Chapter 15 - Methods and Applications of Integration
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Transcript of Chapter 15 - Methods and Applications of Integration
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
2007 Pearson Education Asia
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
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9. Additional Topics in Probability10. Limits and Continuity11. Differentiation12. Additional Differentiation Topics13. Curve Sketching14. Integration
15. Methods and Applications of Integration16. Continuous Random Variables17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL ANALYSIS
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• 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
2007 Pearson Education Asia
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 15: Methods and Applications of Integration
Chapter OutlineChapter Outline
15.4)
15.5)
15.6)
15.7)
2007 Pearson Education Asia
Chapter 15: Methods and Applications of Integration
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
dxxx ln
Cxx
dxx
xxxdxxx
2ln2
122ln ln 2/1
xu ln dxx
dv 1
dxx
du 1 2/12/1 2 xdxxv
2007 Pearson Education Asia
Chapter 15: Methods and Applications of Integration15.1 Integration by Parts
Example 3 – Integration by Parts where u is the Entire IntegrandDetermine
Solution: Let and
Thus,
. ln dyy
yvdydv
CyyCyyy
dyy
yyydyy
1lnln
1ln ln
dyy
du
yu
1
ln
2007 Pearson Education Asia
Chapter 15: Methods and Applications of Integration15.1 Integration by Parts
Example 5 – Applying Integration by Parts TwiceDetermine
Solution: Let and
Thus,
. 122 dxex x
dxxduxu
2
2
2/
12
12
x
x
evdxedv
dxxeex
dxxeexdxex
xx
xxx
2
)2(22
12122
12122122
1
1212
121212
42
22
Cexe
dxexedxxe
xx
xxx
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Chapter 15: Methods and Applications of Integration15.1 Integration by PartsExample 5 – Applying Integration by Parts Twice
Solution (cont’d):
Cxxe
Cexeexdxex
x
xxxx
21
2
422
212
1212122122
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Chapter 15: Methods and Applications of Integration
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 27312
2
. 912
31
2 dxx
x
65
67
2
,3 if and ,3 If3333
12912
AxBxx
Bx
Axx
xx
x
Cxx
xdx
xdx
dxx
x
3ln673ln
65
31
3331
27312 6
765
2
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Chapter 15: Methods and Applications of Integration15.2 Integration by Partial Fractions
Example 3 – An Integral with a Distinct Irreducible Quadratic FactorDetermine by using partial fractions.Solution: Partial fractions:
Equating coefficients of like powers of x, we have
Thus,
dxxxx
x
4223
xCBxxxAx
xxCBx
xA
xxxx
)()1(42
1142
2
22
2 ,4 ,4 CBA
Cxxx
Cxxx
dxxx
xx
dxxxCBx
xA
4
22
2
22
1ln
1ln2ln4
1244
1
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Chapter 15: Methods and Applications of Integration15.2 Integration by Partial Fractions
Example 5 – An Integral Not Requiring Partial FractionsFind
Solution: This integral has the form
Thus,
. 13
322 dx
xxx
Cxxdxxx
x
13ln 13
32 22
. 1 duu
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Chapter 15: Methods and Applications of Integration
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 xdxx
C
buaabua
bbuaduu
ln1 22
C
xxdx
xx
32232ln
91
32 2
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Chapter 15: Methods and Applications of Integration15.3 Integration by Tables
Example 3 – Integration by Tables
Find
Solution: Formula 28 states
Let u = 4x and a = √3, then du = 4 dx.
.316 2
xxdx
Cu
aauaauu
du
22
22ln1
Cx
xxx
dx
4
3316ln31
316
2
2
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Chapter 15: Methods and Applications of Integration15.3 Integration by Tables
Example 5 – Integration by TablesFind
Solution: Formula 42 states
If we let u = 4x, then du = 4 dx. Hence,
. 4ln7 2 dxxx
C
nu
nuuduuu
nnn
2
11
11lnln
Cxx
Cxxx
dxxxdxxx
14ln39
7
94
34ln4
647
44ln447 4ln7
3
33
2
32
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Chapter 15: Methods and Applications of Integration15.3 Integration by Tables
Example 7 – Finding a Definite Integral by Using TablesEvaluate
Solution: Formula 32 states
Letting u = 2x and a2 = 2, we have du = 2 dx.
Thus,
.24
4
12/32
xdx
Caua
uau
du
2222/322
Caua
uau
du
2222/322
621
662
2221
221
24
8
22
4
12/32
4
12/32
u
uu
dux
dx
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Chapter 15: Methods and Applications of Integration
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
37
3121
1
2
1
32
1
2
xdxx
dxxfab
fb
a
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Chapter 15: Methods and Applications of Integration
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 ' yxxyy
xCy
dxx
dyy
xy
dxdy
lnln
11
1
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Chapter 15: Methods and Applications of IntegrationExample 1 – Separation of Variables
Solution (cont’d):
0,
ln
ln
1
1
xCxCy
eey
ey
x
C
xC
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Chapter 15: Methods and Applications of Integration15.5 Differential Equations
Example 3 – Finding the Decay Constant and Half-LifeIf 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
01022.050
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
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Chapter 15: Methods and Applications of Integration
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
ctbeMN
1
tbCMN
1
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Chapter 15: Methods and Applications of Integration15.6 More Applications of Differential Equations
Example 1 – Logistic Growth of Club MembershipSuppose 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?
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Chapter 15: Methods and Applications of Integration15.6 More Applications of Differential EquationsExample 1 – Logistic Growth of Club Membership
Solution: Let N be the number of members enrolled in t years,
Thus,
1511
800501
,0 and 800 When
bbbC
MN
tM
t
5lnln151800200
,200 and 1 When
51
ce
Nt
c
781
151800
451
N
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Chapter 15: Methods and Applications of Integration15.6 More Applications of Differential Equations
Example 3 – Time of MurderA 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 TkdtdTaTk
dtdT
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Chapter 15: Methods and Applications of Integration15.6 More Applications of Differential EquationsExample 3 – Time of Murder
Solution (cont’d):
CktT
dtkT
dT
22ln
22
9ln02231ln,0 and 31 When
CCktT
98ln9ln12230ln
,1 and 30 When
kk
tT
ktTInktT
9229ln22ln Hence,
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Chapter 15: Methods and Applications of Integration15.6 More Applications of Differential EquationsExample 3 - Time of Murder
Solution (cont’d):
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
9/8ln9/15ln
98ln2237ln
, 37 When
tt
T
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Chapter 15: Methods and Applications of Integration
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
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Chapter 15: Methods and Applications of Integration 15.7 Improper Integrals
Example 1 – Improper IntegralsDetermine whether the following improper integrals are convergent or divergent. For any convergent integral, determine its value.
21
210
2lim lim 1 a.
1
2
1
3
13
r
r
r
r
xdxxdxx
1lim lim b.0
00
r
x
rr
x
r
x edxedxe
r
r
r
rxdxxdx
x 12/1
1
2/1
1
2lim lim 1 c.
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Chapter 15: Methods and Applications of Integration15.7 Improper Integrals
Example 3 – Density FunctionIn 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 00 for xkexf
x
11lim1lim
101
00
00
0
kkedxke
dxkedxxfdxxf
rx
r
rx
r
x