472
4
Chapter 6
Further Topics in Integration
This section extends the concept of the definite integral to integrals of the form
Improper
Integrals
GEOMETRIC INTERPRETATION
f(x) dx
a
in which the upper limit of integration is not a finite number. Such integrals are known
as improper integrals and arise in a variety of practical situations.
If f is nonnegative, the improper integral
f(x) dx can be interpreted as the area of
a
yy
;;
;
y
;;y;
yy
the region under the graph of f to the right of x a (Figure 6.21a). Although this
region has infinite extent, its area may be either finite or infinite, depending on how
quickly f(x) approaches zero as x increases.
y
y
y = f(x)
y = f(x)
x
a
(a)
a
x
N
(b)
FIGURE 6.21 Area a
N
f(x) dx lim
Nfi f(x) dx.
a
A reasonable strategy for finding the area of such a region is to first use a definite integral to compute the area from x a to some finite number x N and then
to let N approach infinity in the resulting expression. That is,
N
Total area lim (area from a to N) Nfi lim
Nfi f(x) dx
a
This strategy is illustrated in Figure 6.21b and motivates the following definition of
the improper integral.
The Improper Integral
a
■
If f(x) is continuous for x a, then
N
f(x) dx lim
Nfi a
f(x) dx
Chapter 6 ■ Section 4
Improper Integrals
473
If the limit defining the improper integral is a finite number, the integral is said
to converge. Otherwise the integral is said to diverge. Here are some examples.
EXAMPLE 4.1
Evaluate
1
1
dx .
x2
Solution
First compute the integral from 1 to N and then let N approach infinity. Arrange your
work as follows:
1
Refer to Example 4.1. Place
1
f(x) 2 into Y1 of the
x
equation editor and store the
numerical integral into Y2, using
the interval from x 1 to X. Set
the table feature to start at x 500 increasing in increments of
500. Explain what you observe.
1
dx lim
Nfi x2
N
1
1
1
2 dx lim
Nfi x
x
N
1
lim
Nfi N 1 1
1
EXAMPLE 4.2
Evaluate
1
1
dx .
x
Solution
1
1
dx x
N
lim
Nfi 1
1
dx x
lim
Nfi ln x
N
1
lim
Nfi Notice that the improper integral of the function f(x) ln N 1
in Example 4.1
x2
1
in Example 4.2 diverged. In geometric
x
1
terms, this says that the area to the right of x 1 under the curve y 2 is finite,
x
1
while the corresponding area under the curve y is infinite. The reason for the
x
1
1
difference is that, as x increases, 2 approaches zero more quickly than does (see
x
x
Figure 6.22).
converged, while that of the function f(x) 474
Chapter 6
Further Topics in Integration
y
3
2
1
1
y= x
Finite area
y=
Infinite area
1
FIGURE 6.22 The areas under y 1
x2
x
1
1
and y 2 .
x
x
It can be shown that
lim xm ekx 0
xfi for all m 0 and k 0. This kind of limit occurs frequently in practical problems
involving improper integrals. Here is an example.
EXAMPLE 4.3
Evaluate
xe2x dx.
0
Solution
xe2x dx 0
lim
Nfi lim
Nfi 12 xe
21 xe
N
lim
Nfi 2x
2x
N
0
xe2x dx
0
1
2
N
0
1 2x
e
4
e2x dx
N
0
(integration by parts)
Chapter 6 ■ Section 4
Improper Integrals
lim
Nfi APPLICATIONS OF THE
IMPROPER INTEGRAL
PRESENT VALUE
1
4
21 Ne
2N
1 2N
1
e
0
4
4
475
(since Ne2N fi 0)
The following applications of the improper integral generalize applications of the definite integral that you saw in Sections 2 and 3. In each, the strategy is to use the characterization of the definite integral as the limit of a sum to construct an appropriate
definite integral and then to let the upper limit of integration increase without bound.
As you read these examples, you may want to refer back to the corresponding examples in Sections 2 and 3.
In Example 2.2 of Section 2, you saw that the present value of an investment that
generates income over a finite period of time is given by a definite integral. The present value of an investment that generates income in perpetuity is given by an improper
integral. Here is an example.
EXAMPLE 4.4
A donor wishes to make a gift to a private college from which the college will draw
$7,000 per year in perpetuity to support the operation of its computer center. Assuming that the prevailing annual interest rate will remain fixed at 10% compounded continuously, how much should the donor give the college? That is, what is the present
value of the endowment?
Solution
To find the present value of a gift that generates $7,000 per year for N years, divide
the N-year time interval 0 t N into n subintervals of length nt years and let tj
denote the beginning of the jth subinterval (Figure 6.23).
7,000e –0.1t j ∆nt
7,000 ∆nt
t
0
tj
∆nt
N
FIGURE 6.23 The present value of the money generated during the jth subinterval.
476
Chapter 6
Further Topics in Integration
Then
Amount generated during 7,000 t
n
jth subinterval
and
Present value of amount
generated during
jth subinterval
Hence,
Present value of lim
7,000e0.1tj nt
nfi j1
N-year gift
7,000e0.1tj nt
n
N
7,000e0.1t dt
0
To find the present value of the total gift, take the limit of this integral as N
approaches infinity. That is,
Present value of lim
Nfi total gift
lim
Nfi N
7,000e0.1t dt
0
70,000e0.1t
0
0.1N
lim 70,000(e
Nfi N
1)
$70,000
NUCLEAR WASTE
The next example is similar in structure to the problem involving survival and renewal
functions in Example 3.2 of Section 3.
EXAMPLE 4.5
It is estimated that t years from now, a certain nuclear power plant will be producing
radioactive waste at the rate of f(t) 400t pounds per year. The waste decays
exponentially at the rate of 2% per year. What will happen to the accumulation of
radioactive waste from the plant in the long run?
Solution
To find the amount of radioactive waste present after N years, divide the N-year interval 0 t N into n equal subintervals of length nt years and let tj denote the beginning of the jth subinterval (Figure 6.24). Then,
Amount of waste produced 400t t
j
n
during jth subinterval
Chapter 6 ■ Section 4
Improper Integrals
N – tj years
400tj ∆nt
477
400tj e –0.02(N – t j) ∆nt
t
tj
0
N
∆n t
FIGURE 6.24 Radioactive waste generated during the jth subinterval.
Since the waste decays exponentially at the rate of 2% per year, and since there are
N tj years between times t tj and t N, it follows that
Amount of waste produced 400tje0.02(Ntj) nt
during jth subinterval
still present at t N
Thus,
n
Amount of waste present lim
400tje0.02(Ntj) nt
nfi j1
after N years
N
400te0.02(Nt) dt
0
400e 0.02N
N
te0.02t dt
0
The amount of radioactive waste present in the long run is the limit of this expression as N approaches infinity. That is,
Amount of waste
present in the lim 400e0.02N
Nfi long run
N
te0.02t dt
0
lim 400e0.02N(50te0.02t 2,500e0.02t)
Nfi N
0
lim 400e0.02N(50Ne0.02N 2,500e0.02N 2,500)
Nfi lim 400(50N 2,500 2,500e0.02N)
Nfi That is, in the long run, the accumulation of radioactive waste from the plant will
increase without bound.
478
PROBABILITY DENSITY
FUNCTIONS
Chapter 6
Further Topics in Integration
Improper integrals also appear in the study of probability and statistics, which is an
important tool in certain areas of the social, managerial, and life sciences. We close
this section with a brief introduction to this application.
For example, the life span of a lightbulb selected at random from a manufacturer’s stock is a quantity that cannot be predicted with certainty. The process of
selecting a bulb is called a random experiment, and the life span X of the bulb is a
continuous random variable. Other examples of continuous random variables
include the time a randomly selected motorist spends waiting at a traffic light, the
weight of a randomly selected person, or the time it takes a randomly selected person to learn a particular task.
The probability of an event that can result from a random experiment is a number between 0 and 1 that specifies the likelihood of the event. For instance, in our
lightbulb example, one possible event is that a bulb selected randomly from the manufacturer’s stock has life span between 20 and 35 hours. If X is the random variable
denoting the life span of a randomly selected bulb, then this event can be described
by the inequality 20 X 35 and its probability denoted by P(20 X 35). Similarly, the probability that the bulb will burn for at least 50 hours is denoted by
P(X 50) or P(50 X ).
A probability density function for a continuous random variable X is a function
f with the property that f(x) 0 for all real x and that the area under the graph of f
from x a to x b gives the probability P(a X b). The graph of a possible
probability density function for the life span of a light bulb is sketched in Figure 6.25.
Its shape reflects the fact that most bulbs burn out relatively quickly. For example,
the probability that a bulb will fail within the first 40 hours is represented by the area
under the curve between x 0 and x 40. This is much greater than the area under
the curve between x 80 and x 120, which represents the probability that the bulb
will fail between its 80th and 120th hours of use.
yy
;;
yy
;;;;
yy
y
P(0 ≤ X ≤ 40)
P(80 ≤ X ≤ 120)
0
40
80
120
x
FIGURE 6.25 A possible probability density function for the life span of a lightbulb.
The basic property of probability density functions can be restated in terms of
the integrals you would use to compute the appropriate areas.
Chapter 6 ■ Section 4
Improper Integrals
479
Probability Density Functions ■ A probability density function
for the continuous random variable X is a function f(x) that satisfies the following three conditions:
1. f(x) 0 for all real x
2. The total area under the graph of f(x) is 1
3. The probability that X lies in the interval a X b is given by the
integral
b
P(a X b) f(x) dx
a
The values of a and b in this formula need not be finite, and if either is infinite,
the corresponding probability is given by an improper integral. For example, the probability that X a is
P(X a) P(a X ) f(x) dx
a
The second condition in the definition of probability density function follows
from the fact that the event X is certain to occur. This condition can
also be expressed as
f(x) dx 1
where the improper integral on the left is defined as
0
f(x) dx lim
Nfi N
N
f(x) dx lim
Nfi f(x) dx
0
and converges if and only if both limits exist.
How to determine the appropriate probability density function for a particular random variable is a central problem in probability theory that is beyond the scope of
this book. We restrict our attention to examining two problems illustrating particular
probability density functions that prove useful in a variety of applications.
UNIFORM DENSITY FUNCTIONS
A uniform probability density function (Figure 6.26) is constant over a bounded
interval A x B and zero outside that interval. A random variable that has a
uniform density function is said to be uniformly distributed. Roughly speaking,
for a uniformly distributed random variable, all values in some bounded interval
are “equally likely.” More precisely, a continuous random variable is uniformly
480
B
x
FIGURE 6.26 A uniform density
function.
Further Topics in Integration
distributed if the probability that its value will be in a particular subinterval of the
bounded interval is equal to the probability that it will be in any other subinterval that
has the same length. An example of a uniformly distributed random variable is the
waiting time of a motorist at a traffic light that remains red for, say, 40 seconds at a
time. This random variable has a uniform distribution because all waiting times
between 0 and 40 seconds are equally likely.
y
A
Chapter 6
If k is the constant value of a uniform density function f(x) on the interval A x B, the value of k is determined by the requirement that the total area under the
graph of f be equal to 1. In particular,
1
B
A
and so
B
f(x) dx f(x) dx [since f(x) 0 outside the interval A x B]
A
B
k dx kx
k(B A)
A
k
1
BA
This observation leads to the following formula for a uniform density function.
Uniform Density Function
1
f(x) B A
0
if A x B
otherwise
Here is a typical application involving a uniform density function.
EXAMPLE 4.6
A certain traffic light remains red for 40 seconds at a time. You arrive (at random) at
the light and find it red. Use an appropriate uniform density function to find the probability that you will have to wait at least 15 seconds for the light to turn green.
Solution
Let X denote the random variable that measures the time (in seconds) that you must
wait. Since all waiting times between 0 and 40 are “equally likely,” X is uniformly
distributed over the interval 0 x 40. The corresponding uniform density function is
Chapter 6 ■ Section 4
Improper Integrals
481
1
if 0 x 40
f(x) 40
0
otherwise
and the desired probability is
40
P(15 X 40) 1
x
dx 40
40
15
EXPONENTIAL DENSITY
FUNCTIONS
40
15
40 15 5
40
8
An exponential probability density function is a function f(x) that is zero for x 0 and decreases exponentially for x 0. That is,
f(x) Aekx
0
for x 0
for x 0
where A and k are positive constants.
The value of A is determined by the requirement that the total area under the
graph of f be equal to 1. Thus,
1
lim
f(x) dx Nfi and so
0
A
ekx
k
N
0
Aekx dx lim
Nfi N
lim
Nfi Ak e
kN
Aekx dx
0
A
A
k
k
Ak
This calculation leads to the following general formula for an exponential density
function. The corresponding graph is shown in Figure 6.27.
Exponential Density Function
f(x) y
k
x
FIGURE 6.27 An exponential
density function.
kekx
0
if x 0
if x 0
A random variable that has an exponential density function is said to be exponentially distributed. As you can see from the graph in Figure 6.27, the value of an
exponentially distributed random variable is much more likely to be small than large.
Such random variables include the life span of electronic components, the duration
of telephone calls, and the interval between the arrivals of successive planes at an airport. Here is an example.
482
Chapter 6
Further Topics in Integration
EXAMPLE 4.7
Refer to Example 4.7. Graph
the function by storing Y1 .5e^(.5X) (X 0) in the
function list. Use the viewing
window [10, 10]1 by [0.1,
1]0.1. Explain why the curve is
not graphed for X 0.
Let X be a random variable that measures the duration of telephone calls in a certain
city and suppose that a probability density function for X is
0.5e0.5x
if x 0
if x 0
0
f(x) where x denotes the duration (in minutes) of a randomly selected call.
(a) Find the probability that a randomly selected call will last between 2 and 3
minutes.
(b) Find the probability that a randomly selected call will last at least 2 minutes.
Solution
3
(a) P(2 X 3) 0.5e0.5x dx e0.5x
2
3
2
1.5
e
1
e
0.1447
(b) There are two ways to compute this probability. The first method is to evaluate
an improper integral.
P(X 2) P(2 X ) Nfi 0.5e0.5x dx
2
N
lim
0.5e0.5x dx lim
Nfi 2
0.5N
lim (e
Nfi 1
e
1
)e
e0.5x
N
2
0.3679
The second method is to compute 1 minus the probability that x is less than 2.
That is,
P(X 2) 1 P(X 2)
2
1
0.5e0.5x dx 1 e0.5x
0
1
1 (e
1
1) e
0.3679
2
0
Chapter 6 ■ Section 4
Improper Integrals
P . R . O . B . L . E . M . S
483
6.4
In Problems 1 through 24, evaluate the given improper integral.
1.
1
3.
1
5.
3
7.
3
9.
1
dx
x3
2.
1
dx
x
4.
1
dx
2x 1
6.
1
dx
(2x 1)2
8.
1
13.
1
15.
1
17.
21.
2
23.
0
5e2x dx
x2/3 dx
1
dx
2x 1
3
3
ex dx
0
10.
e1x dx
1
x2
dx
(x3 2)2
12.
x
x 2
3
dx
1
2
14.
x2
dx
x3 2
xex dx
2
0
x
e
dx
x
16.
2xe3x dx
18.
xex dx
0
xe1x dx
0
5xe10x dx
20.
0
x3/2 dx
1
0
19.
1
0
11.
1
1
dx
x ln x
22.
x2ex dx
24.
2
ln x
dx
x
1
dx
xln x
x3ex dx
2
0
PRESENT VALUE
OF AN INVESTMENT
25. An investment will generate $2,400 per year in perpetuity. If the money is
dispensed continuously throughout the year and if the prevailing annual interest
rate remains fixed at 12% compounded continuously, what is the present value of
the investment?
PRESENT VALUE
26. An investment will generate income continuously at the constant rate of Q dollars
per year in perpetuity. Assuming a fixed annual interest rate of r (expressed as a
484 Chapter 6
Further Topics in Integration
decimal) compounded continuously, use an improper integral to show that the
Q
present value of the investment is dollars.
r
PRESENT VALUE OF
RENTAL PROPERTY
27. It is estimated that t years from now an apartment complex will be generating profit
for its owner at the rate of f(t) 10,000 500t dollars per year. If the profit is
generated in perpetuity and the prevailing annual interest rate remains fixed at 10%
compounded continuously, what is the present value of the apartment complex?
PRESENT VALUE
OF A FRANCHISE
28. The management of a national chain of fast-food outlets is selling a permanent
franchise in Seattle, Washington. Past experience in similar localities suggests
that t years from now, the franchise will be generating profit at the rate of f(t) 12,000 900t dollars per year. If the prevailing interest rate remains fixed at 10%
compounded continuously, what is the present value of the franchise?
PRESENT VALUE
29. In t years, an investment will be generating f(t) A Bt dollars per year, where A
and B are constants. If the income is generated in perpetuity and the prevailing
annual interest rate of r (expressed as a decimal) compounded continuously does not
B
A
change, show that the present value of this investment is 2 .
r
r
NUCLEAR WASTE
30. A certain nuclear power plant produces radioactive waste at the rate of 600 pounds
per year. The waste decays exponentially at the rate of 2% per year. How much
radioactive waste from the plant will be present in the long run?
HEALTH CARE
31. The fraction of patients who will still be receiving treatment at a certain health clinic
t months after their initial visit is f(t) et/20. If the clinic accepts new patients at
the rate of 10 per month, approximately how many patients will be receiving
treatment at the clinic in the long run?
POPULATION GROWTH
32. Demographic studies conducted in a certain city indicate that the fraction of the
residents that will remain in the city for at least t years is f(t) et/20. The current
population of the city is 200,000, and it is estimated that new residents will be
arriving at the rate of 100 people per year. If this estimate is correct, what will
happen to the population of the city in the long run?
MEDICINE
33. A hospital patient receives intravenously 5 units of a certain drug per hour. The drug
is eliminated exponentially, so that the fraction that remains in the patient’s body for
t hours is f(t) et/10. If the treatment is continued indefinitely, approximately how many units of the drug will be in the patient’s body in the long run?
In Problems 34 through 41, f(x) is a probability density function for a particular
random variable X. Use integration to find the indicated probabilities.
1
if 2 x 5
34. f(x) 3
0 otherwise
(a) P(2 X 5)
(b) P(3 X 4)
(c) P(X 4)
x
if 0 x 2
35. f(x) 2
0 otherwise
(a) P(0 X 2)
(b) P(1 X 2)
(c) P(X 1)
Chapter 6 ■ Section 4
Improper Integrals
3
(4x x2)
37. f(x) 32
0
(a) P(0 X 4)
(b) P(1 X 2)
(c) P(X 1)
1 x/10
e
if x 0
39. f(x) 10
0
if x 0
(a) P(0 X +)
(b) P(X 2)
(c) P(X 5)
1 x/2
xe
if x 0
f(x) 4
41.
0
if x 0
1
(4 x) if 0 x 4
36. f(x) 8
0
otherwise
(a) P(0 X 4)
(b) P(2 X 3)
(c) P(X 1)
3
if x 1
38. f(x) x4
0
if x 1
(a) P(1 X +)
(b) P(1 X 2)
(c) P(X 2)
2xex
40. f(x) 0
2
if x 0
if x 0
(a) P(X 0)
(b) P(1 X 2)
(c) P(X 2)
485
if 0 x 4
otherwise
(a) P(0 X +)
(b) P(2 X 4)
(c) P(X 6)
PRODUCT RELIABILITY
42. The life span of the lightbulbs manufactured by a certain company is measured by a
random variable X with probability density function f(x) 0.01e0.01x, where x
denotes the life span (in hours) of a randomly selected bulb.
(a) What is the probability that the life span of a randomly selected bulb will be
between 50 and 60 hours?
(b) What is the probability that the life span of a randomly selected bulb will be
less than or equal to 60 hours?
(c) What is the probability that the life span of a randomly selected bulb will be
greater than 60 hours?
PRODUCT RELIABILITY
43. The life of an appliance is measured by a random variable X with probability density
function f(x) 0.2e0.2x, where x denotes the life (in months) of a randomly selected appliance.
(a) What is the probability that a randomly selected appliance will last between
10 and 15 months?
(b) What is the probability that a randomly selected appliance will last less than
8 months?
(c) What is the probability that a randomly selected appliance will last longer than
1 year?
DURATION OF
TELEPHONE CALLS
44. The duration of telephone calls in a certain city is measured by a random variable X
with probability density function f(x) 0.5e0.5x, where x denotes the duration (in
minutes) of a randomly selected call.
486
Chapter 6
Further Topics in Integration
(a) What percentage of the calls can be expected to last between 2 and 3 minutes?
(b) What percentage of the calls can be expected to last 2 minutes or less?
(c) What percentage of the calls can be expected to last more than 2 minutes?
AIRPLANE ARRIVALS
45. The time interval between the arrivals of successive planes at a certain airport
is measured by a random variable X with probability density function f(x) 0.2e0.2x, where x is the time (in minutes) between the arrivals of a randomly
selected pair of successive planes.
(a) What is the probability that two successive planes selected at random will
arrive within 5 minutes of one another?
(b) What is the probability that two successive planes selected at random will
arrive more than 6 minutes apart?
TRAFFIC FLOW
46. A certain traffic light remains red for 45 seconds at a time. You arrive (at random) at
the light and find it red. Use an appropriate uniform density function to find the
probability that the light will turn green within 15 seconds.
COMMUTING
47. During the morning rush hour, commuter trains from Long Island to Manhattan run
every 20 minutes. You arrive (at random) at the station during the rush hour and find
no train at the platform. Assuming that the trains are running on schedule, use an
appropriate uniform density function to find the probability that you will have to
wait at least 8 minutes for your train.
MOVIE THEATERS
48. A 2-hour movie runs continuously at a local theater. You leave for the theater
without first checking the show times. Use an appropriate uniform density function
to find the probability that you will arrive at the theater within 10 minutes of (before
or after) the start of the film.
EXPERIMENTAL PSYCHOLOGY
49. Suppose the length of time that it takes a laboratory rat to traverse a certain maze is
measured by a random variable X that is exponentially distributed with density
function
1 x/3
e
f(x) 3
0
if x 0
if x 0
where x is the number of minutes a randomly selected rat spends in the maze.
Find the probability that a randomly selected rat will take more than 3 minutes
to traverse the maze.
CUSTOMER SERVICE
50. Suppose the time X a customer must spend waiting in line at a certain bank is a
random variable that is exponentially distributed with density function
1 x/4
e
f(x) 4
0
if x 0
if x 0
where x is the number of minutes a randomly selected customer is in line. Find
the probability that a customer will have to stand in line at least 8 minutes.
Chapter 6 ■ Section 4
SPY STORY
Improper Integrals
487
51. Still groggy from his encounter with the camel in Problem 43 of Section 5.1, our spy
stumbles into a trap set by Scélérat and his men. He fights back, but eventually runs
out of bullets and is forced to use his last weapon, a special stun grenade.
Pulling the pin from the grenade, he recalls fondly the time exactly 1 year ago
that he received it along with a pair of brass knuckles from his superior, “N”, for
Valentine’s Day. As the grenade arcs through the air toward the enemy, he also
recalls, somewhat less fondly, that this particular kind of grenade has a 1-year
warranty and that its life span X is a random variable exponentially distributed
with density function
f(t) 0.08e0.08t
0
if t 0
if t 0
where t is the number of months since the grenade left the munitions factory.
Assuming the grenade was new when the spy received it and that it had been chosen randomly from the factory stock, what is the probability that the spy will
expire before the warranty?
ENDOWMENT
52. A wealthy patron of a small private college wishes to endow a chair in mathematics
with a gift of G thousand dollars. Suppose the mathematician who occupies the
chair is to receive $70,000 per year in salary and benefits. If money costs 8% per
year compounded continuously, what is the smallest possible value for G?
CAPITALIZED COST OF AN ASSET
53. The capitalized cost of an asset is the sum of the original cost of the asset and the
present value of maintaining the asset. Suppose a company is considering the
purchase of two different machines. Machine 1 costs $10,000 and t years from now
will cost M1(t) 1,000(1 0.06t) dollars to maintain. Machine 2 costs only $8,000,
but its maintenance cost at time t is M2(t) 1,100 dollars.
(a) If the cost of money is 9% per year compounded continuously, what is the
capitalized cost of each machine? Which one should the company purchase?
(b) Research various methods used by economists to make comparisons between
competing assets. Write a paragraph comparing these methods.
LEGISLATIVE TURNOVER
54. A mathematical model in political science* asserts that the length of time served
(continuously) by a legislator is a random variable that is exponentially distributed
with density function
N(t) cect
0
for t 0
for t 0
where t is the number of years of continuous service and c is a positive constant
that depends on the nature and character of the legislative body.
(a) For the U.S. House of Representatives, it was found that c 0.0866. Find the
probability that a randomly selected House member will serve at least 6 years.
* Thomas W. Casstevens, “Exponential Models for Legislative Turnover,” UMAP Modules 1978: Tools
for Teaching, Consortium for Mathematics and Its Applications, Inc., Lexington, MA, 1979.
488
Chapter 6
Further Topics in Integration
(b) For the British House of Commons, it was found that c 0.135. What is the
probability that a member of this body will serve at least 6 years?
(c) The UMAP module cited in this problem also shows how the density function
given above can be used to estimate how many members of the Soviet Central Committee may have been purged by Nikita Khrushchev in the period
1956–1961. Read this module and write a paragraph on whether you think the
method of analysis used by the author is valid.
CHAPTER SUMMARY AND REVIEW PROBLEMS
IMPORTANT TERMS, SYMBOLS,
AND FORMULAS
Definite integral:
b
f(x) dx lim [f(x1) . . . f(xn)] x
nfi a
Fundamental theorem of calculus
b
f(x) dx F(b) F(a)
where F(x) f(x)
a
Area under a curve (left-hand figure):
b
Area of R f(x) dx
a
y
y
y = f (x)
y = f (x)
R
y = g (x)
R
x
a
x
b
a
Area between two curves (right-hand figure):
b
Area of R a
Amount of an income stream
[ f(x) g(x)] dx
b