Chapter14 integration-151007043436-lva1-app6892

INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 14Chapter 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
∫∫∫ du
u
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
( ) ( ) xxxxxxx
dx
d
dy ∆+−=∆−+−= 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
2
300
300
31 







+
−==
t
tPP

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,
( )
( ) ( )
dt
t
dt
t
tPdPP 4
3
2
300
300
3
300
300
3'
+
=







+
−==≈∆
Solution:
.2500ifFind 2
pq
dq
dp
−=
p
p
dp
dqdq
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 =+=∫ 'onlyif

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:
dx∫5
Cxdx +=∫ 55
Find .
Solution:
dxx∫7
C
x
dxx +=∫ 2
7
7
2
CtC
t
dxtdx
t
+=+== ∫∫
−
2
2/1
1
a.
2/1
2/1
C
x
C
x
dx
x
+−=+





+−
=
+−
∫ 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
Cxe
xx
dxexx
x
x
x
++−=
+−+−=
−+−∫
10
4
7
9
10
10
4
7
5/9
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.
dx
x
x
∫
−
2
3
1
b.
( )( )
( ) ( )
C
xxx
Cx
xx
dx
xx
+−+=
+





−+=
+−
∫
212
5
9
3
2
5
3
2
6
1
6
312
a.
23
23
( )
C
x
x
dxxx
dx
x
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
( ) ( ) CxxCx
x
dxxy +−=+−=−= ∫ 444
2
848 2
2
( ) ( )
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.
164100 2/3
≤≤= xx
dx
dy

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

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

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,
( ) ( )[ ]
Cq
qq
.
dqqq..qc
++





−=
+−= ∫
2.0
2
25
3
002.0
0000010
2.02500200000010
23
2
( )
( ) 67.5416$10000
40002.0
2
25
3
0020
0000010
23
=
++





−=
c
q
qq.
.qc

14.4 More Integration Formulas14.4 More Integration Formulas
Power Rule for Integration
Integrating Natural Exponential Functions
Integrals Involving Logarithmic Functions
1if
1
1
−≠+
+
=∫
+
nC
n
u
dxu
n
n
∫ += Cedue uu
0forln
1
≠+=∫ xCxdx
x

14.4 More Integration Formulas
Basic Integration Formulas

Example 1 - Applying the Power Rule for Integration
Find the integral of
Solution:
( ) ( ) ( ) C
x
C
u
duudxx +
+
=+==+ ∫∫ 21
1
21
1a.
2121
2020
( ) dxx
20
1a. ∫ +
( ) dxxx∫ +
332
73b.
( ) ( ) ( ) C
x
C
u
duudxxx +
+
=+==+ ∫∫ 4
7
4
73
434
3332
dxxduxu 23
37Letb. =⇒+=

Example 3 - Adjusting for du
Find
Solution:
CyC
y
dyya. +=+=∫
3/4
33/4
33
4
63
3/4
66
( )
dx
xx
xx
b. ∫ ++
+
424
3
73
32
( )dxxxduxxu 6473Let 324
+=⇒++=
( )
C
xx
C
udu
u +
++
−=+
−
⋅=





−
−
∫ 324
3
4
736
1
32
1
2
dyya. ∫3 6
( )
dx
xx
xx
b. ∫ ++
+
424
3
73
32

Example 5 - Integrals Involving Exponential Functions
Find
Solution:
dxxex
∫2a.
xdxduxu 2Leta. 2
=⇒=
( ) dxex xx 32 3
1b. +
∫ +
[ ]
Cedue
xdxedxxe
xu
xx
+==
=
∫
∫∫
2
2
22
( )dxxduxxu 333Letb. 23
+=⇒+=
( )
Ce
Cduedxex
xx
uxx
+=
+=+
+
+
∫∫
3
32
3
3
3
1
3
1
1

Example 7 - Integrals Involving Exponential Functions
Find
Solution:
( ) .
73
32
24
3
dx
xx
xx
∫ ++
+
( )dxxxduxxu 6473Let 324
+=⇒++=
( )
( ) Cxx
CxxCudx
xx
xx
+++=
+++=+=
++
+
∫
73ln
2
1
73ln
2
1
ln
2
1
73
32
24
24
24
3

14.5 Techniques of Integration14.5 Techniques of Integration
Example 1 - Preliminary Division before Integration
Find
Cx
x
dx
x
xdx
x
xx
++=





+=
+
∫∫ ln
2
1
a.
2
2
3
Cx
xx
dx
x
xxdx
x
xxx
++++=






+
++=
+
+++
∫∫
12ln
2
1
23
12
1
12
132
b.
23
2
23

14.5 Techniques of Integration
Example 3 - An Integral Involving bu
Find
Solution:
.23
dxx
∫
−
( )( ) dxduxu 2ln32lnLet −=⇒−=
( )( )
( )( ) ( )
CCe
Ceduedxedx
xx
uuxx
+−=+−=
+−=−==
−−
−−
∫∫∫
332ln
32ln3
2
2ln
1
2ln
1
2ln
1
2ln
1
2
• General formula for integrating bu
is
Cb
b
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.
( )dxxf
b
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
4
8
6
12188
22
4
2
n
nnnnn
n
n
n
nn
k
fx
n
kfS
n
k
n
k
n
( )( )
3
16
3
8
8
121
3
4
8limlim 2
=−=










 ++
−=
∞→∞→ n
nn
S
n
n
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
1
1
2
9
15
2
1933
1
−





+=−
+
=











= ∑= nn
n
nn
kfS
n
k
n
( )
2
21
2
9
15
1
1
2
9
limlim5
3
0
−==





−





+==−
∞→∞→∫ n
Sdxx
n
n
n

14.7 The Fundamental Theorem of14.7 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,
( ) ( ) ( )aFbFdxxf
b
a
−=∫
( ) ( )∫∫ −=
a
b
b
a
dxxfdxxf
( ) 0=∫
b
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.aiswhere kdxxfkdxxkf
b
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:
( ) .63
3
1
2
∫−
+− dxxx
( )
( ) ( ) ( ) ( )
48
16
2
1
136
2
3
3
6
2
63
2
3
2
3
3
1
2
3
3
1
2
=






−+
−
−−−





+−=






+−=+−
−−
∫ x
x
xdxxx

Example 3 - Evaluating Definite Integrals
Find
Solution:
( )[ ]∫ ++
2
1
323/1
14a. dxttt
( )[ ] ( ) ( )
( ) ( ) 8
585
2625
8
1
123
4
1
2
1
414a.
3443/4
2
1
2/12
3
4
3/42
1
323/1
+=−+−=







 +






+=++∫
tt
dxttt
[ ] ( ) ( )1
3
1
3
1
3
1
b. 3031
0
3
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 += q
dq
dc
( ) ( ) ( ) [ ]
112020803200
23.026.080100
100
80
2
100
80
=−=
+=+=− ∫ qqdqqcc

14.8 Approximate Integration14.8 Approximate Integration
Trapezoidal Rule
• To find the area of a trapezoidal area, we have
( ) ( ) ( ) ( ) ( )( ) ( )[ ]
( ) ./where
12222
2
nb-ah
bfhnafhafhafaf
h
dxxf
b
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,
dx
x∫ +
1
0
2
1
1
2.0
5
01
=
−
=
−
=
n
ab
h

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
( ) 784.08373.7
2
2.0
1
1
1
0
2
≈≈
+∫ dx
x

Simpson’s Rule
• Approximating the graph of f by parabolic segments
gives
( ) ( ) ( ) ( ) ( ) ( )( )[ ]
( ) even.isand/where
14224
3
nnabh
bfhnafhafhafaf
h
dxxf
b
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.
( )dttl
mx
x
∫
+1

Life table:

Solution: We want to estimate
Thus
The terms to be added are
By the trapezoidal rule,
( ) .
35
20
dttl∫
5
3
2035
=
−
=
−
=
n
ab
h
( )
( ) ( )
( ) ( )
( )
sum
l
l
l
l
=
=
==
==
=
90,7755
964,9735
700,1966230,982302
254,197627,982252
857,9820
( ) ( ) 5.937,476,1775,590
3
5
35
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
( ) ( ) areadxxfxxf
b
a
=→∆ ∫∑
Find the area of the region bounded by the curve
and the x-axis.
2
6 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
9
18
3
8
3
4
12
32
66
2
3
322
3
2
=




 −
−−−−





−−=






−−=−−=
−−
∫
xx
xdxxxarea

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
[ ] ( )1
2
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.
( ) ( )dxxfdxcP
d
c
∫=≤≤
( )4
1
0. ≤≤ xPa
( )2
1
. ≥xPb

14.9 Area
Solution:
a.
b.
( ) ( )
32
5
32
6
60
4/1
0
32
4/1
0
2
4
1
=





−=
−=≤≤ ∫
xx
dxxxxP
( ) ( )
2
1
32
6
6
1
2/1
32
1
2/1
2
2
1
=





−=
−=≥ ∫
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 ∆−
1or0 == xx
( ) 6
1
22/3
1
0
22/31
0
=











−=−= ∫
xx
dxxxarea

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
9and1,2,5ofrightFor
1and9,2,5ofleftFor
xyxy
xyxy
lowerupper
lowerupper
−=+=
+=−=
[ ] ( ) ( )[ ]
( )
[ ] ( ) ( )[ ]
( ) xx
xxxyyxx
xx
xxxyyxx
lowerupper
lowerupper
∆−=
−−+=∆−==
∆−=
+−−=∆−==
82
91,3to2From
28
19,2to0From
2
22
2
22
( ) ( ) 3
46
8228
3
2
2
2
0
2
=−+−= ∫∫ dxxdxxarea

14.10 Area between Curves
Example 5 - Advantage of Horizontal Elements
Find 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
9
2
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.
( )[ ]dqpqfCS
q
∫ −=
0
0
0
( )[ ]dqqgpPS
q
∫ −−=
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
( ) 706001.010Thus
600
05.01001.010
0
0
=+=
==
−=+
p
qq
qq
( )[ ] 000,18$
2
1.060
600
0
2
0
0
0
=





−=−= ∫
q
qdqpgpPS
q
( )[ ] 9000$
2
05.030
600
0
2
0
0
0
=





−=−= ∫
q
qdqpqfCS
q

Chapter14 integration-151007043436-lva1-app6892

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