Introductory maths analysis chapter 15 official

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 15 Chapter 15 Methods and Applications of IntegrationMethods and Applications of Integration


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 develop and apply the formula for integration by parts.

• To show how to integrate a proper rational function.

• To illustrate the use of the table of integrals.

• To develop the concept of the average value of a function.

• To solve a differential equation by using the method of separation of variables.

• To develop the logistic function as a solution of a differential equation.

• To define and evaluate improper integrals.

Chapter 15: Methods and Applications of Integration

Chapter ObjectivesChapter Objectives


Integration by Parts

Integration by Partial Fractions

Integration by Tables

Average Value of a Function

Differential Equations

More Applications of Differential Equations

Improper Integrals

15.1)

15.2)

15.3)


Chapter OutlineChapter Outline

15.4)

15.5)

15.6)

15.7)



15.1 Integration by Parts15.1 Integration by Parts

Example 1 – Integration by Parts

Formula for Integration by Parts

Find by integration by parts.

Solution: Let and

Thus,

duvuvdvu

dxx

x

ln

Cxx

dxx

xxxdxx

x

2ln2

1

22ln ln 2/1

xu ln dxx

dv 1

dxx

du1

2/12/1 2 xdxxv



15.1 Integration by Parts

Example 3 – Integration by Parts where u is the Entire Integrand

Determine

Solution: Let and

Thus,

. ln dyy

yv

dydv

Cyy

Cyyy

dyy

yyydyy

1ln

ln

1

ln ln

dyy

du

yu

1

ln




Example 5 – Applying Integration by Parts Twice

Determine

Solution: Let and

Thus,

. 122 dxex x

dxxdu

xu

2

2

2/

12

12

x

x

ev

dxedv

dxxeex

dxxeex

dxex

xx

xxx

2

)2(22

12122

12122122

1

1212

121212

42

22

Cexe

dxexe

dxxe

xx

xxx




Example 5 – Applying Integration by Parts Twice

Solution (cont’d):

Cxxe

Cexeex

dxex

x

xxxx

2

1

2

422

212

1212122122



15.2 Integration by Partial Fractions15.2 Integration by Partial Fractions

Example 1 – Distinct Linear Factors

• Express the integrand as partial fractions

Determine by using partial fractions.

Solution: Write the integral as

Partial fractions:

Thus,

dxx

x

273

122

. 9

12

3

12

dxx

x

65

67

2

,3 if and ,3 If

3333

12

9

12

AxBx

x

B

x

A

xx

x

x

x

Cxx

x

dx

x

dxdx

x

x

3ln6

73ln

6

5

3

1

333

1

273

12 67

65

2



15.2 Integration by Partial Fractions

Example 3 – An Integral with a Distinct Irreducible Quadratic Factor

Determine by using partial fractions.Solution: Partial fractions:

Equating coefficients of like powers of x, we have

Thus,

dxxxx

x

4223

xCBxxxAx

xx

CBx

x

A

xxx

x

)()1(42

11

42

2

22

2 ,4 ,4 CBA

C

x

xx

Cxxx

dxxx

x

xdx

xx

CBx

x

A

4

22

2

22

1ln

1ln2ln4

1

244

1



15.2 Integration by Partial Fractions

Example 5 – An Integral Not Requiring Partial Fractions

Find

Solution: This integral has the form

Thus,

. 13

322

dxxx

x

Cxxdxxx

x

13ln 13

32 22

. 1

duu



15.3 Integration by Tables15.3 Integration by Tables

Example 1 – Integration by Tables

• In the examples, the formula numbers refer to the Table of Selected Integrals given in Appendix B of the book.

Find

Solution: Formula 7 states

Thus,

.

32

2 x

dxx

C

bua

abua

bbua

duu

ln1

22

C

xxdx

x

x

32

232ln

9

1

32 2



15.3 Integration by Tables


Find


Let u = 4x and a = √3, then du = 4 dx.

.316 2

xx

dx

Cu

aau

aauu

du

22

22ln

1

Cx

x

xx

dx

4

3316ln

3

1

316

2

2





Find


If we let u = 4x, then du = 4 dx. Hence,

. 4ln7 2 dxxx

C

n

u

n

uuduuu

nnn

2

11

11

lnln

Cxx

Cxxx

dxxxdxxx

14ln39

7

9

4

3

4ln4

64

7

44ln44

7 4ln7

3

33

2

32




Example 7 – Finding a Definite Integral by Using TablesEvaluate


Letting u = 2x and a2 = 2, we have du = 2 dx.

Thus,

.24

4

12/32

x

dx

Caua

u

au

du

2222/322

Caua

u

au

du

2222/322

62

1

66

2

222

1

22

1

24

8

22

4

12/32

4

12/32

u

u

u

du

x

dx



15.4 Average Value of a Function15.4 Average Value of a Function

Example 1 – Average Value of a Function

• The average value of a function f (x) is given by

Find the average value of the function f(x)=x2 over the interval [1, 2].

Solution:

dxxfab

fb

a

1

3

7

312

1

1

2

1

32

1

2

xdxx

dxxfab

fb

a



15.5 Differential Equations15.5 Differential Equations

Example 1 – Separation of Variables

• We will use separation of variables to solve differential equations.

Solve

Solution: Writing y’ as dy/dx, separating variables and integrating,

.0, if ' yxx

yy

xCy

dxx

dyy

x

y

dx

dy

lnln

11

1



Example 1 – Separation of Variables


0,

ln

ln

1

1

xCx

Cy

e

ey

ey

x

C

xC



15.5 Differential Equations

Example 3 – Finding the Decay Constant and Half-Life

If 60% of a radioactive substance remains after 50 days, find the decay constant and the half-life of the element.Solution:

Let N be the size of the population at time t, tλeNN 0

01022.0

50

6.0ln

6.0

6.0 and 50 When50

00

0

λ

eNN

NNtλ

days. 82.672ln

is life half the and Thus, 01022.00

λeNN t



15.6 More Applications of Differential Equations15.6 More Applications of Differential Equations

Logistic Function

• The function

is called the logistic function or the Verhulst–Pearl logistic function.

Alternative Form of Logistic Function

ctbe

MN

1

tbC

MN

1



15.6 More Applications of Differential Equations

Example 1 – Logistic Growth of Club Membership

Suppose the membership in a new country club is to be a maximum of 800 persons, due to limitations of the physical plant. One year ago the initial membership was 50 persons, and now there are 200. Provided that enrollment follows a logistic function, how many members will there be three years from now?




Example 1 – Logistic Growth of Club Membership

Solution: Let N be the number of members enrolled in t years,

Thus,

1511

80050

1

,0 and 800 When

bbbC

MN

tM

t

5lnln151

800200

,200 and 1 When

51

ce

Nt

c

781

151

8004

51

N




Example 3 – Time of Murder

A wealthy industrialist was found murdered in his home. Police arrived on the scene at 11:00 P.M. The temperature of the body at that time was 31◦C, and one hour later it was 30◦C. The temperature of the room in which the body was found was 22◦C. Estimate the time at which the murder occurred.

Solution: Let t = no. of hours after the body was discovered andT(t) = temperature of the body at time t.By Newton’s law of cooling,

22 Tkdt

dTaTk

dt

dT




Example 3 – Time of Murder


CktT

dtkT

dT

22ln

22

9ln02231ln

,0 and 31 When

CCk

tT

9

8ln9ln12230ln

,1 and 30 When

kk

tT

ktT

InktT

9

229ln22ln Hence,




Example 3 - Time of Murder


Accordingly, the murder occurred about 4.34 hours before the time of discovery of the body (11:00 P.M.). The industrialist was murdered at about 6:40 P.M.

34.4

9/8ln

9/15ln

9

8ln2237ln

, 37 When

tt

T



15.7 Improper Integrals15.7 Improper Integrals

• The improper integral is defined as

• The improper integral is defined as

dxxfa

dxxfdxxfr

ar

a

lim

dxxfdxxfdxxf 0

0

dxxf



15.7 Improper Integrals

Example 1 – Improper Integrals

Determine whether the following improper integrals are convergent or divergent. For any convergent integral, determine its value.

2

1

2

10

2lim lim

1 a.

1

2

1

3

13

r

r

r

r

xdxxdx

x

1lim lim b.0

00

r

x

rr

x

r

x edxedxe

r

r

r

rxdxxdx

x1

2/1

1

2/1

1

2lim lim 1

c.



15.7 Improper Integrals

Example 3 – Density Function

In statistics, a function f is called a density function if f(x) ≥ 0 and .

Suppose is a density function. Find k.

Solution:

1

dxxf

elsewhere 0

0 for xkexf

x

11lim1lim

101

0

0

00

0

kkedxke

dxkedxxfdxxf

rx

r

rx

r

x

Introductory maths analysis chapter 15 official

Education

x x x cxx x dx x dx

c x xdx x x

parts integration

axbx x b x

dx x dv

partial fractions integration

cba c x

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