Chapter 15 - Methods and Applications of Integration

INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 15Chapter 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
dx
x
xln
∫
( )( ) ( )
( )[ ] Cxx
dx
x
xxxdx
x
x
+−=






−= ∫∫
2ln2
1
22ln
ln 2/1
xu ln= dx
x
dv
1
=
dx
x
du
1
= 2/12/1
2xdxxv == ∫
−

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
dy
y
yyydyy
+−=
+−=






−= ∫∫
1ln
ln
1
lnln
dy
y
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
dxxe
ex
dxx
eex
dxex
x
x
xx
x
2
)2(
22
12
122
12122
122
∫
∫∫
+
+
++
+
−=
−=
1
1212
1212
12
42
22
C
exe
dx
exe
dxxe
xx
xx
x
+−=
−=
++
++
+
∫∫

Example 5 – Applying Integration by Parts Twice
Solution (cont’d):
Cxx
e
C
exeex
dxex
x
xxx
x
+





+−=
++−=
+
+++
+
∫
2
1
2
422
2
12
1212122
122

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,
dx
x
x
273
12
2∫ −
+
.
9
12
3
1
2
dx
x
x
∫ −
+
( )( ) ( ) ( )
6
5
6
7
2
,3ifand,3If
3333
12
9
12
=−===
−
+
+
=
−+
+
=
−
+
AxBx
x
B
x
A
xx
x
x
x
Cxx
x
dx
x
dx
dx
x
x
+





−++=






−
+
+
=
−
+
∫ ∫∫
3ln
6
7
3ln
6
5
3
1
333
1
273
12 6
7
6
5
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,
dx
xxx
x
∫ ++
−− 42
23
( )
xCBxxxAx
xx
CBx
x
A
xxx
x
)()1(42
11
42
2
22
++++=−−
++
+
+=
++
−−
2,4,4 ==−= CBA
( ) C
x
xx
Cxxx
dx
xx
x
x
dx
xx
CBx
x
A
+







 ++
=
++++−=






++
+
+
−
=
++
+
+ ∫∫
4
22
2
22
1
ln
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
32
2
dx
xx
x
∫ ++
+
Cxxdx
xx
x
+++=
++
+
∫ 13ln
13
32 2
2
.
1
du
u∫

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
a
bua
bbua
duu
+





+
++=
+
∫ ln
1
22
( )
C
x
xdx
x
x
+





+
++=
+
∫ 32
2
32ln
9
1
32
2

15.3 Integration by Tables
Find
Let u = 4x and a = √3, then du = 4 dx.
.
316 2∫ +xx
dx
C
u
aau
aauu
du
+
−+
=
+
∫
22
22
ln
1
C
x
x
xx
dx
+
−+
=
+
∫ 4
3316
ln
3
1
316
2
2

Find
If we let u = 4x, then du = 4 dx. Hence,
( ) .4ln7 2
dxxx∫
( )
C
n
u
n
uu
duuu
nn
n
+
+
−
+
=
++
∫ 2
11
11
ln
ln
( ) ( ) ( )( )
( ) ( ) ( )
( )( ) Cx
x
C
xxx
dxxxdxxx
+−=
+







−=
= ∫∫
14ln3
9
7
9
4
3
4ln4
64
7
44ln4
4
7
4ln7
3
33
2
3
2

Example 7 – Finding a Definite Integral by Using
Tables
Evaluate
Letting u = 2x and a2
= 2, we have du = 2 dx.
Thus,
( )
.
24
4
1
2/32∫ +x
dx
( )
C
aua
u
au
du
+
±
±
=
±
∫ 2222/322
( )
C
aua
u
au
du
+
±
±
=
±
∫ 2222/322
( ) ( ) 62
1
66
2
222
1
22
1
24
8
2
2
4
1
2/32
4
1
2/32
−=





+
=
+
=
+
∫∫ 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:
( ) dxxf
ab
f
b
a
1
∫−
=
( )
3
7
312
1
1
2
1
32
1
2
=





=
−
=
−
=
∫
∫
x
dxx
dxxf
ab
f
b
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' >−= yx
x
y
y
xCy
dx
x
dy
y
x
y
dx
dy
lnln
11
1 −=
−=
−=
∫∫

Example 1 – Separation of Variables
0,
ln
ln
1
1
>=
=
= −
xC
x
C
y
e
e
y
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, tλ
eNN −
= 0
( ) 01022.0
50
6.0ln
6.0
6.0and50When
50
00
0
≈−=
=
==
−
λ
eNN
NNt
λ
days.82.67
2ln
islifehalftheandThus, 01022.0
0 ≈≈ −
λ
eNN 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
ct
be
M
N −
+
=
1
t
bC
M
N
+
=
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,
( )
15
11
800
50
1
,0and800When
=⇒
+
=⇒
+
=
==
b
bbC
M
N
tM
t
5lnln
151
800
200
,200and1When
5
1
=−=⇒
+
=
==
−
c
e
Nt
c
( )
781
151
800
4
5
1
≈
+
=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 and
T(t) = temperature of the body at time t.
By Newton’s law of cooling,
( ) ( )22−=⇒−= Tk
dt
dT
aTk
dt
dT

Example 3 – Time of Murder
( ) CktT
dtk
T
dT
+=−
=
− ∫∫
22ln
22
( ) ( ) 9ln02231ln
,0and31When
=⇒+=−
==
CCk
tT
( ) ( )
9
8
ln9ln12230ln
,1and30When
=⇒+=−
==
kk
tT
( ) kt
T
InktT =
−
⇒+=−
9
22
9ln22lnHence,

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
8
ln2237ln
,37When
−≈=⇒





=−
=
tt
T

15.7 Improper Integrals15.7 Improper Integrals
• The improper integral is defined as
• The improper integral is defined as
( ) dxxf
a
∫
∞
( ) ( ) dxxfdxxf
r
a
r
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
1
0
2
limlim
1
a.
1
2
1
3
1
3
=+−=





−==
−
∞→
−
∞→
∞
∫∫
r
r
r
r
x
dxxdx
x
[ ] 1limlimb.
0
00
===
−∞→−∞→
∞−
∫∫ r
x
r
r
x
r
x
edxedxe
[ ] ∞===
∞→
−
∞→
∞
∫∫
r
r
r
r
xdxxdx
x
1
2/1
1
2/1
1
2limlim
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
( )


 ≥
=
−
elsewhere0
0for xke
xf
x
( ) ( )
[ ] 11lim1lim
101
0
0
00
0
=⇒=⇒=
=+⇒=+
−
∞→
−
∞→
∞
−
∞
∞−
∫
∫∫∫
kkedxke
dxkedxxfdxxf
rx
r
r
x
r
x

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