integration pearson

INTRODUCTORY MATHEMATICAL ANALYSISINTRODUCTORY MATHEMATICAL ANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 14 Chapter 14 IntegrationIntegration


INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra

1. Applications and More Algebra

2. Functions and Graphs

3. Lines, Parabolas, and Systems

4. Exponential and Logarithmic Functions

5. Mathematics of Finance

6. Matrix Algebra

7. Linear Programming

8. Introduction to Probability and Statistics


9. Additional Topics in Probability

10. Limits and Continuity

11. Differentiation

12. Additional Differentiation Topics

13. Curve Sketching

14. Integration15. Methods and Applications of Integration

16. Continuous Random Variables

17. Multivariable Calculus

INTRODUCTORY MATHEMATICAL ANALYSIS


To define the differential.

To define the anti-derivative and the indefinite integral.

To evaluate constants of integration.

To apply the formulas for .

To handle more challenging integration problems.

To evaluate simple definite integrals.

To apply Fundamental Theorem of Integral Calculus.

Chapter 14: Integration

Chapter ObjectivesChapter Objectives

duudueduu nn 1 and ,


To use Trapezoidal rule or Simpsons rule.

To use definite integral to find the area of the region.

To find the area of a region bounded by two or more curves.

To develop concepts of consumers surplus and producers surplus.

Chapter 14: Integration Chapter Objectives


Differentials

The Indefinite Integral

Integration with Initial Conditions

More Integration Formulas

Techniques of Integration

The Definite Integral

The Fundamental Theorem of Integral Calculus

14.1)

14.2)

14.3)


Chapter OutlineChapter Outline

14.4)

14.5)

14.6)

14.7)


Approximate Integration

Area

Area between Curves

Consumers and Producers Surplus

14.8)

14.9)

14.10)


Chapter OutlineChapter Outline

14.11)



14.1 Differentials14.1 Differentials

Example 1 Computing a Differential

The differential of y, denoted dy or d(f(x)), is given by ( ) ( )dxxfdyxxfdy '' ==

Find the differential of and evaluate it when x = 1 and x = 0.04.Solution: The differential is

When x = 1 and x = 0.04,

432 23 += xxxy

( ) ( ) xxxxxxxdxddy +=+= 343432 223

( ) ( )[ ]( ) 08.004.031413 2 =+=dy


Chapter 14: Integration14.1 Differentials

Example 3 - Using the Differential to Estimate a Change in a QuantityA governmental health agency examined the records of a group of individuals who were hospitalized with a particular illness. It was found that the total proportion P that are discharged at the end of t days of hospitalization is given by

Use differentials to approximate the change in the proportion discharged if t changes from 300 to 305.

( ) ( )3

2300300

31

+==

ttPP


Chapter 14: Integration14.1 DifferentialsExample 3 - Using the Differential to Estimate a Change in a Quantity

Example 5 - Finding dp/dq from dq/dp

Solution: We approximate P by dP,

( ) ( ) ( ) dttdtttPdPP 43

2 300300

3300

3003'

+=

+==

Solution:

.2500 if Find 2pqdqdp

=

pp

dpdqdq

dpp

pdpdq 2

2

25001

2500

==

=



14.2 The Infinite Integral14.2 The Infinite Integral An antiderivative of a function f is a function F

such that .

In differential notation, .

Integration states that

Basic Integration

Properties:

( ) ( )xfxF ='( )dxxfdF =

( ) ( ) ( ) ( )xfxFCxFdxxf =+= 'only if


Chapter 14: Integration14.2 The Infinite Integral

Example 1 - Finding an Indefinite Integral

Example 3 - Indefinite Integral of a Constant Times a Function

Example 5 - Finding Indefinite Integrals

Find .Solution:

dx5Cxdx += 55

Find .Solution:

dxx7Cxdxx += 277

2

CtCtdxtdxt

+=+== 22/11 a.2/1

2/1

Cx

Cxdxx

+=+

+=

+ 2133 121136161 b.


Find .

Solution:


Example 7 - Indefinite Integral of a Sum and Difference( )dxexx x + 11072 35 4( )( ) ( ) ( )

Cexx

Cxexx

dxexx

x

x

x

++=

++=

+

1047

910

104

75/9

2

11072

45/9

45/9

35 4


Find

Solution:


Example 9 - Using Algebraic Manipulation to Find an Indefinite Integral

( )( )dxxx + 6 312 a.dx

xx 23 1 b.

( )( )

( ) ( )

Cxxx

Cxxx

dxxx

++=

+

+=

+

2125

9

32

53

261

6312

a.

23

23 ( )C

xx

dxxx

dxx

x

++=

=

12

1 b.

2

2

2

3



14.3 Integration with Initial Conditions14.3 Integration with Initial Conditions

Example 1 - Initial-Condition Problem

Use initial conditions to find the constant, C.

If y is a function of x such that y = 8x 4 and y(2) = 5, find y.Solution: We find the integral,

Using the condition,

The equation is

( ) ( ) CxxCxxdxxy +=+== 4442848 22

( ) ( )3

24245 2

=

+=

CC

344 2 = xxy



14.3 Integration with Initial Conditions Example 3 - Income and EducationFor a particular urban group, sociologists studied the current average yearly income y (in dollars) that a person can expect to receive with x years of education before seeking regular employment. They estimated that the rate at which income changes with respect to education is given by

where y = 28,720 when x = 9. Find y.

164 100 2/3 = xxdxdy



14.3 Integration with Initial Conditions Example 3 - Income and Education

Solution: We have

When x = 9,

Therefore,

Cxdxxy +== 2/52/3 40100( )

000,19940720,28 2/5

=

+=

CC

000,1940 2/5 += xy



14.3 Integration with Initial Conditions Example 5 - Finding Cost from Marginal CostIn the manufacture of a product, fixed costs per week are $4000. (Fixed costs are costs, such as rent and insurance, that remain constant at all levels of production during a given time period.) If the marginal-cost function is

where c is the total cost (in dollars) of producing q pounds of product per week, find the cost of producing 10,000 lb in 1 week.

( ) 2.02500200000010 2 += qq..dqdc



14.3 Integration with Initial Conditions Example 5 - Finding Cost from Marginal Cost

Solution: The total cost c is

When q = 0, c = 4000.Cost of 10,000 lb in one week,

( ) ( )[ ]Cqqq.

dqqq..qc

++

=

+= 2.0

225

3002.0

0000010

2.02500200000010

23

2

( )( ) 67.5416$10000

40002.02

253

00200000010

23

=

++

=

c

qqq..qc



14.4 More Integration Formulas14.4 More Integration Formulas Power Rule for Integration

Integrating Natural Exponential Functions

Integrals Involving Logarithmic Functions

1 if 1

1

++

= + nCnudxun

n

+= Cedue uu

0 for ln1

+= xCxdxx



14.4 More Integration Formulas Basic Integration Formulas



14.4 More Integration Formulas Example 1 - Applying the Power Rule for Integration

Find the integral of

Solution:

( ) ( ) ( ) CxCuduudxx ++=+==+ 211211 a.21212020

( ) dxx 201 a. +( ) dxxx + 332 73 b.

( ) ( ) ( ) CxCuduudxxx ++=+==+ 4 7473434

3332

dxxduxu 23 37 Let b. =+=



14.4 More Integration Formulas Example 3 - Adjusting for du

Find

Solution:

CyCydyya. +=+= 3/433/433 4633/466 ( ) dxxx

xxb. ++

+424

3

73

32

( )dxxxduxxu 6473 Let 324 +=++=

( ) CxxCuduu +

++=+

=

324

34

736

132

12

dyya. 3 6 ( ) dxxxxxb.

++

+424

3

73

32



14.4 More Integration Formulas Example 5 - Integrals Involving Exponential Functions

FindSolution:

dxxex2 a.xdxduxu 2 Let a. 2 ==

( ) dxex xx 32 31 b. + +

[ ]Cedue

xdxedxxexu

xx

+==

=

2

2

22

( )dxxduxxu 333 Let b. 23 +=+=( )

Ce

Cduedxex

xx

uxx

+=

+=+

+

+ 3

32

3

3

3131

1



14.4 More Integration Formulas Example 7 - Integrals Involving Exponential Functions

Find

Solution:

( ).

733224

3

dxxx

xx ++ +

( )dxxxduxxu 6473 Let 324 +=++=( )

( ) CxxCxxCudx

xxxx

+++=

+++=+=++

+73ln

21

73ln21

ln21

7332

24

2424

3



14.5 Techniques of Integration14.5 Techniques of IntegrationExample 1 - Preliminary Division before Integration

Find

Cxxdxx

xdxx

xx++=

+=+ ln21 a.

2

2

3

Cxxx

dxx

xxdxx

xxx

++++=

+

++=+

+++ 12ln

21

23

121

12132

b.

23

223


Chapter 14: Integration14.5 Techniques of Integration

Example 3 - An Integral Involving bu

FindSolution:

.23 dxx ( )( ) dxduxu 2ln32ln Let ==

( ) ( )

( ) ( ) ( ) CCe

Ceduedxedx

xx

uuxx

+=+=

+===

332ln

32ln3

22ln

12ln

1

2ln1

2ln1

2

General formula for integrating bu is

Cbb

dub uu += ln1



14.6 The Definite Integral14.6 The Definite Integral

Example 1 - Computing an Area by Using Right-Hand Endpoints

For area under the graph from limit a b,

x is called the variable of integration and f (x) is the integrand.

( )dxxfba

Find the area of the region in the first quadrant bounded by f(x) = 4 x2 and the lines x = 0 and y = 0.Solution: Since the length of [0, 2] is 2, x = 2/n.


Chapter 14: Integration14.6 The Definite IntegralExample 1 - Computing an Area by Using Right-Hand Endpoints

Summing the areas, we get

We take the limit of Sn as n:

Hence, the area of the region is 16/3.

( )( ) ( )( )

++=

++=

=

= ==

23

1

2

1

12134

86

12188

224

2

nnnnnn

nn

n

nnkfx

nkfS

n

k

n

kn

( )( )3

1638

8121

34

8limlim 2 ==

++

= n

nnSnnn


Chapter 14: Integration14.6 The Definite Integral

Example 3 - Integrating a Function over an IntervalIntegrate f (x) = x 5 from x = 0 to x = 3.

Solution:

( )15

11

29

152

1933

1

+=

+=

= =

nnn

nnkfS

n

kn

( )221

29

151

129

limlim53

0

==

+==

nSdxx nnn



14.7 The Fundamental Theorem of 14.7 The Fundamental Theorem of Integral CalculusIntegral Calculus

Fundamental Theorem of Integral Calculus

If f is continuous on the interval [a, b] and F is any antiderivative of f on [a, b], then

Properties of the Definite Integral

If a > b, then

If limits are equal,

( ) ( ) ( )aFbFdxxfba

=

( ) ( ) = ab

b

a

dxxfdxxf

( ) 0=ba

dxxf


Chapter 14: Integration14.7 The Fundamental Theorem of Integral Calculus

Properties of the Definite Integral1. is the area bounded by the graph f(x).

2.

3.

4.

5.

( )ba

dxxf

( ) ( ) constant. a is where kdxxfkdxxkf ba

b

a =

( ) ( )[ ] ( ) ( ) = ba

b

a

b

a

dxxgdxxfdxxgxf

( ) ( ) = ba

b

a

dttfdxxf

( ) ( ) ( ) += cb

b

a

c

a

dxxfdxxfdxxf



Example 1 - Applying the Fundamental Theorem

Find

Solution:

( ) .6331

2

+ dxxx

( )( ) ( ) ( ) ( )

48

1621

13623

3

62

63

23

23

3

1

23

3

1

2

=

+

+=

+=+

xxxdxxx



Example 3 - Evaluating Definite IntegralsFind

Solution:

( )[ ] ++21

323/1 14 a. dxttt

( )[ ] ( ) ( )( ) ( )

8585

262581

123

41

21

414 a.

3443/4

2

1

2/12

34

3/42

1

323/1

+=+=

+

+=++ ttdxttt

[ ] ( ) ( )131

31

31

b. 3031

03

1

0

3==

= eeeedte tt

10

3 b. dte t



Example 5 - Finding a Change in Function Values by Definite Integration

The Definite Integral of a Derivative

The Fundamental Theorem states that( ) ( ) ( )afbfdxxfa

b

= '

A manufacturers marginal-cost function is . If production is presently set at q = 80 units per week, how much more would it cost to increase production to 100 units per week?Solution: The rate of change of c is dc/dq is

26.0 += qdqdc

( ) ( ) ( ) [ ]112020803200

23.026.080100100

802

100

80

==

+=+= qqdqqcc



14.8 Approximate Integration14.8 Approximate IntegrationTrapezoidal Rule

To find the area of a trapezoidal area, we have

( ) ( ) ( ) ( ) ( )( ) ( )[ ]( ) ./ where

122222

nb-ah

bfhnafhafhafafhdxxfb

a

=

++++++++


Chapter 14: Integration14.8 Approximate Integration

Example 1 - Trapezoidal Rule

Use the trapezoidal rule to estimate the value of

for n = 5. Compute each term to four decimal places, and round the answer to three decimal places.

Solution: With n = 5, a = 0, and b = 1,

dxx +

1

021

1

2.05

01=

=

=

nabh


Chapter 14: Integration14.8 Approximate IntegrationExample 1 - Trapezoidal Rule

Solution: The terms to be added are

Estimate for the integral is

( ) ( )( ) ( )

( ) ( )( ) ( )( ) ( )

( ) ( )sum

fbffhaffhaffhaffhaf

faf

=

==

==+

==+

==+

==+

==

8373.7 0.50001

2195.18.02424706.16.02327241.14.02229231.12.022 0000.10

( ) 784.08373.722.0

111

02 + dxx



Simpsons Rule

Approximating the graph of f by parabolic segments gives

( ) ( ) ( ) ( ) ( ) ( )( )[ ]( ) even. is and / where

142243

nnabh

bfhnafhafhafafhdxxfb

a

=

++++++++



Example 3 - Demography

A function often used in demography (the study of births, marriages, mortality, etc., in a population) is the life-table function, denoted l. In a population having 100,000 births in any year of time, l(x) represents the number of persons who reach the age of x in any year of time. For example, if l(20) = 98,857, then the number of persons who attain age 20 in any year of time is 98,857.


Chapter 14: Integration14.8 Approximate IntegrationExample 3 - Demography

Suppose that the function l applies to all people born over an extended period of time. It can be shown that, at any time, the expected number of persons in the population between the exact ages of x and x + m, inclusive, is given by

The following table gives values of l(x) for males and females in the United States. Approximate the number of women in the 2035 age group by using the trapezoidal rule with n = 3.

( )dttlmxx+1



Life table:



Solution: We want to estimate Thus

The terms to be added are

By the trapezoidal rule,

( ) .3520

dttl5

32035

=

=

=

nabh

( )( ) ( )

( ) ( )( )

suml

ll

l

=

=

==

==

=

90,7755 964,9735 700,1966230,982302254,197627,982252

857,9820

( ) ( ) 5.937,476,1775,5903535

20

= dttl



14.9 Area14.9 Area

Example 1 - Using the Definite Integral to Find Area

The width of the vertical element is x. The height is the y-value of the curve.

The area is defined as

( ) ( ) areadxxfxxf ba

= Find the area of the region bounded by the curve

and the x-axis.26 xxy =


Chapter 14: Integration14.9 AreaExample 1 - Using the Definite Integral to Find Area

Solution:

Summing the areas of all such elements from x = 3 to x = 2,

( )( )326 2 +== xxxxy

areadxyxy =

2

3

( )

6125

337

29

1838

34

12

3266

2

3

322

3

2

=

=

==

xxxdxxxarea


Chapter 14: Integration14.9 Area

Example 3 - Finding the Area of a Region

Find the area of the region between the curve y = ex and the x-axis from x = 1 to x = 2.

Solution: We have

[ ] ( )12121

=== eeedxearea xx


Chapter 14: Integration14.9 Area

Example 5 - Statistics ApplicationIn statistics, a (probability) density function f of a variable x, where x assumes all values in the interval [a, b], has the following properties:

The probability that x assumes a value between c and d, which is written P(c x d), where a c d b, is represented by the area of the region bounded by the graph of f and the x-axis between x = c and x = d.

( )( ) 1 (ii)

0 (i)b

a

=

dxxfxf


Chapter 14: Integration14.9 AreaExample 5 - Statistics Application

Hence

For the density function f(x) = 6(x x2), where 0 x 1, find each of the following probabilities.

( ) ( )dxxfdxcP dc=

( )410 . xPa( )21 . xPb


Chapter 14: Integration14.9 AreaExample 5 - Statistics Application

Solution:

a.

b.

( ) ( )

325

326

60

4/1

0

32

4/1

0

241

=

=

= xx

dxxxxP

( ) ( )

21

326

6

1

2/1

32

1

2/1

221

=

=

= xx

dxxxxP



14.10 Area between Curves14.10 Area between Curves

Example 1 - Finding an Area between Two Curves

Vertical Elements The area of the element is

Find the area of the region bounded by the curves y = x and y = x.Solution: Eliminating y by substitution,

[ ] .xyy lowerupper

1 or 0 == xx

( )61

22/3

1

0

22/31

0

=

== xxdxxxarea


Chapter 14: Integration14.10 Area between Curves

Example 3 - Area of a Region Having Two Different Upper Curves

Find the area of the region between the curves y = 9 x2 and y = x2 + 1 from x = 0 to x = 3.Solution: The curves intersect when

219 22

=+=

xxx

( )( ) 22

22

9 and 1 ,2,5 of right For

1 and 9 ,2,5 of left For

xyxy

xyxy

lowerupper

lowerupper

=+=

+==

[ ] ( ) ( )[ ]( )

[ ] ( ) ( )[ ]( ) xx

xxxyyxxxx

xxxyyxx

lowerupper

lowerupper

=

+====

+===

82

91 ,3 to 2 From

28

19 ,2 to 0 From

2

22

2

22

( ) ( )346

82283

2

22

0

2=+= dxxdxxarea


Chapter 14: Integration14.10 Area between Curves

Example 5 - Advantage of Horizontal ElementsFind the area of the region bounded by the graphs of y2 = x and x y = 2.

Solution: The intersection points are (1,1) and (4, 2).The total area is

( )29

22

1

2=+=

dyyyarea



14.11 Consumers and Producers Surplus14.11 Consumers and Producers Surplus

Example 1 - Finding Consumers Surplus and Producers Surplus

Consumers surplus, CS, is defined as

Producers surplus, PS, is defined as

The demand function for a product iswhere p is the price per unit (in dollars) for q units. The supply function is . Determine consumers surplus and producers surplus under market equilibrium.

( )[ ]dqpqfCSq

= 00

0

( )[ ]dqqgpPSq

= 00

0

( ) qqfp 05.0100 ==( ) qqgp 1.010 +==


Chapter 14: Integration14.11 Consumers and Producers SurplusExample 1 - Finding Consumers Surplus and Producers Surplus

Solution:Find the equilibrium point (p0, q0),

Consumers surplus is

Producers surplus is

( ) 706001.010 Thus600

05.01001.010

0

0

=+=

==

=+

pqq

qq

( )[ ] 000,18$2

1.060600

0

2

00

0

=

== qqdqpgpPS

q

( )[ ] 9000$2

05.030600

0

2

00

0

=

== qqdqpqfCS

q

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