value of f is often denoted by fave.

. fave = f(x) dx.

Revision questionsShort answer1 Sketch a graph of an antiderivative of each of the following.

a b

2 The graph of y = f ′(x) is shown. Sketch the graph of y = f(x) if f(0) = 2.


⌠⌡

1b a–------------

⌠⎮⌡a

b

1b a–------------

⌠⎮⌡a

b

x

y

2–2–4 40 x

y

2–2–4 40

x

y

y = f ′(x)

2–2–4 40

y

x

y=

2

2

4

–4

–2

–2–4 4

1

0

2x – 3


334


4 Find the exact area of the region enclosed by the curve with equation y = sin 2x + cos 2x,

the coordinate axes and the line .

5 Find, correct to 2 decimal places, the area enclosed by the curve with equation y = tan x,

the x-axis and the lines and .

6 Find the exact area of the region enclosed by the curves y = x2 and y = 4 − x2.

7 Calculate the area of the shaded region correct to3 decimal places.

8 The gradient of a curve at any point is given by the rule . If the curve

passes through the point (1, 5), find the equation of the curve.

9 Find the rule for f(x) if

a f ′(x) = 3x2 − 10x and f(4) = −9 b f ′(x) = 4ex − 2 and f(3) = 4e − 1.

10 The position x cm of an object relative to the origin at time t seconds is given by x = 2e0.5t − 1, t ≥ 0.

a State the initial position of the object.

b How far has it travelled in the first 3 seconds, correct to the nearest centimetre?

c i Find the average velocity of the object over the first 3 seconds, correct to 1 decimal place.

ii Find the velocity of the object when t = 3.

11 The velocity–time graph for a moving body is shown at right.a Find an expression for v in terms of t.b What is the velocity of the object after 10 seconds?c At what time is the body stationary?d What is the distance travelled by the object in 50 seconds?e If the body starts 10 m to the right of the origin, what is its position after 50 seconds?

12 a Find the equation of the straight line with axes intercepts (−3, 0) and (2, 0).

b Find the exact value of .

c Find the area of the shaded region.

x 5π8

------=

x π4---–= x π

3---=

x

y

0

–1

1 y=cos 2x

y=–sin x

–π –2π

2π π

dydx------ 6

3x 4–( )2----------------------=

t (s)

v (m/s)

10 20 30 40 500

10

5

–5

–10

–15

x

y

y2 = 9 − x

2 4 6 8 100

4

2

–2

–4

⌠⌡0

99 x– xd


Applications of integration 8chap

ter

335

13 a Find the exact value of (ex − 1) dx.

b Use a well scaled graph to show the areafound by the integral in part a.

c Find the derivative of (x + 1) loge (x + 1).

d Hence find the exact area of the shaded region.

e What do you notice about your results from part a and part d?

Extended response1 The entrance to a disused train tunnel of

constant cross-sectional area is shown in the diagram. The arch has the shape of a parabola.

a Find the rule for this parabola, expressingyour answer in the form y = ax2 + bx + c, where a, b and c are real numbers.

b Find the exact cross-sectional area of the tunnel.

Children like to play in this tunnel but it can be quite dangerous because of flooding in wet weather. The local council wishes to fill in the tunnel using concrete.

c If the tunnel is 8 metres long, find the volume of concrete required.

The council gets a quote for filling the tunnel with concrete and finds that it is too expensive. The engineer investigates possible alternatives and determines that access to the tunnel can be prevented if the tunnel is completely filled with concrete to a height of 3 metres and then wire mesh used to cover the remaining area at the entrance and exit.

d What volume of concrete will be required if the council agrees to the engineer’s proposal?

e What area of mesh is required?

2 The rate of change of height of a chair on a rotating Ferris wheel for small children is given

by , where h is the height of the chair above the ground at time t seconds

after the ride commences.

a Sketch a graph of against t.

b By how much does the height of the chair change in the first 15 seconds?

At the start of the ride the chair is at a height of 1 metre above the ground.

c Find an expression for h in terms of t.

d What is the maximum height above the ground reached by the chair.

e For how long is the chair above a height of 2 metres during one revolution?

x

y

0.5 1 1.5 20

1.5

1

0.5

y=loge(x+1)y=1

⌠⌡0

1

x (m)

y (m)

2 4 6 8 10 120

3

4

5

6

2

1

dhdt-------

π60------

πt60------sin=

dhdt-------


336


3 Dr CAS is running late for school. He parks his car and starts running along the footpath towards the school gate. All of a sudden he has a horrible thought; he has lefthis calculator in the car. He turns around and runs back in the direction of the car.On his way back to the car, he remembers that he packed his calculator in his bag the night before. He turns around and starts running in the direction of the school gate.Dr CAS’s velocity, v m/s, at time t seconds after he gets out of his car is given by the

rule .

a What is Dr CAS’s velocity 2 seconds after he starts running?

b At what time(s) is he stationary?

c How far has Dr CAS run in the direction of the gate before he turns around for the first time?

Dr CAS hears the school bell ring 10 seconds after he arrives and he is still not within the school grounds.

d How far has he run in this time?

e How far is he from his car?

f What is his average velocity over the 10 seconds?

g If the school gate is 15 metres from where he parked his car, how far is he from the gate when he hears the bell ring?

4 Shizuka loves to make her own bread. When she removes a loaf of pumpkin bread from the oven, its temperature is 180°C. The loaf immediately starts to cool and the rate of change

of temperature of the loaf is given by , where T is the temperature of the

bread in degrees Celsius, t is the time in minutes that the loaf has been out of the oven and k is a real constant.

a Find .

b If the bread cools to 28°C in 10 minutes, find an expression for T in terms of t, giving the value of k correct to 1 decimal place and any other constants correct to the nearest whole number.

c What is the temperature of the bread after 5 minutes? Give your answer correct to 2 decimal places.

d How long does it take for the bread to cool to 33°C? Give your answer to the nearest second.

e The temperature of the bread eventually approaches the temperature of the kitchen. What is room temperature in the kitchen?

v110------ t3 14t2 45t+–( ) 0 t 10≤ ≤,=

dTdt------- 48e kt––=

⌠⌡ 48e kt–– td

CD8.5–8.7A

trio of testsC

hapter tests


337



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