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INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
2007 Pearson Education Asia
Chapter 14 Chapter 14 IntegrationIntegration
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 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
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• 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 ,
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• 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 14: Integration
Chapter Objectives
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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 14: Integration
Chapter OutlineChapter Outline
14.4)
14.5)
14.6)
14.7)
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Approximate Integration
Area
Area between Curves
Consumers’ and Producers’ Surplus
14.8)
14.9)
14.10)
Chapter 14: Integration
Chapter OutlineChapter Outline
14.11)
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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.
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Find .
Solution:
Chapter 14: Integration
14.2 The Infinite Integral
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
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Find
Solution:
Chapter 14: Integration
14.2 The Infinite Integral
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
2007 Pearson Education Asia
Chapter 14: Integration
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
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Chapter 14: Integration
14.4 More Integration Formulas
Basic Integration Formulas
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Chapter 14: Integration
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.
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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.
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
14.7 The Fundamental Theorem of Integral Calculus
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
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Chapter 14: Integration
14.7 The Fundamental Theorem of Integral Calculus
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
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Chapter 14: Integration
14.7 The Fundamental Theorem of Integral Calculus
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
14.8 Approximate Integration
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
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Chapter 14: Integration
14.8 Approximate Integration
Simpson’s Rule
• Approximating the graph of f by parabolic segments gives
even. is and / where
142243
nnabh
bfhnafhafhafafh
dxxfb
a
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Chapter 14: Integration
14.8 Approximate Integration
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.
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Chapter 14: Integration
14.8 Approximate Integration
Example 3 - Demography
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
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Chapter 14: Integration
14.8 Approximate Integration
Example 3 - Demography
Life table:
2007 Pearson Education Asia
Chapter 14: Integration
14.8 Approximate Integration
Example 3 - Demography
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
14.9 Area
Example 5 - Statistics Application
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
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Chapter 14: Integration
14.9 Area
Example 5 - Statistics Application
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
2007 Pearson Education Asia
Chapter 14: Integration
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
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Chapter 14: Integration
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
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Chapter 14: Integration
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
000,18$2
1.060600
0
2
0
0
0
qqdqpgpPS
q
9000$2
05.030600
0
2
0
0
0
qqdqpqfCS
q