Antiderivatives and the Rules of Integration

1
6
Integration

• Antiderivatives and the Rules of Integration
• Integration by Substitution
• Area and the Definite Integral
• The Fundamental Theorem of Calculus
• Evaluating Definite Integrals
• Area Between Two Curves
• Applications of the Definite Integral to Business
  and Economics

2
6.1

• Antiderivatives and the Rules of Integration

3
Antiderivatives

• Recall the Maglev problem discussed in chapter 2.
  
• The question asked then was
• If we know the position of the maglev at any time
  t, can we find its velocity at that time?
• The position was described by f(t), and the
  velocity by f '(t).
• Now, in Chapters 6 and 7 we will consider
  precisely the opposite problem
• If we know the velocity of the maglev at any time
  t, can we find its position at that time?
• That is, knowing its velocity function f '(t),
  can we find its position function f(t)?

4
Antiderivatives

• To solve this kind of problems, we need the
  concept of the antiderivative of a function.
• A function F is an antiderivative of f on an
  interval I if F '(t) f(t) for
  all of t in I.

5
Examples

• Let
• Show that F is an antiderivative of
• Solution
• Differentiating the function F, we obtain
• and the desired result follows.

Example 1, page 398
6
Examples

• Let F(x) x, G(x) x 2, H(x) x C, where C
  is a constant.
• Show that F, G, and H are all antiderivatives of
  the function f defined by f(x) 1.
• Solution
• Since
• we see that F, G, and H are indeed
  antiderivatives of f.

Example 2, page 398
7
Theorem 1

• Let G be an antiderivative of a function f.
• Then, every antiderivative F of f must be of the
  form
• F(x) G(x) C
• where C is a constant.

8
Example

• Prove that the function G(x) x2 is an
  antiderivative of the function f(x) 2x.
• Write a general expression for the
  antiderivatives of f.
• Solution
• Since G'(x) 2x f(x), we have shown that G(x)
  x2 is an antiderivative of f(x) 2x.
• By Theorem 1, every antiderivative of the
  function f(x) 2x has the form F(x)
  x2 C, where C is a constant.

Example 3, page 399
9
The Indefinite Integral

• The process of finding all the antiderivatives of
  a function is called antidifferentiation or
  integration.
• We use the symbol ?, called an integral sign, to
  indicate that the operation of integration is to
  be performed on some function f.
• Thus,
• where C and K are arbitrary constants.

10
Basic Integration Rules

• Rule 1 The Indefinite Integral of a Constant

11
Example

• Find each of the following indefinite integrals
• a. b.
• Solution
• Each of the integrals had the form f(x) k,
  where k is a constant.
• Applying Rule 1 in each case yields
• a. b.

Example 4, page 400
12
Basic Integration Rules

• From the rule of differentiation,
• we obtain the following rule of integration
• Rule 2 The Power Rule

13
Examples

• Find the indefinite integral

Example 5, page 401
14
Examples

• Find the indefinite integral

Example 5, page 401
15
Examples

• Find the indefinite integral

Example 5, page 401
16
Basic Integration Rules

• Rule 3 The Indefinite Integral of a Constant
  Multiple of a Function
• where c is a constant.

17
Examples

• Find the indefinite integral

Example 6, page 402
18
Examples

• Find the indefinite integral

Example 6, page 402
19
Basic Integration Rules

• Rule 4 The Sum Rule

20
Examples

• Find the indefinite integral

Example 7, page 402
21
Basic Integration Rules

• Rule 5 The Indefinite Integral of the
  Exponential Function

22
Examples

• Find the indefinite integral

Example 8, page 403
23
Basic Integration Rules

• Rule 6 The Indefinite Integral of the Function
  f(x) x1

24
Examples

• Find the indefinite integral

Example 9, page 403
25
Differential Equations

• Given the derivative of a function, f ', can we
  find the function f ?
• Consider the function f '(x) 2x 1 from
  which we want to find f(x).
• We can find f by integrating the equation
• where C is an arbitrary constant.
• Thus, infinitely many functions have the
  derivative f ', each differing from the other by
  a constant.

26
Differential Equations

• Equation f '(x) 2x 1 is called a
  differential equation.
• In general, a differential equation involves the
  derivative of an unknown function.
• A solution of a differential equation is any
  function that satisfies the differential
  equation.
• For the case of f '(x) 2x 1, we find that
  f(x) x2 x C gives all the solutions of the
  differential equation, and it is therefore called
  the general solution of the differential
  equation.

27
Differential Equations
f(x) x2 x 3
y

• Different values of C yield different functions
  f(x).
• But all these functions have the same slope for
  any given value of x.
• For example, for any value of C, we always find
  that f '(1) 1.

f(x) x2 x 2
5 4 3 2 1 1
f(x) x2 x 1
f(x) x2 x 0
f '(1) 1
f '(1) 1
f(x) x2 x 1
f '(1) 1
f '(1) 1
x
1 1 2 3
f '(1) 1
28
Differential Equations

• It is possible to obtain a particular solution by
  specifying the value the function must assume for
  a given value of x.
• For example, suppose we know the function f must
  pass through the point (1, 2), which means f(1)
  2.
• Using this condition on the general solution we
  can find the value of C
• f(1) 12 1 C 2
• C 2
• Thus, the particular solution is
• f(x) x2 x 2

29
Differential Equations
y

• Here is the graph of the particular solution of
  f when C 2.
• Note that this graph does go through the
  point (1, 2).

f(x) x2 x 2
5 4 3 2 1 1
(1, 2)
x
1 1 2 3
30
Initial Value Problems

• The problem we just discussed is of a type called
  initial value problem.
• In this type of problem we are required to find a
  function satisfying
• A differential equation.
• One or more initial conditions.

31
Example

• Find the function f if it is known that
• Solution
• Integrating the function f ', we find

Example 10, page 404
32
Example

• Find the function f if it is known that
• Solution
• Using the condition f(1) 9, we have
• Therefore, the required function f is

Example 10, page 404
33
Applied Example Velocity of Maglev

• In a test run of a maglev, data obtained from
  reading its speedometer indicate that the
  velocity of the maglev at time t can be described
  by the velocity function
• Find the position function of the maglev.
• Assume that initially the maglev is located at
  the origin of a coordinate line.

Applied Example 11, page 404
34
Applied Example Velocity of Maglev

• Solution
• Let s(t) denote the position of the maglev at any
  given time t (0 ? t ? 30). Then, s'(t) v(t).
• So, we have the initial value problem
• Integrating the function s', we find

Applied Example 11, page 404
35
Applied Example Velocity of Maglev

• Solution
• Using the condition s(0) 0, we have
• Therefore, the required function s is

Applied Example 11, page 404
36
6.2

• Integration by Substitution

37
Integration by Substitution

• The method of substitution is related to the
  chain rule for differentiating functions.
• It is a powerful tool for integrating a large
  class of functions.

38
How the Method of Substitution Works

• Consider the indefinite integral
• One way to solve this integral is to expand the
  expression and integrate the resulting integrand
  term by term.
• An alternative approach simplifies the integral
  by making a change of variable.
• Write u 2x 4
• with differential du 2dx

39
How the Method of Substitution Works

• Substitute u 2x 4 and du 2dx in the
  original expression
• Now its easy to integrate
• Replacing u by u 2x 4, we obtain

40
How the Method of Substitution Works

• We can verify the result by finding its
  derivative
• The derivative is indeed the original integrand
  expression.

41
The Method of Integration by Substitution

• Step 1 Let u g(x), where g(x) is part of the
  integrand, usually the inside function of the
  composite function f(g(x)).
• Step 2 Find du g'(x)dx.
• Step 3 Use the substitution u g(x) and du
  g'(x)dx to convert the entire integral into one
  involving only u.
• Step 4 Evaluate the resulting integrand.
• Step 5 Replace u by g(x) to obtain the final
  solution as a function of x.

42
Examples

• Find

Solution Step 1 The integrand involves the
composite function with inside function
So, we choose
Example 1, page 413
43
Examples

• Find

Solution Step 2 Find du/dx and solve for du
Example 1, page 413
44
Examples

• Find

Solution Step 3 Substitute u x2 3 and du
2xdx, to obtain an integral involving only u
Example 1, page 413
45
Examples

• Find

Solution Step 4 Evaluate the integral Step
5 Replace u by x2 3 to find the solution
Example 1, page 413
46
Examples

• Find
• Solution
• Let u 3x, so that du 3dx, or dx ? du.
• Substitute to express the integrand in terms of
  u
• Evaluate the integral
• Replace u by 3x to find the solution

Example 4, page 414
47
Examples

• Find
• Solution
• Let u 3x2 1, so that du 6xdx, or xdx
  du.
• Substitute to express the integrand in terms of
  u
• Evaluate the integral
• Replace u by 3x2 1 to find the solution

Example 5, page 415
48
Examples

• Find
• Solution
• Let u ln x, so that du 1/x dx, or dx/x du.
• Substitute to express the integrand in terms of
  u
• Evaluate the integral
• Replace u by ln x to find the solution

Example 6, page 415
49
6.3

• Area and the Definite Integral

50
The Area Under the Graph of a Function

• Let f be a nonnegative continuous function on a,
  b. Then, the area of the region under the graph
  of f is
• where x1, x2, , xn are arbitrary points in the
  n subintervals of a, b of equal width Dx (b
  a)/n.

51
The Definite Integral

• Let f be a continuous function defined on a, b.
  If
• exists for all choices of representative points
  x1, x2, , xn in the n subintervals of a, b
  of equal width Dx (b a)/n, then the limit is
  called the definite integral of f from a to b and
  is denoted by
• Thus,
• The number a is the lower limit of integration,
  and the number b is the upper limit of
  integration.

52
Integrability of a Function

• Let f be a continuous on a, b. Then, f is
  integrable on a, b that is, the definite
  integral
• exists.

53
Geometric Interpretation of the Definite Integral

• If f is nonnegative and integrable on a, b,
  then
• is equal to the area of the region under the
  graph of f on a, b.

54
Geometric Interpretation of the Definite Integral

• The definite integral is equal to the area of the
  region under the graph of f on a, b

y
x
a
b
55
Geometric Interpretation of the Definite Integral

• If f is continuous on a, b, then
• is equal to the area of the region above a, b
  minus the region below a, b.

56
Geometric Interpretation of the Definite Integral

• The definite integral is equal to the area of the
  region above a, b minus the region below a,
  b

y
R3
R1
x
a
b
R2
57
6.4

• The Fundamental Theorem of Calculus

58
Theorem 2The Fundamental Theorem of Calculus

• Let f be continuous on a, b. Then,
• where F is any antiderivative of f that is, F
  '(x) f(x).

59
Example

• Let R be the region under the graph of f(x) x
  on the interval 1, 3.
• Use the fundamental theorem of calculus to find
  the area A of R and verify your result by
  elementary means.

Example 1, page 431
60
Example

• Solution
• The graph shows the region to be evaluated.
• Since f is nonnegative on 1, 3, the area of R
  is given by the definite integral of f from 1 to
  3.

y
4 3 2 1
x 3
R
x 1
x
1 2 3 4
Example 1, page 431
61
Example

• Solution
• By the fundamental theorem of calculus, we have
• Thus, the area A of region R is 4 square units.
• Note that the constant of integration C dropped
  out.
• This is true in general.

Example 1, page 431
62
Example

• Solution
• Using elementary means, note that area A is equal
  
• to R1 (base ? height) plus R2 ( ? base ?
  height).
• Thus, A R1 R2 2(1) (2)(2) 4

y
4 3 2 1
2
R2
R1
x
1 2 3 4
Example 1, page 431
2
63
Example

• Find the area of the region under the graph of y
  x2 1 from x 1 to x 2.
• Solution
• Note below that the full region R under
  consideration lies above the x axis.

y
f(x) x2 1
5 4 3 2
R
x
2 1 1 2
Example 3, page 432
64
Example

• Find the area of the region under the graph of y
  x2 1 from x 1 to x 2.
• Solution
• Using the fundamental theorem of calculus, we
  find that the required area is
• or 6 square units.

Example 3, page 432
65
Net Change Formula

• The net change in a function f over an interval
  a, b is given by
• provided f ' is continuous on a, b.

66
Applied Example Population Growth in Clark County

• Clark County, Nevada, (dominated by Las Vegas) is
  the fastest growing metropolitan area in the
  United States.
• From 1970 through 2000, the population was
  growing at a rate of
• people per decade, where t 0 corresponds to
  the beginning of 1970.
• What was the net change in population over the
  decade from 1980 to 1990?

Applied Example 6, page 433
67
Applied Example Population Growth in Clark County

• Solution
• The net change in population over the decade from
  1980 to 1990 is given by P(2) P(1) , where P
  denotes the population in the county at time t.
• But P ' R, and so the net change in population
  is

Applied Example 6, page 433
68
Applied Example Assembly Time of Workers

• An efficiency study conducted for Elektra
  Electronics showed that the rate at which Space
  Commander walkie-talkies are assembled by the
  average worker t hours after starting work at
  800 a.m. is given by the function
• Determine how many walkie-talkies can be
  assembled by the average worker in the first hour
  of the morning shift.

Applied Example 8, page 435
69
Applied Example Assembly Time of Workers

• Solution
• Let N(t) denote the number of walkie-talkies
  assembled by the average worker t hours after
  starting work in the morning shift.
• Then, we have
• Therefore, the number of units assembled by the
  average worker in the first hour of the morning
  shift is
• or 20 units.

Applied Example 8, page 435
70
6.5

• Evaluating Definite Integrals

71
Properties of the Definite Integral

• Let f and g be integrable functions, then

72
Examples Using the Method of Substitution

• Evaluate
• Solution
• First, find the indefinite integral
• Let u 9 x2 so that

Example 1, page 442
73
Examples Using the Method of Substitution

• Evaluate
• Solution
• First, find the indefinite integral
• Then, integrate by substitution using xdx du

Example 1, page 442
74
Examples Using the Method of Substitution

• Evaluate
• Solution
• Using the results, we evaluate the definite
  integral

Example 1, page 442
75
Examples Using the Method of Substitution

• Evaluate
• Solution
• Let u x3 1 so that

Example 3, page 444
76
Examples Using the Method of Substitution

• Evaluate
• Solution
• Find the lower and upper limits of integration
  with respect to u
• When x 0, the lower limit is u (0)3 1 1.
• When x 1, the upper limit is u (1)3 1 2.
• Substitute x2dx 2du, along with the limits of
  integration

Example 3, page 444
77
Examples Using the Method of Substitution

• Find the area of the region R under the graph of
  
• f(x) e(1/2)x from x 1 to x 1.
• Solution
• The graph shows region R

y
3 2 1
f(x) e(1/2)x
R
x
2 1 1 2
Example 4, page 444
78
Examples Using the Method of Substitution

• Find the area of the region R under the graph of
  
• f(x) e(1/2)x from x 1 to x 1.
• Solution
• Since f(x) is always greater than zero, the area
  is given by
• To evaluate this integral, we substitute
• so that

Example 4, page 444
79
Examples Using the Method of Substitution

• Find the area of the region R under the graph of
  
• f(x) e(1/2)x from x 1 to x 1.
• Solution
• When x 1 , u , and when x 1, u .
• Substitute dx 2du, along with the limits of
  integration
• or approximately 2.08 square units.

Example 4, page 444
80
Average Value of a Function

• Suppose f is integrable on a, b.
• Then, the average value of f over a, b is

81
Applied Example Automobile Financing

• The interest rates changed by Madison Finance on
  auto loans for used cars over a certain 6-month
  period in 2008 are approximated by the function
• where t is measured in months and r(t) is the
  annual percentage rate.
• What is the average rate on auto loans extended
  by Madison over the 6-month period?

Applied Example 6, page 446
82
Applied Example Automobile Financing

• Solution
• The average rate over the 6-month period is given
  by
• or 9 per year.

Applied Example 6, page 446
83
6.6

• Area Between Two Curves

84
The Area Between Two Curves

• Let f and g be continuous functions such that
  f(x) ? g(x) on the interval a, b.
• Then, the area of the region bounded above by
  y f(x) and below by y g(x) on a,
  b is given by

85
Examples

• Find the area of the region bounded by the
  x-axis, the graph of y x2 4x 8, and the
  lines x 1 and x 4.
• Solution
• The region R is being bounded above by the graph
  f(x) 0 and below by the graph of g(x) y x2
  4x 8 on 1, 4

y
2 2 4 6
x
4 8 12
R
x 4
x 1
y x2 4x 8
Example 1, page 454
86
Examples

• Find the area of the region bounded by the
  x-axis, the graph of y x2 4x 8, and the
  lines x 1 and x 4.
• Solution
• Therefore, the area of R is given by

Example 1, page 454
87
Examples

• Find the area of the region bounded by
• f(x) 2x 1, g(x) x2 4, x 1, and x
  2.
• Solution
• Note that the graph of f always lies above that
  of g for all x in the interval 1, 2

y
y 2x 1
4 2 2 4
y x2 4
R
x
4 2 2 4
x 2
x 1
Example 2, page 454
88
Examples

• Find the area of the region bounded by
• f(x) 2x 1, g(x) x2 4, x 1, and x
  2.
• Solution
• Since the graph of f always lies above that of g
  for all x in the interval 1, 2, the required
  area is given by

Example 2, page 454
89
Examples

• Find the area of the region that is completely
  enclosed by the graphs of f(x) 2x 1 and
  g(x) x2 4.
• Solution
• First, find the points of intersection of the two
  curves.
• To do this, you can set g(x) f(x) and solve for
  x
• so, the graphs intersect at x 1 and at x 3.

Example 3, page 455
90
Examples

• Find the area of the region that is completely
  enclosed by the graphs of f(x) 2x 1 and
  g(x) x2 4.
• Solution
• The graph of f always lies above that of g for
  all x in the interval 1, 3 between the two
  intersection points

y
y 2x 1
(3 , 5)
4 2 4
y x2 4
R
x
4 2 2 4
( 1 , 3)
Example 3, page 455
91
Examples

• Find the area of the region that is completely
  enclosed by the graphs of f(x) 2x 1 and
  g(x) x2 4.
• Solution
• Since the graph of f always lies above that of g
  for all x in the interval 1, 3, the required
  area is given by

Example 3, page 455
92
Examples

• Find the area of the region bounded by
• f(x) x2 2x 1, g(x) ex 1, x
  1, and x 1.
• Solution
• Note that the graph of f always lies above that
  of g for all x in the interval 1, 1

y
3 2 1 3
y x2 2x 1
x 1
x
3 2 2 3
R
x 1
y ex 1
Example 4, page 455
93
Examples

• Find the area of the region bounded by
• f(x) x2 2x 1, g(x) ex 1, x
  1, and x 1.
• Solution
• Since the graph of f always lies above that of g
  for all x in the interval 1, 1, the required
  area is given by

Example 4, page 455
94
Examples

• Find the area of the region bounded by
• f(x) x3, the x-axis, x 1, and x 1.
• Solution
• The region being considered is composed of
• two subregions R1 and R2

y
y x3
x 1
1 1
R2
1
x
R1
1
x 1
Example 5, page 456
95
Examples

• Find the area of the region bounded by
• f(x) x3, the x-axis, x 1, and x 1.
• Solution
• To find R1 and R2 consider the x-axis as g(x)
  0.
• Since g(x) ? f(x) on 1, 0, the area of R1 is
  given by

y
y x3
x 1
1 1
1
x
R1
1
x 1
Example 5, page 456
96
Examples

• Find the area of the region bounded by
• f(x) x3, the x-axis, x 1, and x 1.
• Solution
• To find R1 and R2 consider the x-axis as g(x)
  0.
• Since g(x) ? f(x) on 0, 1, the area of R2 is
  given by

y
y x3
x 1
1 1
R2
1
x
1
x 1
Example 5, page 456
97
Examples

• Find the area of the region bounded by
• f(x) x3, the x-axis, x 1, and x 1.
• Solution
• Therefore, the required area R is
• square units.

y
y x3
x 1
1 1
R2
1
x
R1
1
x 1
Example 5, page 456
98
Examples

• Find the area of the region bounded by
• f(x) x3 3x 3 and g(x) x 3.
• Solution
• The region R being considered is composed of
• two subregions R1 and R2

y
y x 3
5 3 1
y x3 3x 3
R1
R2
x
3 1 1 2 3
Example 6, page 457
99
Examples

• Find the area of the region bounded by
• f(x) x3 3x 3 and g(x) x 3.
• Solution
• To find the points of intersection, we solve
  simultaneously the equations y x3 3x 3
  and y x 3.

y
y x 3
(2 , 5)
5 3 1
y x3 3x 3
R1

• So, x 0, x 2, and x 2.
• The points of intersection of the two
  curves are
• ( 2, 1), (0, 3), and (2, 5).

(0 , 3)
R2
( 2, 1)
x
3 1 1 2 3
Example 6, page 457
100
Examples

• Find the area of the region bounded by
• f(x) x3 3x 3 and g(x) x 3.
• Solution
• Note that f(x) ? g(x) for 2, 0, so the area
  of region R1 is
  given by

y
y x 3
(2 , 5)
5 3 1
y x3 3x 3
R1
(0 , 3)
( 2, 1)
x
3 1 1 2 3
Example 6, page 457
101
Examples

• Find the area of the region bounded by
• f(x) x3 3x 3 and g(x) x 3.
• Solution
• Note that g(x) ? f(x) for 0, 2, so the area of
  region R2 is given by

y
y x 3
(2 , 5)
5 3 1
y x3 3x 3
(0 , 3)
R2
( 2, 1)
x
3 1 1 2 3
Example 6, page 457
102
Examples

• Find the area of the region bounded by
• f(x) x3 3x 3 and g(x) x 3.
• Solution
• Therefore, the required area R is
• square units.

y
y x 3
(2 , 5)
5 3 1
y x3 3x 3
R1
(0 , 3)
R2
( 2, 1)
x
3 1 1 2 3
Example 6, page 457
103
6.7

• Applications of the Definite Integral to Business
  and Economics

104
Consumers and Producers Surplus

• Suppose p D(x) is the demand function that
  relates the price p of a commodity to the
  quantity x demanded of it.
• Now suppose a unit market price p has been
  established, along with a corresponding quantity
  demanded x.
• Those consumers who would be willing to pay a
  unit price higher than p for the commodity would
  in effect experience a savings.
• This difference between what the consumer would
  be willing to pay and what they actually have to
  pay is called the consumers surplus.

105
Consumers and Producers Surplus

• To derive a formula for computing the consumers
  surplus, divide the interval 0,x into n
  subintervals, each of length ?x x/n, and denote
  the right endpoints of these intervals by x1, x2,
  , xn x

p
D(x1)
D(x2)
D(x3)
D(x4)
D(xn 1)
p
p p
D(xn) p
x
x1
x2
xn 1
xn
x3
x4
106
Consumers and Producers Surplus

• There are consumers who would pay a price of at
  least D(x1) for the first ?x units instead of the
  market price of p.
• The savings to these consumers is approximated by
• which is the area of the rectangle r1

p
D(x1)
r1
p p
p
x
x1
107
Consumers and Producers Surplus

• Similarly, the savings the consumer experiences
  for the consecutive increments of ?x are depicted
  by the areas of rectangles r2, r3, r4,
  , rn

p
r1
r2
r3
r4
rn 1
rn
p p
p
x
xn
108
Consumers and Producers Surplus

• Adding r1 r2 r3 rn, and letting n
  approach infinity, we obtain the consumers
  surplus CS formula
• where D(x) is the demand function, p is the
  unit market price, and x is the quantity
  demanded.

p
CS
p D(x)
p
x
x
109
Consumers and Producers Surplus

• Similarly, we can derive a formula for the
  producers surplus.
• Suppose p S(x) is the supply function that
  relates the price p of a commodity to the
  quantity x supplied of it.
• Again, suppose a unit market price p has been
  established, along with a corresponding quantity
  supplied x.
• Those sellers who would be willing to sell at
  unit price lower than p for the commodity would
  in effect experience a gain or profit.
• This difference between what the seller would be
  willing to sell for and what they actually can
  sell for is called the producers surplus.

110
Consumers and Producers Surplus

• Geometrically, the producers surplus is given by
  the area
• of the region bounded above the straight line p
  p and below the supply curve p S(x) from x
  0 to x x

p
p S(x)
p
PS
x
x
111
Consumers and Producers Surplus

• The producers surplus PS is given by
• where S(x) is the supply function, p is the
  unit market price, and x is the quantity
  supplied.

p
p S(x)
p
PS
x
x
112
Example

• The demand function for a certain make of
  10-speed bicycle is given by
• where p is the unit price in dollars and x is
  the quantity demanded in units of a thousand.
• The supply function for these bicycles is given
  by
• where p stands for the price in dollars and x
  stands for the number of bicycles that the
  supplier will want to sell.
• Determine the consumers surplus and the
  producers surplus if the market price of a
  bicycle is set at the equilibrium price.

Example 1, page 466
113
Example

• Solution
• To find the equilibrium point, equate S(x) and
  D(x) to solve the system of equations and find
  the point of intersection of the demand and
  supply curves
• Thus, x 625/2 or x 300. The first number is
  discarded for being negative, so the solution is
  x 300.

Example 1, page 466
114
Example

• Solution
• Substitute x 300 to find the equilibrium value
  of p
• Thus, the equilibrium point is (300, 160).
• That is, the equilibrium quantity is 300,000
  bicycles, and the equilibrium price is 160 per
  bicycle.

Example 1, page 466
115
Example

• Solution
• To find the consumers surplus, we set x 300
  and p 160 in the consumers surplus formula
• or 18,000,000.

Example 1, page 466
116
Example

• Solution
• To find the producers surplus, we setx 300
  andp 160 in the producers surplus formula
• or 11,700,000.

Example 1, page 466
117
Example

• Solution
• Consumers surplus and producers surplus when
  the market is in equilibrium

p
250 200 150 100 50
CS PS
p p 160
x
100 200 300 400 500
Example 1, page 466
118
Accumulated or Total Future Value of an Income
Stream

• The accumulated, or total, future value after T
  years of an income stream of R(t) dollars per
  year, earning interest rate of r per year
  compounded continuously, is given by

119
Applied Example Income Stream

• Crystal Car Wash recently bought an automatic
  car-washing machine that is expected to generate
  40,000 in revenue per year, t years from now,
  for the next 5 years.
• If the income is reinvested in a business earning
  interest at the rate of 12 per year compounded
  continuously, find the total accumulated value of
  this income stream at the end of 5 years.

Applied Example 2, page 468
120
Applied Example Income Stream

• Solution
• We are required to find the total future value of
  the given income stream after 5 years.
• Setting R(t) 40,000, r 0.12, and T 5 in the
  accumulated income stream formula we get
• or approximately 274,040.

Applied Example 2, page 468
121
Present Value of an Income Stream

• The present value of an income stream of R(t)
  dollars in a year, earning interest at the rate
  of r per year compounded continuously, is given by

122
Applied Example Investment Analysis

• The owner of a local cinema is considering two
  alternative plans for renovating and improving
  the theater.
• Plan A calls for an immediate cash outlay of
  250,000, whereas plan B requires an immediate
  cash outlay of 180,000.
• It has been estimated that adopting plan A would
  result in a net income stream generated at the
  rate of
• f(t) 630,000
• dollars per year, whereas adopting plan B would
  result in a net income stream generated at the
  rate of
• g(t) 580,000
• for the next three years.
• If the prevailing interest rate for the next five
  years is 10 per year, which plan will generate a
  higher net income by the end of year 3?

123
Applied Example Investment Analysis

• Solution
• We can find the present value of the net income
  NI for plan A setting R(t) 630,000, r 0.1,
  and T 3, using the present value formula
• or approximately 1,382,845.

Applied Example 2, page 468
124
Applied Example Investment Analysis

• Solution
• To find the present value of the net income NI
  for plan B setting R(t) 580,000, r 0.1, and T
  3, using the present value formula
• or approximately 1,323,254.

Applied Example 2, page 468
125
Applied Example Investment Analysis

• Solution
• Thus, we conclude that plan A will generate a
  higher present value of net income by the end of
  the third year (1,382,845), than plan B
  (1,323,254).

Applied Example 2, page 468
126
Amount of an Annuity

• The amount of an annuity is
• where P, r, T, and m are as defined earlier.

127
Applied Example IRAs

• On January 1, 1990, Marcus Chapman deposited
  2000 into an Individual Retirement Account (IRA)
  paying interest at the rate of 10 per year
  compounded continuously.
• Assuming that he deposited 2000 annually into
  the account, how much did he have in his IRA at
  the beginning of 2006?

Applied Example 4, page 470
128
Applied Example IRAs

• Solution
• We set P 2000, r 0.1, T 16, and m 1 in
  the amount of annuity formula, obtaining
• Thus, Marcus had approximately 79,061 in his
  account at the beginning of 2006.

Applied Example 4, page 470
129
Present Value of an Annuity

• The present value of an annuity is given by
• where P, r, T, and m are as defined earlier.

130
Applied Example Sinking Funds

• Tomas Perez, the proprietor of a hardware store,
  wants to establish a fund from which he will
  withdraw 1000 per month for the next ten years.
• If the fund earns interest at a rate of 6 per
  year compounded continuously, how much money does
  he need to establish the fund?

Applied Example 5, page 471
131
Applied Example Sinking Funds

• Solution
• We want to find the present value of an annuity
  with P 1000, r
  0.06, T 10, and m 12.
• Using the present value of an annuity formula, we
  find
• Thus, Tomas needs approximately 90,238 to
  establish the fund.

Applied Example 5, page 471
132
End of Chapter