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. Antiderivatives and the Rules of Integration
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:
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) = x–1

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 ′(1) = 1
f(x) = x2 – x + 0
f ′(1) = 1
f(x) = x2 – x – 1
f ′(1) = 1
f ′(1) = 1 x –
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 Equationsy
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 –

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. Integration by Substitution
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 it’s 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 = – ⅓ 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. Area and the Definite Integral
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. The Fundamental Theorem of Calculus
6.4
The Fundamental Theorem of Calculus

58. Theorem 2 The 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
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
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 –
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 8:00 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. Evaluating Definite Integrals
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 –
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. Area Between Two Curves
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 –
– 4
– 8
– 12 x 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 –
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 –
(– 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 – 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
x = 1
y = x3 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
x = 1
y = x3 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
x = 1
y = x3 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
x = 1
y = x3 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
– –
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
– –
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
– –
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
– –
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
– –
Example 6, page 457

103. Applications of the Definite Integral to Business and Economics
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)
D(xn) =`p `p
p =`p x x1 x2 x3 x4
xn –1 xn

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 set`x = 300 and`p = 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
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