Introductory maths analysis chapter 14 official

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 Probability

10. Limits and Continuity

11. Differentiation

12. Additional Differentiation Topics

13. Curve Sketching

14. Integration

15. 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

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 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

xxxxxxxdx

ddy 343432 223

08.004.031413 2 dy



14.1 Differentials

Example 3 - Using the Differential to Estimate a Change in a Quantity

A 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

2300

30031

ttPP



14.1 Differentials

Example 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

tPdPP4

3

2 300

3003

300

3003'

Solution:

.2500 if Find 2pqdq

dp

p

p

dpdqdq

dp

p

p

dp

dq 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



14.2 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:

dxx7C

xdxx 2

77

2

CtCt

dxtdxt

22/1

1 a.

2/12/1

Cx

Cx

dxx

2

13

3 12

1

136

1

6

1 b.


Find .

Solution:



Example 7 - Indefinite Integral of a Sum and Difference

dxexx x 11072 35 4

Cexx

Cxexx

dxexx

x

x

x

104

7

9

10

104

75/9

2

11072

45/9

45/9

35 4


Find

Solution:



Example 9 - Using Algebraic Manipulation to Find an Indefinite Integral

dx

xx

6

312 a.

dxx

x

2

3 1 b.

Cxxx

Cxxx

dxxx

212

5

9

32

53

26

1

6

312 a.

23

23

Cx

x

dxxx

dxx

x

1

2

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

CxxCxx

dxxy 4442

848 22

3

24245 2

C

C

344 2 xxy



14.3 Integration with Initial Conditions Example 3 - Income and Education

For 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 xxdx

dy



14.3 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

C

C

000,1940 2/5 xy



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

In 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..dq

dc



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

2

25

3

002.00000010

2.02500200000010

23

2

67.5416$10000

40002.02

25

3

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

nCn

udxu

nn

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:

C

xC

uduudxx

21

1

211 a.

21212020

dxx20

1 a.

dxxx 332 73 b.

C

xC

uduudxxx

4

7

473

4343332

dxxduxu 23 37 Let b.



14.4 More Integration Formulas Example 3 - Adjusting for du

Find

Solution:

CyCy

dyya. 3/433/4

33

4

63

3/466

dxxx

xxb.

424

3

73

32

dxxxduxxu 6473 Let 324

Cxx

Cudu

u

324

34

736

1

32

1

2

dyya. 3 6 dxxx

xxb.

424

3

73

32



14.4 More Integration Formulas Example 5 - Integrals Involving Exponential Functions

Find

Solution:

dxxex2 a.

xdxduxu 2 Let a. 2

dxex xx 32 3

1 b.

Cedue

xdxedxxe

xu

xx

2

2

22

dxxduxxu 333 Let b. 23

Ce

Cduedxex

xx

uxx

3

32

3

3

3

13

11



14.4 More Integration Formulas Example 7 - Integrals Involving Exponential Functions

Find

Solution:

.

73

3224

3

dxxx

xx

dxxxduxxu 6473 Let 324

Cxx

CxxCudxxx

xx

73ln2

1

73ln2

1ln

2

1

73

32

24

2424

3



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

Find

Cxx

dxx

xdxx

xx

ln

2

1 a.

2

2

3

Cxxx

dxx

xxdxx

xxx

12ln2

1

23

12

1

12

132 b.

23

223



14.5 Techniques of Integration

Example 3 - An Integral Involving bu

Find

Solution:

.23 dxx

dxduxu 2ln32ln Let

CCe

Ceduedxedx

xx

uuxx

332ln

32ln3

22ln

1

2ln

1

2ln

1

2ln

12

• General formula for integrating bu is

Cbb

dub uu ln

1



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.



14.6 The Definite Integral

Example 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

121

3

48

6

12188

224

2

n

nnnnn

nn

n

nn

kfx

nkfS

n

k

n

kn

3

16

3

88

121

3

48limlim

2

n

nnS

nn

n



14.6 The Definite Integral

Example 3 - Integrating a Function over an Interval

Integrate f (x) = x − 5 from x = 0 to x = 3.

Solution:

15

11

2

915

2

1933

1

nn

n

nnkfS

n

kn

2

21

2

915

11

2

9limlim5

3

0

nSdxx

nn

n



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



14.7 The Fundamental Theorem of Integral Calculus

Properties of the Definite Integral

1. 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

162

1136

2

33

62

63

23

23

3

1

23

3

1

2

x

xxdxxx




Example 3 - Evaluating Definite Integrals

Find

Solution:

2

1

323/1 14 a. dxttt

8

5852625

8

1123

4

1

2

1414 a.

3443/4

2

1

2/12

34

3/42

1

323/1

ttdxttt

13

1

3

1

3

1 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

afbfdxxfa

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 qdq

dc

112020803200

23.026.080100100

802

100

80

qqdqqcc



14.8 Approximate Integration14.8 Approximate Integration

Trapezoidal Rule

• To find the area of a trapezoidal area, we have

./ where

122222

nb-ah

bfhnafhafhafafh

dxxfb

a



14.8 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

n

abh




Example 1 - Trapezoidal Rule

Solution: The terms to be added are

Estimate for the integral is

sum

fbf

fhaf

fhaf

fhaf

fhaf

faf

8373.7

0.50001

2195.18.0242

4706.16.0232

7241.14.0222

9231.12.022

0000.10

784.08373.72

2.0

1

11

02

dx

x




Simpson’s Rule

• Approximating the graph of f by parabolic segments gives

even. is and / where

142243

nnabh

bfhnafhafhafafh

dxxfb

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.





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

3

2035

n

abh

sum

l

l

l

l

90,7755

964,9735

700,1966230,982302

254,197627,982252

857,9820

5.937,476,1775,5903

535

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



14.9 Area

Example 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

6

125

3

37

2

918

3

8

3

412

3266

2

3

322

3

2

xxxdxxxarea



14.9 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

12

1

2

1

eeedxearea xx



14.9 Area

Example 5 - Statistics Application

In 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



14.9 Area


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



14.9 Area


Solution:

a.

b.

32

5

326

60

4/1

0

32

4/1

0

241

xx

dxxxxP

2

1

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

6

1

22/3

1

0

22/31

0

xxdxxxarea



14.10 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

2

19 22

x

xx

22

22

9 and 1 ,2,5 of right For

1 and 9 ,2,5 of left For

xyxy

xyxy

lowerupper

lowerupper

xx

xxxyyxx

xx

xxxyyxx

lowerupper

lowerupper

82

91 ,3 to 2 From

28

19 ,2 to 0 From

2

22

2

22

3

468228

3

2

22

0

2 dxxdxxarea



14.10 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

2

92

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 is

where 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

0

0

dqqgpPSq

0

0

0

qqfp 05.0100

qqgp 1.010



14.11 Consumers’ and Producers’ Surplus

Example 1 - Finding Consumers’ Surplus and Producers’ Surplus

Solution:

Find the equilibrium point (p0, q0),

Consumers’ surplus is

Producers’ surplus is

706001.010 Thus

600

05.01001.010

0

0

p

qq

qq

000,18$2

1.060600

0

2

0

0

0

qqdqpgpPS

q

9000$2

05.030600

0

2

0

0

0

qqdqpqfCS

q

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