6 Antiderivatives and the Rules of Integration Integration by Substitution Area and the Definite...

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Antiderivatives and the Rules of IntegrationAntiderivatives and the Rules of Integration Integration by SubstitutionIntegration by Substitution Area and the Definite IntegralArea and the Definite Integral The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus Evaluating Definite IntegralsEvaluating Definite Integrals Area Between Two CurvesArea Between Two Curves Applications of the Definite Integral to Applications of the Definite Integral to

Business and EconomicsBusiness and Economics

IntegrationIntegration

6.16.1Antiderivatives and the Rules of IntegrationAntiderivatives and the Rules of Integration

–1 1 2 3

5

4

3

2

1

–1

x

y

AntiderivativesAntiderivatives

Recall the Recall the MaglevMaglev problem discussed in problem discussed in chapter 2chapter 2. . The question asked then was: The question asked then was:

✦ If we know the If we know the positionposition of the maglev at any of the maglev at any timetime tt, , can we find its can we find its velocityvelocity at that time? at that time?

✦ The The positionposition was described by was described by ff((tt)), and the , and the velocityvelocity by by ff ′′((tt)). .

Now, in Now, in Chapters 6 and 7Chapters 6 and 7 we will consider precisely we will consider precisely the the opposite problemopposite problem::✦ If we know the If we know the velocityvelocity of the maglev at any of the maglev at any timetime tt, ,

can we find its can we find its positionposition at that time? at that time?

✦ That is, knowing its That is, knowing its velocity functionvelocity function ff ′′((tt)), can we , can we find its find its position functionposition function ff((tt))??

AntiderivativesAntiderivatives

To solve this kind of problems, we need the concept of the To solve this kind of problems, we need the concept of the antiderivative of a function.antiderivative of a function.

✦ A function A function FF is an is an antiderivativeantiderivative of of ff on an on an interval interval II if if FF ′′((tt)) == ff((tt)) for all of for all of tt in in II..

ExamplesExamples

LetLet

Show that Show that FF is an is an antiderivativeantiderivative of of

SolutionSolution DifferentiatingDifferentiating the function the function FF, we obtain , we obtain

and the and the desired resultdesired result follows. follows.

3 213( ) 2 1F x x x x 3 213( ) 2 1F x x x x

2( ) 4 1f x x x 2( ) 4 1f x x x

2( ) 4 1 ( )F x x x f x 2( ) 4 1 ( )F x x x f x

Example 1, page 398

ExamplesExamples

Let Let FF((xx) = ) = xx, , GG((xx) = ) = x x + 2+ 2, , HH((xx) = ) = x x ++ C C, where , where CC is a is a constant.constant.

Show that Show that FF, , GG, and , and HH are all are all antiderivativesantiderivatives of the of the function function f f defined by defined by ff((xx) = 1) = 1..

SolutionSolution SinceSince

we see that we see that FF, , GG, and , and HH are indeed are indeed antiderivativesantiderivatives of of ff..

( ) ( ) 1 ( )

( ) ( 2) 1 ( )

( ) ( ) 1 ( )

dF x x f x

dxd

G x x f xdxd

H x x C f xdx

( ) ( ) 1 ( )

( ) ( 2) 1 ( )

( ) ( ) 1 ( )

dF x x f x

dxd

G x x f xdxd

H x x C f xdx

Example 2, page 398

Theorem 1Theorem 1

Let Let GG be an be an antiderivativeantiderivative of a function of a function ff.. Then, every antiderivative Then, every antiderivative FF of of ff must be must be

of the form of the form

FF((xx) = ) = GG((xx) + ) + CC

where where CC is a constant. is a constant.

ExampleExample

ProveProve that the function that the function GG((xx) = ) = xx22 is an is an antiderivativeantiderivative of the of the function function ff((xx) = 2) = 2xx..

Write a general expression for the antiderivatives of Write a general expression for the antiderivatives of ff..

SolutionSolution Since Since GG′′((xx) = 2) = 2x x == f f((xx)), we have shown that , we have shown that GG((xx) = ) = xx22 is an is an

antiderivative of antiderivative of ff((xx) = 2) = 2xx.. By By Theorem 1Theorem 1, , every antiderivativeevery antiderivative of the function of the function

ff((xx) = 2) = 2xx has the form has the form FF((xx) = ) = xx22 + + CC, where , where CC is a is a constantconstant..

Example 3, page 399

The Indefinite IntegralThe Indefinite Integral

The process of finding all the antiderivatives of a function The process of finding all the antiderivatives of a function is called is called antidifferentiationantidifferentiation or or integrationintegration..

We use the symbol We use the symbol ∫∫, called an , called an integral signintegral sign, to indicate , to indicate that the operation of integration is to be performed on that the operation of integration is to be performed on some function some function ff. .

Thus,Thus,

where where CC and and KK are arbitrary constants. are arbitrary constants.

21 2dx x C x dx x K and 21 2dx x C x dx x K and

Basic Integration RulesBasic Integration Rules

Rule 1:Rule 1: The Indefinite Integral of a Constant The Indefinite Integral of a Constant

( , a constant) kkdx kx C ( , a constant) kkdx kx C

ExampleExample

Find each of the following indefinite integrals:Find each of the following indefinite integrals:

a.a. b.b.

SolutionSolution Each of the integrals had the Each of the integrals had the formform ff((xx) = ) = kk, where , where kk is a is a

constantconstant.. Applying Applying Rule 1Rule 1 in each case yields: in each case yields:

a.a. b.b.

2dx2dx 2dx 2dx

2 2dx x C 2 2dx x C 2 2dx x C 2 2dx x C

Example 4, page 400


From the From the rule of differentiationrule of differentiation, ,

we obtain the following we obtain the following rule of integrationrule of integration::

✦ Rule 2:Rule 2: The Power Rule The Power Rule

1n ndx nx

dx 1n nd

x nxdx

11( 1)

1 n nx dx x C n

n

11( 1)

1 n nx dx x C n

n

ExamplesExamples

Find the indefinite integral:Find the indefinite integral:

3x dx 3x dx 41

4x C 41

4x C

Example 5, page 401

ExamplesExamples


3/2x dx 3/2x dx 5/2

52

5/2

1

2

5

x C

x C

5/2

52

5/2

1

2

5

x C

x C

Example 5, page 401

ExamplesExamples


3/2

1dx

x 3/2

1dx

x3/2

1/2

12

1/2

1/2

1

2

2

x dx

x C

x C

Cx

3/2

1/2

12

1/2

1/2

1

2

2

x dx

x C

x C

Cx

Example 5, page 401


✦ Rule 3:Rule 3: The Indefinite Integral of a Constant The Indefinite Integral of a Constant Multiple of a FunctionMultiple of a Function

where where cc is a constant. is a constant.

( ) ( )cf x dx c f x dx ( ) ( )cf x dx c f x dx

ExamplesExamples


32t dt 32t dt 3

4

4

4

2

12

4

12

21

2

t dt

t K

t K

t C

3

4

4

4

2

12

4

12

21

2

t dt

t K

t K

t C

Example 6, page 402

ExamplesExamples


23x dx 23x dx 2

1

3

3( 1)

3

x dx

x C

Cx

2

1

3

3( 1)

3

x dx

x C

Cx

Example 6, page 402


✦ Rule 4:Rule 4: The Sum Rule The Sum Rule

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

f x g x dx f x dx g x dx


( ) ( ) ( ) ( )

( ) ( ) ( ) ( )



ExamplesExamples


5 3/2 1/23 4 2x x x dx 5 3/2 1/23 4 2x x x dx 5 3/2 1/2

5 3/2 1/2

3 4 2

3 4 2

x dx x dx x dx

x dx x dx x dx

5 3/2 1/2

5 3/2 1/2

3 4 2

3 4 2

x dx x dx x dx

x dx x dx x dx

6 5/2 1/2

6 5/2 1/2

1 23 4 2 2

6 5

1 84

2 5

x x x C

x x x C

6 5/2 1/2

6 5/2 1/2

1 23 4 2 2

6 5

1 84

2 5

x x x C

x x x C

Example 7, page 402


✦ Rule 5:Rule 5: The Indefinite Integral of the The Indefinite Integral of the Exponential FunctionExponential Function

x xe dx e C x xe dx e C

ExamplesExamples


3(2 )xe x dx 3(2 )xe x dx 3

3

4

2

2

12

4

x

x

x

e dx x dx

e dx x dx

e x C

3

3

4

2

2

12

4

x

x

x

e dx x dx

e dx x dx

e x C

Example 8, page 403


✦ Rule 6:Rule 6: The Indefinite Integral of the FunctionThe Indefinite Integral of the Function ff((xx) = ) = xx–1–1

1 1ln ( 0) x dx dx x C x

x 1 1

ln ( 0) x dx dx x C xx

ExamplesExamples


2

3 42x dx

x x 2

3 42x dx

x x 2

2

3 42

12 3 4

xdx dx dxx x

xdx dx x dxx

2

2

3 42

12 3 4

xdx dx dxx x

xdx dx x dxx

2 1

2

12 3ln 4( 1)

2

43ln

x x x C

x x Cx

2 1

2

12 3ln 4( 1)

2

43ln

x x x C

x x Cx

Example 9, page 403

Differential EquationsDifferential Equations

Given the Given the derivative of a functionderivative of a function, , ff ′′, can we find the , can we find the functionfunction ff ??

Consider the function Consider the function ff ′(′(xx)) = 2= 2xx – 1 – 1 from which we want from which we want to find to find ff((xx))..

We can find We can find ff by by integratingintegrating the equation: the equation:

where where CC is an is an arbitrary constantarbitrary constant.. Thus, Thus, infinitely many functionsinfinitely many functions have the derivative have the derivative ff ′′, each , each

differing from the other by a differing from the other by a constantconstant..

2( ) (2 1)f x dx x dx x x C 2( ) (2 1)f x dx x dx x x C


EquationEquation ff ′(′(xx)) = 2= 2xx – 1 – 1 is called a is called a differential equationdifferential equation.. In general, a differential equation involves In general, a differential equation involves the derivative the derivative

of an unknown functionof an unknown function.. A A solutionsolution of a differential equation is of a differential equation is any function that any function that

satisfies the differential equationsatisfies the differential equation.. For the case of For the case of ff ′(′(xx)) = 2= 2xx – 1 – 1, we find that , we find that ff((xx)) = = xx22 – – xx + + CC

gives gives allall the solutionsthe solutions of the differential equation, and it is of the differential equation, and it is therefore called the therefore called the general solutiongeneral solution of the differential of the differential equation. equation.

Differential Equations

Different values of C yield different functions f(x).

But all these functions have the same slope for any given value of x.

For example, for any value of C, we always find that f ′(1) = 1.

–1 1 2 3

5

4

3

2

1

–1f ′(1) = 1

f(x) = x2 – x – 1

x

y

f ′(1) = 1

f(x) = x2 – x + 0

f ′(1) = 1

f(x) = x2 – x + 1

f ′(1) = 1

f(x) = x2 – x + 2

f ′(1) = 1

f(x) = x2 – x + 3


It is possible to obtain a It is possible to obtain a particular solutionparticular solution by specifying by specifying the value the function must assume for a given value of the value the function must assume for a given value of xx..

For example, suppose we know the function For example, suppose we know the function f f must pass must pass through the pointthrough the point (1, 2)(1, 2), which means , which means ff(1) = 2(1) = 2..

Using this condition on the general solution we can find Using this condition on the general solution we can find the value of the value of CC::

ff(1) = 1(1) = 122 – 1 + – 1 + CC = 2= 2

CC = 2= 2 Thus, the Thus, the particular solutionparticular solution is is

ff((xx) = ) = xx22 – – xx + 2 + 2

(1, 2)

Differential Equations

Here is the graph of the Here is the graph of the particular solution particular solution of of ff when when CC = 2 = 2..

Note that this graph Note that this graph does go throughdoes go through the the point point (1, 2)(1, 2). .

–1 1 2 3

5

4

3

2

1

–1

x

y

f(x) = x2 – x + 2

Initial Value ProblemsInitial Value Problems

The problem we just discussed is of a type called The problem we just discussed is of a type called initial value probleminitial value problem..

In this type of problem we are required to find a In this type of problem we are required to find a function satisfying function satisfying

1.1. A differential equation.A differential equation.

2.2. One or more initial conditions.One or more initial conditions.

ExampleExample

Find the function Find the function ff if it is known that if it is known that

SolutionSolution IntegratingIntegrating the function the function ff ′′, we find, we find

2( ) 3 4 8 (1) 9 andf x x x f 2( ) 3 4 8 (1) 9 andf x x x f

2

3 2

( ) ( )

(3 4 8)

2 8

f x f x dx

x x dx

x x x C

2

3 2

( ) ( )

(3 4 8)

2 8

f x f x dx

x x dx

x x x C

Example 10, page 404

ExampleExample

Find the function Find the function ff if it is known that if it is known that

SolutionSolution Using the Using the conditioncondition ff(1) = 9(1) = 9, we have, we have

Therefore, the Therefore, the required functionrequired function ff is is

2( ) 3 4 8 (1) 9 andf x x x f 2( ) 3 4 8 (1) 9 andf x x x f

3 2( ) 2 8f x x x x C 3 2( ) 2 8f x x x x C 3 29 (1) (1) 2(1) 8(1)

9 7

2

f C

C

C

3 29 (1) (1) 2(1) 8(1)

9 7

2

f C

C

C

3 2( ) 2 8 2f x x x x 3 2( ) 2 8 2f x x x x

Example 10, page 404

Applied Example:Applied Example: Velocity of Maglev Velocity of Maglev

In a test run of a maglev, data obtained from reading its In a test run of a maglev, data obtained from reading its speedometer indicate that the velocity of the maglev at speedometer indicate that the velocity of the maglev at time time tt can be described by the can be described by the velocity functionvelocity function

Find the Find the position functionposition function of the maglev. of the maglev. Assume that initially the maglev is located at the Assume that initially the maglev is located at the originorigin of of

a coordinate line.a coordinate line.

( ) 8 (0 30) v t t t ( ) 8 (0 30) v t t t

Applied Example 11, page 404


SolutionSolution Let Let ss((tt)) denote the denote the positionposition of the maglev at any given of the maglev at any given

time time tt (0 (0 tt 30) 30). Then, . Then, ss′′((tt) = ) = vv((tt)). . So, we have the So, we have the initial value probleminitial value problem

IntegratingIntegrating the function the function ss′′, we find, we find

( ) 8

(0) 0

s t t

s

( ) 8

(0) 0

s t t

s

2

( ) ( )

8

4

s t s t dt

tdt

t C

2

( ) ( )

8

4

s t s t dt

tdt

t C



SolutionSolution Using the Using the conditioncondition ss(0) = 0(0) = 0, we have, we have

Therefore, the Therefore, the required functionrequired function ss is is

2( ) 4s t t C 2( ) 4s t t C

0 (0) 4(0)

0 0

0

s C

C

C

0 (0) 4(0)

0 0

0

s C

C

C

2( ) 4 (0 30) s t t t 2( ) 4 (0 30) s t t t


6.26.2Integration by SubstitutionIntegration by Substitution

2 2

1 1 1 1 1( )

3 1 3 1 6 6

xdx xdx du du

x x u u

2 2

1 1 1 1 1( )

3 1 3 1 6 6

xdx xdx du du

x x u u

5 5 52(2 4) (2 4) (2 )x dx x dx u du 5 5 52(2 4) (2 4) (2 )x dx x dx u du

3 1 1

3 3x u ue dx e du e du

3 1 1


Integration by SubstitutionIntegration by Substitution

The method of substitution is related to the The method of substitution is related to the chain rulechain rule for differentiating functions.for differentiating functions.

It is a It is a powerful toolpowerful tool for integrating a large class of for integrating a large class of functions.functions.

How the Method of Substitution WorksHow the Method of Substitution Works

Consider the indefinite integralConsider the indefinite integral

One way to solve this integral is to One way to solve this integral is to expand the expressionexpand the expression and integrate the resulting integrand term by term.and integrate the resulting integrand term by term.

An An alternative approachalternative approach simplifiessimplifies the integral by making the integral by making a a change of variablechange of variable..

Write Write uu = 2= 2xx + 4 + 4

with differential with differential dudu = 2= 2dxdx

52(2 4)x dx 52(2 4)x dx


SubstituteSubstitute uu = 2 = 2xx + 4 + 4 andand dudu = 2 = 2dxdx in the in the original expressionoriginal expression::

Now it’s easy to Now it’s easy to integrateintegrate::

ReplacingReplacing u u by by uu = 2 = 2xx + 4 + 4, we obtain:, we obtain:

5 5 52(2 4) (2 4) (2 )x dx x dx u du 5 5 52(2 4) (2 4) (2 )x dx x dx u du

5 61

6u du u C 5 61

6u du u C

5 6

6

12(2 4)

61

(2 4)6

x dx u C

x C

5 6

6

12(2 4)

61

(2 4)6

x dx u C

x C


We can We can verifyverify the result by the result by finding its derivativefinding its derivative::

The derivative is indeed The derivative is indeed the original integrand expressionthe original integrand expression..

6 5

5

1 1(2 4) 6 (2 4) (2)

6 6

2(2 4)

dx C x

dx

x

6 5

5

1 1(2 4) 6 (2 4) (2)

6 6

2(2 4)

dx C x

dx

x

The Method of Integration by SubstitutionThe Method of Integration by Substitution

Step 1Step 1 Let Let uu = = g g((xx)), where , where gg((xx)) is part of the is part of the integrandintegrand, , usually the “inside function” of the composite usually the “inside function” of the composite function function ff((gg((xx))))..

Step 2Step 2 Find Find dudu = = gg′′((xx))dxdx..

Step 3Step 3 Use the Use the substitutionsubstitution uu = = g g((xx)) and and dudu = = gg′′((xx))dxdx to to convert the convert the entireentire integralintegral into one involving into one involving onlyonly uu..

Step 4Step 4 EvaluateEvaluate the resulting the resulting integrandintegrand..

Step 5Step 5 ReplaceReplace uu by by gg((xx)) to obtain the to obtain the final solutionfinal solution as as a function of a function of xx..

ExamplesExamples

Find Find

SolutionSolution

Step 1Step 1 The integrand involves the The integrand involves the composite functioncomposite function

with “inside function” with “inside function”

So, we chooseSo, we choose

2 42 ( 3)x x dx 2 42 ( 3)x x dx

2 4( 3)x 2 4( 3)x

2( ) 3g x x 2( ) 3g x x

2 3u x 2 3u x

Example 1, page 413

ExamplesExamples

Find Find

SolutionSolution

Step 2Step 2 Find Find du/dxdu/dx and and solvesolve for for dudu::

2 42 ( 3)x x dx 2 42 ( 3)x x dx

2 3u x 2 3u x

2

2

dux

dxdu xdx

2

2

dux

dxdu xdx

Example 1, page 413

ExamplesExamples

Find Find

SolutionSolution

Step 3Step 3 SubstituteSubstitute uu = = x x22 + 3 + 3 and and dudu = = 22xdxxdx, to obtain an , to obtain an integral involving integral involving onlyonly uu: :

2 42 ( 3)x x dx 2 42 ( 3)x x dx

2 4 2 4

4

2 ( 3) ( 3) (2 )x x dx x xdx

u du

2 4 2 4

4

2 ( 3) ( 3) (2 )x x dx x xdx

u du

Example 1, page 413

ExamplesExamples

Find Find

SolutionSolution

Step 4Step 4 Evaluate Evaluate the integral: the integral:

Step 5Step 5 ReplaceReplace uu by by x x22 + 3 + 3 to find the to find the solutionsolution::

2 42 ( 3)x x dx 2 42 ( 3)x x dx

4 51

5u du u C 4 51

5u du u C

2 4 2 512 ( 3) ( 3)

5x x dx x C 2 4 2 51

2 ( 3) ( 3)5

x x dx x C

Example 1, page 413

ExamplesExamples

FindFind

SolutionSolution Let Let u u = –3= –3xx, so that , so that dudu = –3 = –3dxdx, or , or dx dx = –= – ⅓⅓ dudu.. SubstituteSubstitute to express the integrand in terms of to express the integrand in terms of uu::

EvaluateEvaluate the integral: the integral:

ReplaceReplace uu by by –3–3xx to find the to find the solutionsolution::

3xe dx 3xe dx

3 1 1


3 1 1


1 1

3 3u ue du e C

1 1

3 3u ue du e C

3 31

3x xe dx e C 3 31

3x xe dx e C

Example 4, page 414

ExamplesExamples

FindFind

SolutionSolution Let Let u u = 3= 3xx22 + 1+ 1, so that , so that dudu = 6 = 6xdxxdx, or , or xdx xdx = = dudu.. SubstituteSubstitute to express the integrand in terms of to express the integrand in terms of uu::


ReplaceReplace uu by by 33xx22 + 1+ 1 to find the to find the solutionsolution::

23 1

xdx

x 23 1

xdx

x

2 2

1 1 1 1 1( )

3 1 3 1 6 6

xdx xdx du du

x x u u

2 2

1 1 1 1 1( )

3 1 3 1 6 6

xdx xdx du du

x x u u

1 1 1ln

6 6 du u C

u

1 1 1ln

6 6 du u C

u

22

1ln 3 1

3 1 6

xdx x C

x

22

1ln 3 1

3 1 6

xdx x C

x

1

6

Example 5, page 415

ExamplesExamples

FindFind

SolutionSolution Let Let u u = ln = ln xx, so that , so that dudu = 1/ = 1/xx dxdx, or , or dx/x dx/x = = dudu.. SubstituteSubstitute to express the integrand in terms of to express the integrand in terms of uu::


ReplaceReplace uu by by ln ln xx to find the to find the solutionsolution::

2(ln )

2

xdx

x2(ln )

2

xdx

x

22 2(ln ) 1 1

(ln ) ( )2 2 2

x dxdx x u du

x x

22 2(ln ) 1 1

(ln ) ( )2 2 2

x dxdx x u du

x x

2 3 31 1 1 1

2 2 3 6u du u C u C 2 3 31 1 1 1

2 2 3 6u du u C u C

23(ln ) 1

(ln )2 6

xdx x C

x

23(ln ) 1

(ln )2 6

xdx x C

x

Example 6, page 415

6.36.3Area and the Definite IntegralArea and the Definite Integral

1 2 3( )b

aA f x dx R R R 1 2 3( )

b

aA f x dx R R R

x

y

( )y f x ( )y f x

a b

R1

R2

R3

The Area Under the Graph of a FunctionThe Area Under the Graph of a Function

Let Let ff be a nonnegative continuous function on be a nonnegative continuous function on [[aa, , bb]]. . Then, the Then, the area of the region under the grapharea of the region under the graph of of ff is is

where where xx11, , xx22, … , , … , xxnn are arbitrary points in the are arbitrary points in the nn

subintervals of subintervals of [[aa, , bb]] of equal width of equal width xx = ( = (bb – – aa)/)/nn..

1 2lim ( ) ( ) ... ( )nnA f x f x f x x

1 2lim ( ) ( ) ... ( )nn

A f x f x f x x

The Definite IntegralThe Definite Integral

LetLet ff be a continuous function defined onbe a continuous function defined on [[aa, , bb]].. If If

exists for all choices of representative pointsexists for all choices of representative points xx11, , xx22, … , , … , xxnn in in

thethe nn subintervals ofsubintervals of [[aa, , bb]] of equal widthof equal width xx = ( = (bb – – aa)/)/nn, then , then the limit is called the the limit is called the definite integraldefinite integral of of ff fromfrom aa toto bb and is and is denoted bydenoted by

Thus,Thus,

The numberThe number aa is the is the lower limit of integrationlower limit of integration, and the , and the numbernumber bb is the is the upper limit of integrationupper limit of integration..

1 2lim ( ) ( ) ... ( )nnf x x f x x f x x

1 2lim ( ) ( ) ... ( )nn

f x x f x x f x x

1 2( ) lim ( ) ( ) ... ( )b

na nf x dx f x x f x x f x x

1 2( ) lim ( ) ( ) ... ( )

b

na nf x dx f x x f x x f x x

( )b

af x dx ( )

b

af x dx

Integrability of a FunctionIntegrability of a Function

Let Let ff be a be a continuouscontinuous on on [[aa, , bb]].. Then, Then, ff is is integrableintegrable on on [[aa, , bb]];; that is, the definite integral that is, the definite integral

exists.exists.

( )b

af x dx ( )

b

af x dx

Geometric Interpretation of the Definite IntegralGeometric Interpretation of the Definite Integral

If If ff is nonnegative and integrable on is nonnegative and integrable on [[aa, , bb]],, then then

is equal to the is equal to the areaarea of the of the region under the graphregion under the graph of of ff on on [[aa, , bb]]..

( )b

af x dx ( )

b

af x dx


The The definite integraldefinite integral is equal to the is equal to the areaarea of the of the region region under the graphunder the graph of of ff on on [[aa, , bb]]::

( )b

aA f x dx ( )

b

aA f x dx

xx

yy

( )y f x ( )y f x

aa bb


If If ff is continuous on is continuous on [[aa, , bb]],, then then

is equal to the is equal to the areaarea of the of the region above region above [[aa, , bb]] minusminus thethe region below region below [[aa, , bb]]..

( )b

af x dx ( )

b

af x dx


The The definite integraldefinite integral is equal to the is equal to the areaarea of the of the region region above above [[aa, , bb]] minusminus thethe region below region below [[aa, , bb]]::

1 2 3( )b

aA f x dx R R R 1 2 3( )

b

aA f x dx R R R

xx

yy

( )y f x ( )y f x

aa bb

RR11

RR22

RR33

6.46.4The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus

–2 –1 1 2

5

4

3

2

1

x

y

f(x) = x2 + 1

R

Theorem 2Theorem 2The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus

Let Let ff be be continuouscontinuous on on [[aa, , bb]].. Then, Then,

where where FF is any is any antiderivativeantiderivative of of ff; that is, ; that is, FF ′′((xx)) == f f((xx))..

( ) ( ) ( )b

af x dx F b F a ( ) ( ) ( )

b

af x dx F b F a

ExampleExample

Let Let RR be the be the region under the graphregion under the graph of of ff((xx) =) = x x on the on the interval interval [1, 3][1, 3]..

Use the Use the fundamental theorem of calculusfundamental theorem of calculus to find the area to find the area A A of of RR and verify your result by elementary means. and verify your result by elementary means.

Example 1, page 431

ExampleExample

SolutionSolution The The graphgraph shows the shows the regionregion to be evaluated. to be evaluated. Since Since ff is is nonnegativenonnegative on on [1, 3][1, 3], the , the areaarea of of RR is given by is given by

the the definite integraldefinite integral of of ff from from 11 to to 33..

xx

( )y f x x ( )y f x x

11 22 33 44

RRx x = 3= 3

x x = 1= 1

44

33

22

11

yy

Example 1, page 431

ExampleExample

SolutionSolution By the By the fundamental theorem of calculusfundamental theorem of calculus, we have, we have

Thus, the Thus, the areaarea AA of of regionregion RR is is 44 square unitssquare units.. Note that the Note that the constant of integrationconstant of integration CC dropped outdropped out..

This is true in general.This is true in general.

3 32

11

1

29 1

2 2

4

A xdx x C

C C

3 32

11

1

29 1

2 2

4

A xdx x C

C C

Example 1, page 431

44

33

22

11

ExampleExample

SolutionSolution Using elementary means, note that Using elementary means, note that areaarea AA is equal is equal

to to RR11 (base (base ☓☓ height) height) plus plus RR22 (( ☓☓ base base ☓☓ height) height)..

Thus, Thus, AA = = RR11 + + RR22 = 2(1) + = 2(1) + ((2)2)((2) = 42) = 4

xx

yy

( )y f x x ( )y f x x

11 22 33 44

RR11

22

22

RR22

1

2

1

2

Example 1, page 431

––22 –1 –1 11 22

55

44

33

22

ExampleExample Find the Find the areaarea of the of the region under the graphregion under the graph of of yy = = xx22 + 1 + 1

from from xx = –1 = –1 to to xx = 2 = 2. .

SolutionSolution Note below that the Note below that the full regionfull region RR under consideration under consideration lies lies

above theabove the xx axisaxis..

xx

yyff((xx)) = = xx22 + 1+ 1

RR

Example 3, page 432

ExampleExample Find the Find the areaarea of the of the region under the graphregion under the graph of of yy = = xx22 + 1 + 1

from from xx = –1 = –1 to to xx = 2 = 2. .

SolutionSolution Using the Using the fundamental theorem of calculusfundamental theorem of calculus, we find that , we find that

the the required area required area isis

or or 66 square unitssquare units..

22 2 3

11

3 3

1( 1)

3

1 1(2) (2) ( 1) ( 1)

3 3

8 12 1

3 36

x dx x x

2

2 2 3

11

3 3

1( 1)

3

1 1(2) (2) ( 1) ( 1)

3 3

8 12 1

3 36

x dx x x

Example 3, page 432

Net Change FormulaNet Change Formula

The The net changenet change in a function in a function f f over an over an intervalinterval [[aa, , bb]] is given byis given by

provided provided f f ′′ is is continuouscontinuous on on [[aa, , bb]]..

( ) ( ) ( )b

af b f a f x dx ( ) ( ) ( )

b

af b f a f x dx

Applied Example:Applied Example: Population Growth in Clark County Population Growth in Clark County

Clark County, Nevada, (dominated by Clark County, Nevada, (dominated by Las VegasLas Vegas) is the ) is the fastest growing metropolitan areafastest growing metropolitan area in the United States. in the United States.

From From 19701970 through through 20002000, the , the populationpopulation was was growinggrowing at a at a raterate of of

people per decadepeople per decade, where , where tt = 0 = 0 corresponds to the corresponds to the beginning of beginning of 19701970..

What was the What was the net change in populationnet change in population over the decade over the decade from from 19801980 to to 19901990??

2( ) 133,680 178,788 234,633 (0 3)R t t t t 2( ) 133,680 178,788 234,633 (0 3)R t t t t


Applied Example:Applied Example: Population Growth in Clark County Population Growth in Clark County

SolutionSolution The The net change in populationnet change in population over the decade from over the decade from 19801980 to to

19901990 is given by is given by PP(2)(2) –– PP(1)(1) , where , where PP denotes the denotes the population in the county at time population in the county at time tt..

But But PP ′′ = R= R, and so the , and so the net change in populationnet change in population is is2 2

1 1(2) (1) ( ) ( )P P P t dt R t dt

2 2

1 1(2) (1) ( ) ( )P P P t dt R t dt

2 2

1

23 2

1

(133,680 178,788 234,633)

44,560 89,394 234,633

t t dt

t t t

2 2

1

23 2

1

(133,680 178,788 234,633)

44,560 89,394 234,633

t t dt

t t t

3 2

3 2

[44,560(2) 89,394(2) 234,633(2)]

[44,560(1) 89,394(1) 23,4633(1)]

278,371

3 2

3 2

[44,560(2) 89,394(2) 234,633(2)]

[44,560(1) 89,394(1) 23,4633(1)]

278,371


Applied Example:Applied Example: Assembly Time of Workers Assembly Time of Workers

An efficiency study conducted for Elektra Electronics An efficiency study conducted for Elektra Electronics showed that the showed that the raterate at which Space Commander walkie- at which Space Commander walkie-talkies are talkies are assembledassembled by the by the average workeraverage worker tt hours hours after after starting workstarting work at at 8:00 a.m.8:00 a.m. is given by the function is given by the function

Determine Determine how manyhow many walkie-talkies can be walkie-talkies can be assembledassembled by by the average worker the average worker in the first hourin the first hour of the morning shift. of the morning shift.

2( ) 3 12 15 (0 4) f t t t t 2( ) 3 12 15 (0 4) f t t t t


Applied Example:Applied Example: Assembly Time of Workers Assembly Time of Workers

SolutionSolution Let Let NN((tt)) denote the number of denote the number of walkie-talkies assembledwalkie-talkies assembled by by

the average worker the average worker tt hours after starting workhours after starting work in the in the morning shift.morning shift.

Then, we haveThen, we have

Therefore, the Therefore, the number of units assemblednumber of units assembled by the average by the average worker in the worker in the first hourfirst hour of the morning shift is of the morning shift is

or or 2020 unitsunits..

1 1 2

0 0(1) (0) ( ) ( 3 12 15)N N N t dt t t dt

1 1 2

0 0(1) (0) ( ) ( 3 12 15)N N N t dt t t dt

13 2

0( 6 15 ) 1 6 15

20

t t t

13 2

0( 6 15 ) 1 6 15

20

t t t

2( ) ( ) 3 12 15N t f t t t 2( ) ( ) 3 12 15N t f t t t


6.56.5Evaluating Definite IntegralsEvaluating Definite Integrals

–2 –1 1 2

3

2

1

x

y

f(x) = e(1/2)x

R

Properties of the Definite IntegralProperties of the Definite Integral

Let Let ff and and gg be be integrable functionsintegrable functions, then, then

1.1.

2.2.

3.3.

4.4.

5.5.

( ) 0a

af x dx ( ) 0

a

af x dx

( ) ( )b a

a bf x dx f x dx ( ) ( )

b a

a bf x dx f x dx

( ) ( ) ( , constant )b b

a acf x dx c f x dx c a ( ) ( ) ( , constant )

b b

a acf x dx c f x dx c a

( ) ( ) ( ) ( )b b b

a a af x g x dx f x dx g x dx ( ) ( ) ( ) ( )

b b b

a a af x g x dx f x dx g x dx

( ) ( ) ( ) ( ) b c b

a a cf x dx f x dx f x dx a c b ( ) ( ) ( ) ( )

b c b

a a cf x dx f x dx f x dx a c b

Examples: Examples: Using the Method of SubstitutionUsing the Method of Substitution

Evaluate Evaluate

SolutionSolution First, find the First, find the indefinite integralindefinite integral::

✦ Let Let uu = 9 + = 9 + xx22 so that so that

4 2

09x x dx

4 2

09x x dx

29I x x dx 29I x x dx

2(9 )

2

1

2

ddu x dx

dxxdx

xdx du

2(9 )

2

1

2

ddu x dx

dxxdx

xdx du

Example 1, page 442


Evaluate Evaluate

SolutionSolution First, find the First, find the indefinite integralindefinite integral::

✦ Then, integrate by Then, integrate by substitution substitution using using xdxxdx = = dudu::

2

1/2

9

1 1

2 2

I x x dx

udu u du

2

1/2

9

1 1

2 2

I x x dx

udu u du

3/2

2 3/2

1

31

(9 )3

u C

x C

3/2

2 3/2

1

31

(9 )3

u C

x C

4 2

09x x dx

4 2

09x x dx

29I x x dx 29I x x dx 1

2

Example 1, page 442


Evaluate Evaluate

SolutionSolution Using the Using the resultsresults, we evaluate the , we evaluate the definite integraldefinite integral::

4 2

09x x dx

4 2

09x x dx

44 2 2 3/2

00

19 (9 )

3x x dx x

44 2 2 3/2

00

19 (9 )

3x x dx x

2 3/2 2 3/2

23

1[(9 (4) ) (9 (0) ) ]

31

(125 27)398

332

2 3/2 2 3/2

23

1[(9 (4) ) (9 (0) ) ]

31

(125 27)398

332

Example 1, page 442


Evaluate Evaluate

SolutionSolution Let Let uu = = xx33 + 1 + 1 so thatso that

21

30 1

xdx

x 21

30 1

xdx

x

3

2

2

( 1)

3

1

3

ddu x dx

dx

x dx

du x dx

3

2

2

( 1)

3

1

3

ddu x dx

dx

x dx

du x dx

Example 3, page 444


Evaluate Evaluate

SolutionSolution Find the Find the lowerlower and and upperupper limits limits of integration of integration with with

respect torespect to uu::✦ When When xx = 0 = 0, the , the lower limitlower limit is is uu = (0) = (0)33 + 1 = 1+ 1 = 1..

✦ When When xx = 1 = 1, the , the upper limitupper limit is is uu = (1) = (1)33 + 1 = 2+ 1 = 2..

SubstituteSubstitute xx22dxdx = = dudu, along with the , along with the limits of integrationlimits of integration::21 1 2 22

3 30 0 1 1

1 1 1 1 1

1 1 3 3

xdx x dx du du

x x u u

21 1 2 22

3 30 0 1 1

1 1 1 1 1

1 1 3 3

xdx x dx du du

x x u u

2

1

1 1 1ln (ln 2 ln1) ln 2

3 3 3u

2

1

1 1 1ln (ln 2 ln1) ln 2

3 3 3u

21

30 1

xdx

x 21

30 1

xdx

x

Example 3, page 444


Find the Find the areaarea of the region of the region RR under the graphunder the graph of of

ff((xx) = ) = ee(1/2)(1/2)xx from from xx = –1 = –1 to to xx = 1 = 1. .

SolutionSolution The graph shows region The graph shows region RR: :

––22 –1 –1 11 22

33

22

11

xx

yy

ff((xx) = ) = ee(1/2)(1/2)xx

RR

Example 4, page 444




SolutionSolution Since Since ff((xx) ) is always is always greater than zerogreater than zero, the , the areaarea is given by is given by

To evaluate this integral, we To evaluate this integral, we substitutesubstitute

so thatso that

1 (1/2)

1

xA e dx

1 (1/2)

1

xA e dx

1

22

du dx

du dx

1

22

du dx

du dx

1

2u x

1

2u x

Example 4, page 444




SolutionSolution When When xx = –1 = –1 , , uu = = – – , and when , and when xx = 1 = 1, , uu = = .. SubstituteSubstitute dxdx = 2 = 2dudu, along with the , along with the limits of integrationlimits of integration::

or approximately or approximately 2.08 2.08 square unitssquare units..

1 1/2 1/2(1/2)

1 1/2 1/22 2x u uA e dx e du e du

1 1/2 1/2(1/2)

1 1/2 1/22 2x u uA e dx e du e du

1/2 1/2 1/2

1/22 2( )

2.08

ue e e

1/2 1/2 1/2

1/22 2( )

2.08

ue e e

1

2

1

2

Example 4, page 444

Average Value of a FunctionAverage Value of a Function

Suppose Suppose f f is integrable on is integrable on [[aa, , bb]]. . Then, the Then, the average valueaverage value of of ff over over [[aa, , bb]] is is

1( )

b

af x dx

b a 1

( )b

af x dx

b a

Applied Example:Applied Example: Automobile Financing Automobile Financing

The The interest ratesinterest rates changed by Madison Finance on auto changed by Madison Finance on auto loans for used cars over a certain loans for used cars over a certain 66-month period-month period in in 20082008 are approximated by the functionare approximated by the function

where where tt is measured in is measured in monthsmonths and and rr((tt)) is the is the annual annual percentage ratepercentage rate..

What is the average rate on auto loans extended by What is the average rate on auto loans extended by Madison over the Madison over the 66-month period-month period??

3 21 7( ) 3 12 (0 6)

12 8r t t t t t 3 21 7( ) 3 12 (0 6)

12 8r t t t t t


Applied Example:Applied Example: Automobile Financing Automobile Financing

SolutionSolution The The average rateaverage rate over the over the 66-month period-month period is given by is given by

or or 9%9% per yearper year..

6 3 2

0

6

4 3 2

0

4 3 2

1 1 73 12

6 0 12 8

1 1 7 312

6 48 24 2

1 1 7 3(6) (6) (6) 12(6)

6 48 24 2

9

t t t dx

t t t t

6 3 2

0

6

4 3 2

0

4 3 2

1 1 73 12

6 0 12 8

1 1 7 312

6 48 24 2

1 1 7 3(6) (6) (6) 12(6)

6 48 24 2

9

t t t dx

t t t t


6.66.6Area Between Two CurvesArea Between Two Curves

5

3

1

y

y = x3 – 3x + 3R1

R2

– 3 – 1 1 2 3x

(2 , 5)y = x + 3

(– 2, 1)

(0 , 3)

The Area Between Two CurvesThe Area Between Two Curves

Let Let f f and and gg be continuous functions such that be continuous functions such that ff((xx) ) g g((xx)) on the interval on the interval [[aa, , bb]]. .

Then, the area of the region bounded above by Then, the area of the region bounded above by yy = = ff((xx) ) and below by and below by yy = = gg((xx)) on on [[aa, , bb]] is is given bygiven by

( ) ( )b

af x g x dx ( ) ( )

b

af x g x dx

ExamplesExamples

Find the Find the areaarea of the region bounded by the of the region bounded by the xx-axis-axis, the , the graphgraph of of yy = – = –xx22 + 4 + 4xx – 8 – 8, and the , and the lineslines xx = –1 = –1 and and xx = 4 = 4..

SolutionSolution The The regionregion RR is being is being bounded abovebounded above by the graph by the graph ff((xx) = 0) = 0

and and belowbelow by the graph of by the graph of gg((xx) =) = yy = – = –xx22 + 4 + 4xx – 8 – 8 on on [–1, 4][–1, 4]::

–– 22 22 44 66

–– 44

–– 88

–– 1212

xx

yy

yy = –= –xx22 + 4 + 4xx – 8 – 8

RRxx = 4= 4xx = –= – 11

Example 1, page 454

ExamplesExamples

Find the Find the areaarea of the region bounded by the of the region bounded by the xx-axis-axis, the , the graphgraph of of yy = – = –xx22 + 4 + 4xx – 8 – 8, and the , and the lineslines xx = –1 = –1 and and xx = 4 = 4..

SolutionSolution Therefore, the Therefore, the areaarea of of RR is given by is given by

4 2

1

4 2

1

( ) ( ) 0 ( 4 8)

( 4 8)

b

af x g x dx x x dx

x x dx

4 2

1

4 2

1

( ) ( ) 0 ( 4 8)

( 4 8)

b

af x g x dx x x dx

x x dx

43 2

1

3 2 3 2

23

12 8

3

1 1(4) 2(4) 8(4) ( 1) 2( 1) 8( 1)

3 3

31

x x x

43 2

1

3 2 3 2

23

12 8

3

1 1(4) 2(4) 8(4) ( 1) 2( 1) 8( 1)

3 3

31

x x x

Example 1, page 454

ExamplesExamples

Find the Find the areaarea of the region of the region boundedbounded by by ff((xx) = 2) = 2xx – 1 – 1, , gg((xx) =) = xx22 – 4 – 4,, xx = 1 = 1, and , and xx = 2 = 2..

SolutionSolution Note that the graph of Note that the graph of ff always lies always lies aboveabove that of that of gg for all for all xx

in the in the intervalinterval [1, 2][1, 2]::

–– 44 – – 2 2 22 44

44

22

–– 22

–– 44

xx

yy

yy = = xx22 – 4 – 4

RR

yy = 2= 2xx – 1 – 1

xx = 2= 2

xx = 1= 1Example 2, page 454

ExamplesExamples

Find the Find the areaarea of the region of the region boundedbounded by by ff((xx) = 2) = 2xx – 1 – 1, , gg((xx) =) = xx22 – 4 – 4,, xx = 1 = 1, and , and xx = 2 = 2..

SolutionSolution Since the graph of Since the graph of ff always lies always lies aboveabove that of that of gg for all for all xx in in

the the intervalinterval [1, 2][1, 2], the required area is given by, the required area is given by

2 2

1( ) ( ) (2 1) ( 4)

b

af x g x dx x x dx

2 2

1( ) ( ) (2 1) ( 4)

b

af x g x dx x x dx

3 2 3 21 1(2) (2) 3(2) (1) (1) 3(1)

3 3

11

3

3 2 3 21 1(2) (2) 3(2) (1) (1) 3(1)

3 3

11

3

22 2 3 2

11

1( 2 3) 3

3x x dx x x x

22 2 3 2

11

1( 2 3) 3

3x x dx x x x

Example 2, page 454

ExamplesExamples Find the Find the areaarea of the region that is of the region that is completely enclosed completely enclosed by by

the graphsthe graphs ofof f f((xx) = 2) = 2xx – 1 – 1 and and gg((xx) =) = xx22 – 4 – 4..SolutionSolution First, find the First, find the points of intersectionpoints of intersection of the of the two curvestwo curves.. To do this, you can set To do this, you can set gg((xx) = ) = ff((xx)) and solve for and solve for xx::

so, the graphs intersect at so, the graphs intersect at xx = – = – 11 and at and at xx = 3 = 3..

2

2

4 2 1

2 3 0

( 1)( 3) 0

x x

x x

x x

2

2

4 2 1

2 3 0

( 1)( 3) 0

x x

x x

x x

Example 3, page 455

ExamplesExamples Find the Find the areaarea of the region that is of the region that is completely enclosed completely enclosed by by

the graphsthe graphs ofof f f((xx) = 2) = 2xx – 1 – 1 and and gg((xx) =) = xx22 – 4 – 4..SolutionSolution The graph of The graph of ff always lies always lies aboveabove that of that of gg for all for all xx in the in the

intervalinterval [–[– 1, 3]1, 3] between the two between the two intersection pointsintersection points::

– 4 – 2 2 4

4

2

– 4

x

y

yy = = xx22 – 4 – 4

R

yy = 2= 2xx – 1 – 1

(– 1 , – 3)

(3 , 5)

Example 3, page 455

ExamplesExamples

Find the Find the areaarea of the region that is of the region that is completely enclosed completely enclosed by by the graphsthe graphs ofof f f((xx) = 2) = 2xx – 1 – 1 and and gg((xx) =) = xx22 – 4 – 4..


the the intervalinterval [–[– 1, 3]1, 3], the required area is given by, the required area is given by

3 2

1( ) ( ) (2 1) ( 4)

b

af x g x dx x x dx

3 2

1( ) ( ) (2 1) ( 4)

b

af x g x dx x x dx

3 2 3 21 1(3) (3) 3(3) ( 1) ( 1) 3( 1)

3 3

210

3

3 2 3 21 1(3) (3) 3(3) ( 1) ( 1) 3( 1)

3 3

210

3

33 2 3 2

11

1( 2 3) 3

3x x dx x x x

3

3 2 3 2

11

1( 2 3) 3

3x x dx x x x

Example 3, page 455

ExamplesExamples

Find the Find the areaarea of the region of the region boundedbounded by by ff((xx) =) = xx22 – 2 – 2x x – 1– 1,, gg((xx) = –) = – eexx – 1– 1, , xx = – = – 11, and , and xx = 1 = 1..

SolutionSolution Note that the graph of Note that the graph of ff always lies always lies aboveabove that of that of gg for all for all xx

in the in the intervalinterval [–[– 1, 1]1, 1]::

–– 33 – – 2 2 22 33

33

22

11

–– 33

xx

yy

yy = = xx22 – 2 – 2x x – 1– 1

RR

yy = –= – eexx – 1– 1

xx = –= – 11

xx = 1= 1

Example 4, page 455

ExamplesExamples

Find the Find the areaarea of the region of the region boundedbounded by by ff((xx) =) = xx22 – 2 – 2x x – 1– 1,, gg((xx) = –) = – eexx – 1– 1, , xx = – = – 11, and , and xx = 1 = 1..


the the intervalinterval [–[– 1, 1]1, 1], the required area is given by, the required area is given by

1 2

1( ) ( ) ( 2 1) ( 1)

b x

af x g x dx x x e dx

1 2

1( ) ( ) ( 2 1) ( 1)

b x

af x g x dx x x e dx

3 2 (1) 3 2 ( 1)1 1(1) (1) ( 1) ( 1)

3 3

2 13.02

3

e e

ee

3 2 (1) 3 2 ( 1)1 1(1) (1) ( 1) ( 1)

3 3

2 13.02

3

e e

ee

11 2 3 2

11

1( 2 )

3x xx x e dx x x e

1

1 2 3 2

11

1( 2 )

3x xx x e dx x x e

Example 4, page 455

ExamplesExamples Find the Find the areaarea of the region of the region boundedbounded by by

ff((xx) =) = xx33,, thethe x x-axis-axis, , xx = – = – 11, and , and xx = 1 = 1..SolutionSolution The region being considered is composed of The region being considered is composed of

two subregionstwo subregions RR11 and and RR22::

1

1

– 1

x

yyy = = xx33

R1

xx = –= – 11

xx = 1= 1

R2– 1

Example 5, page 456


ff((xx) =) = xx33,, thethe x x-axis-axis, , xx = – = – 11, and , and xx = 1 = 1..SolutionSolution To find To find RR11 and and RR22 consider the consider the xx-axis-axis as as gg((xx) = 0) = 0. . SinceSince g g((xx) ) ff((xx)) on on [–[– 1, 0]1, 0], the , the areaarea of of RR11 is given by is given by

1

1

– 1

x

yyy = = xx33

R1

xx = –= – 11

xx = 1= 1

– 1

1

0 03 3

1 1

( ) ( )

(0 )

b

aR g x f x dx

x dx x dx

1

0 03 3

1 1

( ) ( )

(0 )

b

aR g x f x dx

x dx x dx

04

1

1 1 10

4 4 4x

04

1

1 1 10

4 4 4x

Example 5, page 456


ff((xx) =) = xx33,, thethe x x-axis-axis, , xx = – = – 11, and , and xx = 1 = 1..SolutionSolution To find To find RR11 and and RR22 consider the consider the xx-axis-axis as as gg((xx) = 0) = 0. . SinceSince g g((xx) ) ff((xx)) on on [0, 1][0, 1], the , the areaarea of of RR22 is given by is given by

11

11

–– 11

xx

yyyy = = xx33

xx = –= – 11

xx = 1= 1

RR22–– 11

2

1 13 3

0 0

( ) ( )

( 0)

b

aR f x g x dx

x dx x dx

2

1 13 3

0 0

( ) ( )

( 0)

b

aR f x g x dx

x dx x dx

14

0

1 1 10

4 4 4x

14

0

1 1 10

4 4 4x

Example 5, page 456

RR11


ff((xx) =) = xx33,, thethe x x-axis-axis, , xx = – = – 11, and , and xx = 1 = 1..SolutionSolution Therefore, the required area Therefore, the required area RR is is

square units.square units.

11

11

–– 11

xx

yyyy = = xx33

xx = –= – 11

xx = 1= 1

RR22–– 11

1 2

1 1 1

4 4 2R R R 1 2

1 1 1

4 4 2R R R

Example 5, page 456

Find the Find the areaarea of the region of the region boundedbounded by by ff((xx) =) = xx33 – 3 – 3xx + 3 + 3 and and gg((xx) =) = xx + 3 + 3..

SolutionSolution The region The region RR being considered is composed of being considered is composed of

two subregionstwo subregions RR11 and and RR22::

ExamplesExamples

55

33

11

yy

yy = = xx33 – 3 – 3xx + 3 + 3

yy = = xx + 3 + 3

–– 33 – – 1 1 11 22 33xx

RR11

RR22

Example 6, page 457


SolutionSolution To find the To find the points of intersectionpoints of intersection, we , we solve simultaneouslysolve simultaneously

the equations the equations yy = = xx33 – 3 – 3xx + 3 + 3 and and yy = = xx + 3 + 3..

ExamplesExamples

55

33

11

yy

yy = = xx33 – 3 – 3xx + 3 + 3 RR11

RR22

3

3

3 3 3

4 0

( 2)( 2) 0

x x x

x x

x x x

3

3

3 3 3

4 0

( 2)( 2) 0

x x x

x x

x x x

–– 33 – – 1 1 11 22 33xx

So, So, xx = 0 = 0, , xx = – = – 22, and , and x x = 2= 2.. The The points of intersectionpoints of intersection

of the of the two curvestwo curves are are (–(– 2, 1)2, 1), , (0, 3)(0, 3), and , and (2, 5)(2, 5). .

(2 , 5)(2 , 5)yy = = xx + 3 + 3

(–(– 2, 1)2, 1)

(0 , 3)(0 , 3)

Example 6, page 457


SolutionSolution Note that Note that ff((xx) ) gg((xx) ) for for [–[– 2, 0]2, 0], so the area of , so the area of

regionregion RR11 is given by is given by

ExamplesExamples

55

33

11

yy

yy = = xx33 – 3 – 3xx + 3 + 3 RR11

–– 33 – – 1 1 11 22 33xx

(2 , 5)(2 , 5)yy = = xx + 3 + 3

(–(– 2, 1)2, 1)

(0 , 3)(0 , 3)

1

0 3

2

0 3

2

( ) ( )

( 3 3) ( 3)

( 4 )

b

aR f x g x dx

x x x dx

x x dx

1

0 3

2

0 3

2

( ) ( )

( 3 3) ( 3)

( 4 )

b

aR f x g x dx

x x x dx

x x dx

04 2

2

12

4

(4 8) 4

x x

04 2

2

12

4

(4 8) 4

x x

Example 6, page 457


SolutionSolution Note that Note that gg((xx) ) ff((xx) ) for for [0, 2][0, 2], so the area of , so the area of

regionregion RR22 is given by is given by

ExamplesExamples

55

33

11

yy

yy = = xx33 – 3 – 3xx + 3 + 3

RR22

–– 33 – – 1 1 11 22 33xx

(2 , 5)(2 , 5)yy = = xx + 3 + 3

(–(– 2, 1)2, 1)

(0 , 3)(0 , 3)

2

2 3

0

2 3

0

( ) ( )

( 3) ( 3 3)

( 4 )

b

aR g x f x dx

x x x dx

x x dx

2

2 3

0

2 3

0

( ) ( )

( 3) ( 3 3)

( 4 )

b

aR g x f x dx

x x x dx

x x dx

24 2

0

12

4

4 8 4

x x

24 2

0

12

4

4 8 4

x x

Example 6, page 457


SolutionSolution Therefore, the required area Therefore, the required area RR is is

square units.square units.

ExamplesExamples

55

33

11

yy

yy = = xx33 – 3 – 3xx + 3 + 3 RR11

RR22

–– 33 – – 1 1 11 22 33xx

(2 , 5)(2 , 5)yy = = xx + 3 + 3

(–(– 2, 1)2, 1)

(0 , 3)(0 , 3)

1 2 4 4 8R R R 1 2 4 4 8R R R

Example 6, page 457

6.76.7Applications of the Definite Integral Applications of the Definite Integral to Business and Economicsto Business and Economics

100 200 300 400 500x

p

250

200

150

100

50

CS

PS

2( ) 0.001 250p D x x 2( ) 0.001 250p D x x

2( ) 0.0006 .02 100p S x x x 2( ) 0.0006 .02 100p S x x x

p =p = 160

Consumers’ and Producers’ SurplusConsumers’ and Producers’ Surplus

Suppose Suppose p p = = DD((xx)) is the is the demand functiondemand function that relates the that relates the priceprice p p of a commodity to the of a commodity to the quantityquantity xx demanded of it. demanded of it.

Now suppose a unit market Now suppose a unit market priceprice pp has been establishedhas been established, ,

along with a corresponding along with a corresponding quantity demandedquantity demanded xx.. Those consumers who Those consumers who would be willing to paywould be willing to pay a unit price a unit price

higherhigher than than pp for the commodity would in effect for the commodity would in effect experience a experience a savingssavings..

This difference between what the consumer This difference between what the consumer would bewould be willingwilling to payto pay and what they and what they actuallyactually havehave to payto pay is called is called the the consumers’ surplusconsumers’ surplus..


To To derive a formuladerive a formula for computing the for computing the consumers’ consumers’ surplussurplus, divide the interval , divide the interval [0,[0,xx]] into into nn subintervals, each subintervals, each of length of length xx = =x/nx/n, and denote the right endpoints of these , and denote the right endpoints of these intervals by intervals by xx11, , xx22, …, , …, xxnn = =xx::

pp p p ==pp

xx

pp

xx11 xx22 xxnn –1 –1 xxnn xx33 xx44

DD((xx11))

DD((xx22))DD((xx33))DD((xx44))

DD((xxnn –1–1))DD((xxnn) =) =pp

rr11


There are consumers who would pay a price of There are consumers who would pay a price of at leastat least DD((xx11)) for the for the firstfirst xx units units instead of the market priceinstead of the market price of of pp..

The The savingssavings to these consumers is to these consumers is approximatedapproximated by by

which is the which is the areaarea of the of the rectanglerectangle rr11::

p p ==pp

xx

pp

DD((xx11))

xx11

1 1( ) [ ( ) ]D x x p x D x p x 1 1( ) [ ( ) ]D x x p x D x p x

pp

rr11


Similarly, the Similarly, the savingssavings the consumer experiences for the the consumer experiences for the consecutive incrementsconsecutive increments of of xx are depicted by the are depicted by the areasareas of of rectanglesrectangles rr22, , rr33, , rr44, … , , … , rrnn: :

rr22

rrnn –1 –1

p p ==pp

xx

pp

rr33 rr44rrnn

xxnn

pp

xx


Adding Adding rr11 + + rr22 + + rr33 + … + + … + rrnn, and letting , and letting nn approach approach infinityinfinity, we obtain the , we obtain the consumers’ surplusconsumers’ surplus CSCS formula: formula:

where where DD((xx)) is the is the demand functiondemand function,, pp is the unit is the unit market market priceprice, and, and xx is the is the quantity demandedquantity demanded..

0( ) x

CS D x dx p x 0 ( ) xCS D x dx p x

xx

pp

ppp = Dp = D((xx))

CSCS


Similarly, we can derive a Similarly, we can derive a formulaformula for the for the producers’ producers’ surplussurplus..

Suppose Suppose p p = = SS((xx)) is the is the supply functionsupply function that relates the that relates the priceprice p p of a commodity to the of a commodity to the quantityquantity xx suppliedsupplied of it. of it.

Again, suppose a unit market Again, suppose a unit market priceprice pp has been establishedhas been established, ,

along with a corresponding along with a corresponding quantity suppliedquantity supplied xx.. Those sellers who Those sellers who would be willing to sellwould be willing to sell at unit price at unit price

lowerlower than than pp for the commodity would in effect experience for the commodity would in effect experience a a gaingain or or profitprofit..

This difference between what the seller This difference between what the seller would bewould be willingwilling to to sell forsell for and what they and what they actuallyactually cancan sell for sell for is called the is called the producers’ surplusproducers’ surplus..

xx


Geometrically, the producers’ surplus is given by the area Geometrically, the producers’ surplus is given by the area

of the region bounded above the straight line of the region bounded above the straight line p p ==pp and and

below the supply curve below the supply curve p p = = SS((xx)) from from xx = 0 = 0 to to x x ==xx: :

xx

pp

pp

p = Sp = S((xx))

PSPS

Consumers’ and Producers’ SurplusConsumers’ and Producers’ Surplus The The producers’ surplusproducers’ surplus PSPS is given by is given by

where where SS((xx)) is the is the supply functionsupply function,, pp is the unit is the unit market market priceprice, and, and xx is the is the quantity suppliedquantity supplied..

0( ) x

PS p x S x dx 0 ( ) xPS p x S x dx

xx

xx

pp

pp

p = Sp = S((xx))

PSPS

ExampleExample

The The demand functiondemand function for a certain make of for a certain make of 10-speed 10-speed bicyclebicycle is given by is given by

where where pp is the unit is the unit priceprice in dollars and in dollars and xx is the is the quantity quantity demandeddemanded in units of a thousand. in units of a thousand.

The The supply functionsupply function for these bicycles is given by for these bicycles is given by

where where pp stands for the stands for the priceprice in dollars and in dollars and xx stands for the stands for the number of bicycles that the number of bicycles that the suppliersupplier will want to sell. will want to sell.

DetermineDetermine the the consumers’ surplusconsumers’ surplus and the and the producers’ producers’ surplussurplus if the market price of a bicycle is set at the if the market price of a bicycle is set at the equilibrium priceequilibrium price..

2( ) 0.001 250p D x x 2( ) 0.001 250p D x x

2( ) 0.0006 .02 100p S x x x 2( ) 0.0006 .02 100p S x x x

Example 1, page 466

ExampleExample

SolutionSolution To find the To find the equilibrium pointequilibrium point, equate , equate SS((xx)) and and DD((xx)) to solve to solve

the the system of equationssystem of equations and find the and find the point of intersectionpoint of intersection of of the the demand demand and and supply supply curves:curves:

Thus, Thus, xx = – 625/2 = – 625/2 or or xx = 300 = 300. The . The first numberfirst number is is discardeddiscarded for being for being negativenegative, so the , so the solutionsolution is is xx = 300 = 300..

2 2

2

0.0006 .02 100 0.001 250

0.0016 .02 150 0

x x x

x x

2 2

2

0.0006 .02 100 0.001 250

0.0016 .02 150 0

x x x

x x

2

2

16 200 1,500,000 0

2 25 187,500 0

x x

x x

2

2

16 200 1,500,000 0

2 25 187,500 0

x x

x x

(2 625)( 300) 0x x (2 625)( 300) 0x x

Example 1, page 466

ExampleExample

SolutionSolution SubstituteSubstitute xx = 300 = 300 to find the equilibrium value of to find the equilibrium value of pp::

Thus, the Thus, the equilibrium pointequilibrium point is is (300, 160)(300, 160). . That is, the That is, the equilibrium quantityequilibrium quantity is is 300,000300,000 bicycles, and bicycles, and

the the equilibrium priceequilibrium price is is $160$160 per bicycle. per bicycle.

20.001(300) 250 160p 20.001(300) 250 160p

Example 1, page 466

ExampleExample

SolutionSolution To find theTo find the consumers’ surplusconsumers’ surplus, we, we set set xx = 300 = 300 and and p p = 160= 160

in the consumers’ surplus formula:in the consumers’ surplus formula:

or or $18,000,000$18,000,000..

300 2

0( 0.001 250) (160)(300)x dx

300 2

0( 0.001 250) (160)(300)x dx

300

3

0

1250 48,000

3000x x

300

3

0

1250 48,000

3000x x

3300250(300) 48,000 18,000

3000

3300250(300) 48,000 18,000

3000

0( ) x

CS D x dx p x 0 ( ) xCS D x dx p x

Example 1, page 466

ExampleExample

SolutionSolution To find theTo find the producers’ surplusproducers’ surplus, we, we set setxx = 300 = 300 and andp p = 160= 160

in the producers’ surplus formula:in the producers’ surplus formula:

or or $11,700,000$11,700,000..

300 2

0(160)(300) (0.0006 0.02 100)x x dx

300 2

0(160)(300) (0.0006 0.02 100)x x dx

3003 2

048,000 (0.0002 0.01 100 )x x x

3003 2

048,000 (0.0002 0.01 100 )x x x

3 248,000 [0.0002(300) 0.01(300) 100(300)]

11,700

3 248,000 [0.0002(300) 0.01(300) 100(300)]

11,700

0( ) x

PS p x S x dx 0 ( ) xPS p x S x dx

Example 1, page 466

100100 200200 300300 400400 500500

ExampleExample

SolutionSolution Consumers’ surplusConsumers’ surplus and and producers’ surplusproducers’ surplus when the when the

market is in market is in equilibriumequilibrium::

xx

pp

250250

200200

150150

100100

5050

CSCS

PSPS

2( ) 0.001 250p D x x 2( ) 0.001 250p D x x

2( ) 0.0006 .02 100p S x x x 2( ) 0.0006 .02 100p S x x x

p p ==p p = 160= 160

Example 1, page 466

Accumulated or Total Future Value Accumulated or Total Future Value of an Income Streamof an Income Stream

The The accumulatedaccumulated, or total, , or total, future valuefuture value after after TT years years of an of an income streamincome stream of of RR((tt)) dollars per year, earning dollars per year, earning interest rateinterest rate of of rr per year per year compounded continuouslycompounded continuously, , is given byis given by

0( )

TrT rtA e R t e dt 0 ( )TrT rtA e R t e dt

Applied Example:Applied Example: Income Stream Income Stream

Crystal Car Wash recently bought an automatic car-Crystal Car Wash recently bought an automatic car-washing machine that is expected to generate washing machine that is expected to generate $40,000$40,000 in in revenuerevenue per yearper year, , t t years from now, for the next years from now, for the next 55 yearsyears..

If the income is If the income is reinvestedreinvested in a business earning in a business earning interestinterest at at the rate of the rate of 12%12% per year per year compounded continuouslycompounded continuously, find , find the the total accumulated valuetotal accumulated value of this income stream at the of this income stream at the end of end of 55 yearsyears..


Applied Example:Applied Example: Income Stream Income Stream

SolutionSolution We are required to find the We are required to find the total future valuetotal future value of the given of the given

income streamincome stream after after 55 yearsyears.. Setting Setting RR((tt)) = 40,000= 40,000, , rr = 0.12 = 0.12, and , and TT = 5 = 5 in the in the

accumulated income stream accumulated income stream formula we getformula we get

or approximately or approximately $274,040$274,040..

0( )

TrT rtA e R t e dt 0 ( )TrT rtA e R t e dt

50.12(5) 0.12

040,000 te e dt

50.12(5) 0.12

040,000 te e dt

5

0.6 0.12

0

40,000

0.12te e

5

0.6 0.12

0

40,000

0.12te e

0.60.640,000

( 1) 274,039.600.12

ee

0.60.640,000

( 1) 274,039.600.12

ee


Present Value of an Income StreamPresent Value of an Income Stream

The The present valuepresent value of an of an income streamincome stream of of RR((tt)) dollars in a year, earning dollars in a year, earning interestinterest at the rate of at the rate of rr per year per year compounded continuouslycompounded continuously, is given by, is given by

0( )

T rtPV R t e dt0 ( )T rtPV R t e dt

Applied Example:Applied Example: Investment Analysis Investment Analysis

The owner of a local cinema is considering The owner of a local cinema is considering two alternative two alternative plansplans for renovating and improving the theater. for renovating and improving the theater.

Plan APlan A calls for an calls for an immediate cash outlayimmediate cash outlay of of $250,000$250,000, , whereas whereas plan Bplan B requires an requires an immediate cash outlayimmediate cash outlay of of $180,000$180,000..

It has been estimated that adopting It has been estimated that adopting plan Aplan A would result in a would result in a net net income stream generatedincome stream generated at the rate of at the rate of

ff((tt) = 630,000) = 630,000dollars per year, whereas adopting dollars per year, whereas adopting plan Bplan B would result in a would result in a net net income stream generatedincome stream generated at the rate of at the rate of

gg((tt) = 580,000) = 580,000for the next for the next three yearsthree years..

If the prevailing If the prevailing interest rateinterest rate for the next for the next five yearsfive years is is 10%10% per year, per year, which planwhich plan will generate a will generate a higher net incomehigher net income by the by the end of end of year year 33??

Applied Example:Applied Example: Investment Analysis Investment AnalysisSolutionSolution We can find the We can find the present valuepresent value of the of the net incomenet income NI NI for for

plan Aplan A setting setting RR((tt) = 630,000) = 630,000, , rr = 0.1 = 0.1, and , and TT = 3 = 3, using the , using the present value formula:present value formula:

or approximately or approximately $1,382,845$1,382,845..

0( ) 250,000

T rtNI R t e dt 0 ( ) 250,000T rtNI R t e dt 3 0.1

0630,000 250,000te dt

3 0.1

0630,000 250,000te dt

30.1

0

630,000250,000

0.1te

30.1

0

630,000250,000

0.1te

0.36,300,000 6,300,000 250,000

1,382,845

e

0.36,300,000 6,300,000 250,000

1,382,845

e


Applied Example:Applied Example: Investment Analysis Investment AnalysisSolutionSolution To find the To find the present valuepresent value of the of the net incomenet income NI NI for for plan Bplan B

setting setting RR((tt) = 580,000) = 580,000, , rr = 0.1 = 0.1, and , and TT = 3 = 3, using the present , using the present value formula:value formula:

or approximately or approximately $1,323,254$1,323,254..

0( ) 180,000

T rtNI R t e dt 0 ( ) 180,000T rtNI R t e dt 3 0.1

0580,000 180,000te dt

3 0.1

0580,000 180,000te dt

30.1

0

580,000180,000

0.1te

30.1

0

580,000180,000

0.1te

0.35,800,000 5,800,000 180,000

1,323,254

e

0.35,800,000 5,800,000 180,000

1,323,254

e


Applied Example:Applied Example: Investment Analysis Investment AnalysisSolutionSolution Thus, we conclude thatThus, we conclude that plan A plan A will generate a will generate a higherhigher

present value present value ofof net income net income by the end of the third year by the end of the third year (($1,382,845$1,382,845), than ), than plan Bplan B ( ($1,323,254$1,323,254).).


Amount of an AnnuityAmount of an Annuity

The amount of an annuity isThe amount of an annuity is

where where PP, , rr, , TT, and , and mm are as defined earlier. are as defined earlier.

( 1)rTmPA e

r ( 1)rTmP

A er

Applied Example:Applied Example: IRAs IRAs

On On January 1, 1990January 1, 1990, Marcus Chapman deposited , Marcus Chapman deposited $2000$2000 into an into an Individual Retirement Account (IRA)Individual Retirement Account (IRA) paying paying interestinterest at the rate of at the rate of 10%10% per year per year compounded compounded continuouslycontinuously..

Assuming that he Assuming that he depositeddeposited $2000$2000 annuallyannually into the into the account, how much did he have in his account, how much did he have in his IRAIRA at the at the beginning of 2006beginning of 2006??


Applied Example:Applied Example: IRAs IRAs

SolutionSolution We set We set PP = 2000 = 2000, , rr = 0.1 = 0.1, , TT = 16 = 16, and , and mm = 1 = 1 in the in the amount of amount of

annuity formulaannuity formula, obtaining, obtaining

Thus, Marcus had approximately Thus, Marcus had approximately $79,061$79,061 in his account at in his account at the beginning of the beginning of 20062006..

1.62000( 1) ( 1) 79,060.65

0.1rTmP

A e er

1.62000( 1) ( 1) 79,060.65

0.1rTmP

A e er


Present Value of an AnnuityPresent Value of an Annuity

The present value of an annuity is given byThe present value of an annuity is given by

where where PP, , rr, , TT, and , and mm are as defined earlier. are as defined earlier.

(1 )rTmPPV e

r (1 )rTmP

PV er

Applied Example:Applied Example: Sinking Funds Sinking Funds

Tomas Perez, the proprietor of a hardware store, wants to Tomas Perez, the proprietor of a hardware store, wants to establish a fundestablish a fund from which he will from which he will withdrawwithdraw $1000$1000 per per monthmonth for the next for the next ten yearsten years..

If the fund earns If the fund earns interestinterest at a rate of at a rate of 6%6% per yearper year compounded continuouslycompounded continuously, how , how much moneymuch money does he need does he need to to establish the fundestablish the fund??


Applied Example:Applied Example: Sinking Funds Sinking Funds

SolutionSolution We want to find the present value of an annuity with We want to find the present value of an annuity with

PP = 1000 = 1000, , rr = 0.06 = 0.06, , TT = 10 = 10, and , and mm = 12 = 12.. Using the Using the present value of an annuitypresent value of an annuity formula, we find formula, we find

Thus, Tomas Thus, Tomas needsneeds approximately approximately $90,238$90,238 to to establish establish the fundthe fund..

(0.06)(10)12,000(1 ) (1 ) 90,237.70

0.06rTmP

PV e er

(0.06)(10)12,000(1 ) (1 ) 90,237.70

0.06rTmP

PV e er


End of End of Chapter Chapter

6 Antiderivatives and the Rules of Integration Integration by Substitution Area and the Definite...

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