Chapter 14 - Integration

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor 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 Probability10. Limits and Continuity11. Differentiation12. Additional Differentiation Topics13. Curve Sketching

14. Integration15. Methods and Applications of Integration16. Continuous Random Variables17. 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

duu

dueduu nn 1 and ,


• To use Trapezoidal rule or Simpson’s 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

230030031

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,

dtt

dtt

tPdPP 4

3

2 3003003

3003003'

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 2

772

CtCtdxtdxt

22/1

1 a.2/1

2/1

Cx

Cxdxx

2

13

3 121

1361

61 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.

dxx

x

2

3 1 b.

Cxxx

Cxxx

dxxx

2125

9

32

53

261

6312 a.

23

23

Cx

x

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 4442

848 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


Chapter 14: Integration14.3 Integration with Initial Conditions Example 3 - Income and Education

Solution: We have

When x = 9,

Therefore,

Cxdxxy 2/52/3 40100

000,19

940720,28 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

3002.00000010

2.0250020000001023

2

67.5416$10000

40002.02

253

0020000001023

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

nCnudxu

nn

Cedue uu

0 for ln1 xCxdx

x


Chapter 14: Integration14.4 More Integration Formulas Basic Integration Formulas


Chapter 14: Integration14.4 More Integration Formulas Example 1 - Applying the Power Rule for Integration

Find the integral of

Solution:

CxCuduudxx

211

211 a.

21212020

dxx20

1 a.

dxxx 332 73 b.

CxCuduudxxx

47

473

4343332

dxxduxu 23 37 Let b.


Chapter 14: Integration14.4 More Integration Formulas Example 3 - Adjusting for du

Find

Solution:

CyCydyya. 3/433/4

33

463

3/466

dxxx

xxb.

424

3

73

32

dxxxduxxu 6473 Let 324

Cxx

Cuduu

324

34

736

132

12

dyya. 3 6 dxxx

xxb.

424

3

73

32


Chapter 14: Integration14.4 More Integration Formulas Example 5 - Integrals Involving Exponential FunctionsFindSolution:

dxxex2 a.

xdxduxu 2 Let a. 2

dxex xx 32 3

1 b.

Cedue

xdxedxxexu

xx

2

2

22

dxxduxxu 333 Let b. 23

Ce

Cduedxex

xx

uxx

3

32

3

3

31311


Chapter 14: Integration14.4 More Integration Formulas Example 7 - Integrals Involving Exponential FunctionsFind

Solution:

.73

3224

3

dxxx

xx

dxxxduxxu 6473 Let 324

Cxx

CxxCudxxx

xx

73ln21

73ln21ln

21

7332

24

2424

3



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

Find

Cxxdxx

xdxx

xx

ln

21 a.

2

2

3

Cxxx

dxx

xxdxx

xxx

12ln21

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

2ln12

• 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.

dxxfb

a

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

121348

612188

2242

nnnnnn

nn

n

nnkfx

nkfS

n

k

n

kn

3

16388121

348limlim 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:

15112915

21933

1

nnn

nnkfS

n

kn

221

291511

29limlim5

3

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,

aFbFdxxfb

a

a

b

b

a

dxxfdxxf

0b

a

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.

b

a

dxxf

constant. a is where kdxxfkdxxkfb

a

b

a

b

a

b

a

b

a

dxxgdxxfdxxgxf

b

a

b

a

dttfdxxf

c

b

b

a

c

a

dxxfdxxfdxxf



Example 1 - Applying the Fundamental Theorem

Find

Solution:

.633

1

2

dxxx

48

1621136

233

62

63

23

23

3

1

23

3

1

2

xxxdxxx



Example 3 - Evaluating Definite IntegralsFind

Solution:

2

1

323/1 14 a. dxttt

8

585262581123

41

21414 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

1

0

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 afbfdxxf

a

b

'

A manufacturer’s 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

23.026.08010010080

2100

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



Simpson’s 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 20–35 age group by using the trapezoidal rule with n = 3.

dttlmx

x1



Life table:



Solution: We want to estimate Thus

The terms to be added are

By the trapezoidal rule,

.35

20

dttl5

32035

nabh

sum

ll

ll

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

areadxxfxxfb

a

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

2918

38

3412

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

121

2

1

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

dxxf

xf


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.

dxxfdxcPd

c

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

3468228

3

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

292

2

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

0

00

dqqgpPSq

0

00

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

qqdqpgpPSq

9000$2

05.030600

0

2

00

0

qqdqpqfCSq

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