review 5.1-5.3 distance traveled area under a curve antiderivatives calculus

Review 5.1-5.3

Distance Traveled

Area Under a curve

Antiderivatives

Calculus

time

velocity

After 4 seconds, the object has gone 12 feet.

Consider an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance:

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

3t d

If the velocity is not constant,we might guess that the distance traveled is still equalto the area under the curve.

(The units work out.)

211

8V t Example:

We could estimate the area under the curve by drawing rectangles touching at their left corners.

This is called the Left-hand Rectangular Approximation Method (LRAM).

1 11

8

11

2

12

8t v

10

1 11

8

2 11

2

3 12

8Approximate area: 1 1 1 3

1 1 1 2 5 5.758 2 8 4

We could also use a Right-hand Rectangular Approximation Method (RRAM).

11

8

11

2

12

8

Approximate area: 1 1 1 31 1 2 3 7 7.75

8 2 8 4

3

211

8V t

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).

1.031251.28125

1.78125

Approximate area:6.625

2.53125

t v

1.031250.5

1.5 1.28125

2.5 1.78125

3.5 2.53125

In this example there are four subintervals.As the number of subintervals increases, so does the accuracy.

211

8V t

211

8V t

Approximate area:6.65624

t v

1.007810.25

0.75 1.07031

1.25 1.19531

1.382811.75

2.25

2.75

3.25

3.75

1.63281

1.94531

2.32031

2.75781

13.31248 0.5 6.65624

width of subinterval

With 8 subintervals:

The exact answer for thisproblem is .6.6

Circumscribed rectangles are all above the curve:

Inscribed rectangles are all below the curve:

When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.

211

8V t

subinterval

The width of a rectangle is called a subinterval.

Subintervals do not all have to be the same size.

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

(dummy variable)

It is called a dummy variable because the answer does not depend on the variable chosen.

b

af x dx

We have the notation for integration, but we still need to learn how to evaluate the integral.

time

velocity

After 4 seconds, the object has gone 12 feet.

In section 5.1, we considered an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance: 3t d

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

If the velocity varies:

11

2v t

Distance:21

4s t t

(C=0 since s=0 at t=0)

After 4 seconds:1

16 44

s

8s

1Area 1 3 4 8

2

The distance is still equal to the area under the curve!

Notice that the area is a trapezoid.

Fill in the blanks:

a)Integrating velocity gives ____________________

b)Integrating the absolute value of velocity gives ______________

c) New Position = _____________ + _______________

Displacement

Total Distance

Initial Disp. Total Distance

Definition: A function F is called an antiderivative of f on an interval I if F’(x) = f(x) for all x in I.

Example: Let f(x)=x3. If F(x) =1/4 * x4 then F’(x) = f(x)

Theorem: If F is an antiderivative of f on an interval I, then the most generalantiderivative of f on I is

F(x) + Cwhere C is an arbitrary constant.

Antiderivatives

F is an antiderivative of f

Table of particular antiderivatives

Function Antiderivative Function Antiderivative

xr, r -1 xr+1/(r+1) 1/x ln |x|

sin x - cos x ex ex

cos x sin x ax ax / ln a

sec2 x tan x tan-1 x

sin-1 x 2

1

1 x

2

1

1 x

These rules give particular antiderivatives of the listed functions. A general antiderivative can be obtained by adding a constant .

Integrate

Rewriting before integrating

Find the general solution of the equation F’(x) = and

find the particular solution given the point F(1) = 0.

Now plug in (1,0) and solve for C.

0 = -1 + C

C = 1

Final answer.

Exploration

Finding Antiderivatives For each of the following derivatives, describe the original

function .F

(a) '( ) 2F x x

(b) '( )F x x

2(c) '( )F x x

2

1(d) '( )F x

x

3

1(e) '( )F x

x

(f) '( ) cosF x x

2( )F x x

2( ) / 2F x x

3( ) / 3F x x

( ) 1/F x x

2( ) 1/ 2F x x

( ) sinF x x

2( ) 3f x x

Find the function that is the antiderivative of .F f

3( ) + C becauseF x x3

2[ ]3

d x Cx

dx

2

Here are some members of the

family of antiderivatives of ( ) 3 :f x x

31( ) 5F x x 3

2 ( ) 36F x x ...etc

33( )

3

xF x C

2[ ] 2xD x x

The family of all antiderivatives of ( ) 2 is

represented by

f x x

2( )F x x C

Find the general solution of the differential equation

' 2y 2y x C

:Notation

( )y f x dx ( )F x C

Integrand

Variable of

Integration

Constant of

Integration

is the antiderivative (indefinite integral) of ( )

with respect to .

y f x

x

3xdx 3 xdx 2

32

xC

23

2x C

RewriteIntegrate

Simplify

3

1dx

x 3x dx 2

1

2C

x

2

2

xC

xdx 1/ 2x dx 3 / 2

3/ 2

xC

3 / 22

3x C

2sin xdx 2 sin xdx 2 ( cos )x C

2cos x C

( 2)x dx 2

1 222

xC x C

2xdx dx

2

22

xx C

4 2(3 5 )x x x dx May we skip step 2?????

5 3 23 5 1

5 3 2x x x C

5 3 2

3 55 3 2

x x xC

1xdx

x

1/ 2 1/ 2( )x x dx 3 / 2 1/ 2

3/ 2 1/ 2

x xC 3 / 2 1/ 22

23

x x C

Never integrate the numerator and denominator seperately!

Rewrite the Integrand

2 5 3 2x xdx

x

5/2 3/2 1/212 2220

5 3x x x C

2

sin

cos

xdx

x

1 sin

cos cos

xdx

x x

sec tanx xdx sec x C

????

Find the general solution of

2

1'( ) , 0f x F x x

x

2( )F x x dx1

1

xC

1

Cx

Find the particular solution that

satisfies the initial condition

(1) 0F 1

(1)1

F C 0

1C

1( ) 1, 0F x x

x

1F x C

x

Solve the differential equation given the initial condition.

3 1, 2 3f x x f

231

2f x x x

A ball is thrown upward with an initial velocity of

64 feet per second from an initial height of 80 feet.

(a) Find the position function giving the height as a function of the time .s t

0 initial timet

Given Initial Conditions:Acceleration due to gravity: 32 feet per second per second

''( ) 32s t

'(0) 64s (0) &80s

80 ft

(a) Find the position function giving the height as a function of the time .s t

0 initial timet Given Conditions:Acceleration due to gravity: 32 feet per second per second

''( ) 32s t

'(0) 64s (0) &80s

'( )s t ''( ) 32s t dt dt 132t C

164 32(0) C 1 64C

'( ) 32 64s t t

(a) Find the position function giving the height as a function of the time .s t

( ) '( ) ( 32 64)s t s t dt t dt 2

232 642

tt C

2216 64t t C Using (0) 80:s

2

280 16 0 64 0 C 2 80C 2( ) 16 64 80s t t t

(b) When does the ball hit the ground?

2( ) 64 816 0s t t t 0216( 4 5) 0t t

16( 5)( 1) 0t t

5 secondst

80 ft

5t

Basic Integration Rules

These two equations allow you to obtain integration formulas directly from differentiation formulas, as shown in the following summary.

Basic Integration Rulescon

t’d

Before you begin the exercise set, be sure you realize

that one of the most important steps in integration is

rewriting the integrand

in a form that fits the basic integration rules.

Practice Exercises

Original Rewrite Integrate Simplify

2dx

x

22 1t dt

3

2

3xdx

x

3 4x x dx

Computing AntiderivativesProblem

2

cos 3Let f . Find antiderivatives of the function f.

12

xx

x

Solution

2

An antiderivative of the function cos is sin

1and an antiderivative of is arctan .

1

x x

xx sin

Hence an antiderivative of the given function is 3arctan .2

xx

Analyzing the motion of an object using antiderivatives

A particle is moving with the given data. Find the position of the particle.a(t) = 10 + 3t -3t2, v(0) = 2, s(0) = 5

v(t) is the antiderivative of a(t): v’(t) = a(t) = 10 + 3t -3t2

Antidifferentiation gives v(t) = 10t + 1.5t2 – t3 + Cv(0) = 2 implies that C=2; thus, v(t) = 10t + 1.5t2 – t3 + 2

s(t) is the antiderivative of v(t): s’(t) = v(t) = 10t + 1.5t2 – t3 + 2Antidifferentiation gives s(t) = 5t2 + 0.5t3 – 0.25t4 + 2t + Ds(0) = 5 implies that D=5; thus, s(t) = 5t2 + 0.5t3 – 0.25t4 + 2t + 5

Problem

Solution

Some indefinite integrals

1

xx

2 2

1 + C , if n -1. ln(| |)

1 x

a, a

ln(a)

sin( ) cos( ) , cos( ) sin( ) ,

sec ( ) tan( ) , csc ( )

nn

x x

xx dx dx x C

n

e dx e C dx C

x dx x C x dx x C

x dx x C x dx

1 12 2

cot( ) ,

sec( ) tan( ) sec( ) , csc( ) cot( ) csc( ) ,

1 1tan ( ) , sin ( ) ,

1 1

x C

x x dx x C x x dx x C

dx x C dx x Cx x