Smith R., Minton R. Calculus

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7-63 SECTION 7.8

Probability 483

EXAMPLE 8.4 Determining the Coefficient of a pdf

Suppose that the pdf for a random variable has the form f (x) = ce

−3x

for some constant

c, with 0 ≤ x ≤ 1. Find the value of c that makes this a pdf.

Solution To be a pdf, we ﬁrst need that f (x) = ce

−3x

≥ 0, for all x ∈ [0, 1]. (This will

be the case as long as c ≥ 0.) Also, the integral over the domain must equal 1. So, we set

1 =



−3x

dx = c



−



−3x



(1 − e

−3

It now follows that c =

1 − e

−3

≈ 3.1572.



TODAY IN

MATHEMATICS

Persi Diaconis (1945– )

An American statistician who was

one of the first recipients of a

lucrative MacArthur Foundation

Fellowship, often called a “genius

grant.” Diaconis trained on the

violin at Juilliard until age 14,

when he left home to become a

professional magician for 10 years.

His varied interests find

expression in his work, where he

uses all areas of mathematics and

statistics to solve problems from

throughout science and

engineering. He says, “What

makes somebody a good applied

mathematician is a balance

between finding an interesting

real-world problem and finding an

interesting real-world problem

which relates to beautiful

mathematics.”

Given a pdf, it is possible to compute various statistics to summarize the properties

of the random variable. The most common statistic is the mean, the best-known measure

of average value. If you wanted to average test scores of 85, 89, 93 and 93, you would

probably compute the mean, given by

85+89+93+93

= 90.

Notice here that there were three different test scoresrecorded: 85, which has a relative

frequency of

, 89, also with a relative frequency of

and 93, with a relative frequency of

. We can also compute the mean by multiplying each value by its relative frequency and

then summing: (85)

+ (89)

+ (93)

= 90.

Now, suppose we wanted to compute the mean height of the people in the following

table.

Height 63



Number 23 32 61 94 133 153 155 134 96 62 31 26

It would be silly to write out the heights of all 1000 people, add and divide by 1000. It is

much easier to multiply each height by its relative frequency and add the results. Following

this route, the mean m is given by

m = (63)

1000

+ (64)

1000

+ (65)

1000

+ (66)

1000

+ (67)

133

1000

+···+(74)

1000

= 68.523.

If we denote the heights by x

, x

,...,x

and let f (x

) be the relative frequency or proba-

bility corresponding to x = x

, the mean then has the form

m = x

f (x

) + x

f (x

) + x

f (x

) +···+x

f (x

If the heights in our data set were given for every half-inch or tenth-of-an-inch, we would

compute the mean by multiplying each x

by the corresponding probability f (x

)x,where

x is the fraction of an inch between data points. The mean now has the form

m = [x

f (x

) + x

f (x

) + x

f (x

) +···+x

f (x

)]x =



i=1

f (x

)x,

where n is the number of data points. Notice that, as n increases and x approaches 0, the

Riemann sum approaches the integral



xf(x)dx. This gives us the following deﬁnition.

DEFINITION 8.2

The mean μ of a random variable with pdf f on the interval [a, b]isgivenby

μ =



xf(x)dx. (8.2)

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484 CHAPTER 7

Integration Techniques 7-64

Although the mean is commonly used to reportthe averagevalue of a random variable,

it is important to realize that it is not the only measure of average used by statisticians. An

alternative measurement of average is the median, the x-value that divides the probability

in half. (That is, half of all values of the random variable lie at or below the median and

half lie at or above the median.) In example 8.5 and in the exercises, you will explore situa-

tions in which each measure provides a different indication about the average of a random

variable.

EXAMPLE 8.5 Finding the Mean Age and Median Age

of a Group of Cells

Suppose that the age in days of a type of single-celled organism has pdf

f (x) = (ln2)e

−kx

, where k =

ln2. The domain is 0 ≤ x ≤ 2. (The assumption here is

that upon reaching an age of 2 days, each cell divides into two daughter cells.) Find (a)

the mean age of the cells, (b) the proportion of cells that are younger than the mean and

Solution For part (a), we have from (8.2) that the mean is given by

μ ≈



x(ln 2)e

−(ln2)x/2

dx ≈ 0.88539 day,

where we have approximated the value of the integral numerically. Notice that even

though the cells range in age from 0 to 2 days, the mean is not 1. The graph of the pdf in

Figure 7.32 shows that younger ages are more likely than older ages and this causes the

mean to be less than 1.

0.5 1 1.5 2

0.1

0.2

0.3

0.4

0.5

0.6

0.7

FIGURE 7.32

y = (ln 2)e

−(ln2)x/2

For part (b), notice that the proportion of cells younger than the mean is the same as

the probability that a randomly selected cell is younger than the mean. This probability

is given by

P(0 ≤ X ≤ μ) ≈



0.88539

(ln2)e

−(ln2)x/2

dx ≈ 0.52848,

where we have again approximated the value of the integral numerically. Therefore, the

proportion of cells younger than the mean is about 53%. Notice that in this case the

mean does not represent the 50% mark for probabilities. In other words, the mean is not

the same as the median.

To ﬁnd the median in part (c), we must solve for the constant c such that

0.5 =



(ln2)e

−(ln2)x/2

dx.

Since an antiderivative of e

−(ln2)x/2

is −

ln2

−(ln2)x/2

,wehave

0.5 =



(ln2)e

−(ln2)x/2

= ln 2



−

ln2

−(ln2)x/2



=−2e

−(ln2)c/2

+ 2.

Subtracting 2 from both sides, we have

−1.5 =−2e

−(ln2)c/2

so that dividing by −2 yields

0.75 = e

−(ln2)c/2

Taking the natural log of both sides gives us

ln0.75 =−(ln 2)c/2.

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7-65 SECTION 7.8

Probability 485

Finally, solving for c gives us

c =

−2ln0.75

ln2

so that the median is −2ln0.75/ ln 2 ≈ 0.83. We can now conclude that half of the cells

are younger than 0.83 day and half the cells are older than 0.83 day.



EXERCISES 7.8

WRITING EXERCISES

1. In the text, we stated that the probability of tossing two fair

coins and getting two heads is

. If you try this experiment

four times, explain why you will not always get two heads ex-

actly one out of four times. If probability doesn’t give precise

predictions, what is its usefulness? To answer this question,

discuss the information conveyed by knowing that in the above

experiment the probability of getting one head and one tail is

(twice as big as

2. Suppose you toss two coins numerous times (or simulate this

on your calculator or computer). Theoretically, the probability

ofgettingtwoheadsis

.Inthelong run (as the coinsaretossed

more and more often), what proportion of the time should two

heads occur? Try this and discuss how your results compare to

the theoretical calculation.

3. Based on Figures 7.25 and 7.26, describe what you expect the

histogram to look like for larger numbers of coins. Compare

to Figure 7.31.

4. The height of a person is determined by numerous factors,

both hereditary and environmental (e.g., diet). Explain why

this might produce a histogram similar to that produced by

tossing a large number of coins.

In exercises 1–6, show that the given function is a pdf on the

indicated interval.

1. f (x) = 4x

, [0, 1] 2. f (x) =

, [0, 2]

3. f (x) = x + 2x

, [0, 1] 4. f (x) = cosx, [0,π/2]

5. f (x) =

sin x, [0,π] 6. f (x) = e

−x/2

, [0, ln4]

............................................................

In exercises 7–12, ﬁnd a value of c for which f (x) is a pdf on the

indicated interval.

7. f (x) = cx

, [0, 1] 8. f (x) = cx + x

, [0, 1]

9. f (x) = ce

−4x

, [0, 1] 10. f (x) = 2ce

−cx

, [0, 2]

11. f (x) =

1 + x

, [0, 1] 12. f (x) =

√

1 − x

, [0, 1]

............................................................

In exercises 13–16, use the pdf in example 8.2 to ﬁnd the proba-

bility that a randomly selected American male has height in the

indicated range.

13. Between 5





and 6



14. Between 6





and 6





15. Between 7



and 10



16. Between 2



and 5



............................................................

In exercises 17–20, ﬁnd the indicated probabilities, given that

the lifetime of a lightbulb is exponentially distributed with pdf

f (x)  6e

−6x

(with x measured in years).

17. The lightbulb lasts less than 3 months.

18. The lightbulb lasts less than 6 months.

19. The lightbulb lasts between 1 and 2 years.

20. The lightbulb lasts between 3 and 10 years.

............................................................

In exercises 21–24, suppose the lifetime of an organism has pdf

f (x)  4xe

−2x

(with x measured in years).

21. Find the probability that the organism lives less than 1 year.

22. Find the probability that the organism lives between 1 and

2 years.

23. Find the mean lifetime (0 ≤ x ≤ 10).

24. Graph the pdf and compare the maximum value of the pdf to

the mean.

............................................................

In exercises 25–30, ﬁnd (a) the mean and (b) the median of the

random variable with the given pdf.

25. f (x) = 3x

, [0, 1] 26. f (x) = 4x

, 0 ≤ x ≤ 1

27. f (x) =

4/π

1 + x

, [0, 1] 28. f (x) =

2/π

√

1 − x

, [0, 1]

29.

f (x) =

sin x, [0,π] 30. f (x) = cos x, [0,π/2]

............................................................

31. For f (x) = ce

−4x

, ﬁnd c so that f (x) is a pdf on the interval

[0, b] for b > 0. What happens to c as b →∞?

32. For the pdf of exercise 31, ﬁnd the mean exactly (use a

CAS for the antiderivative). As b increases, what happens to

the mean?

33. Repeat exercises 31 and 32 for f (x) = ce

−6x

34. Based on the results of exercises 31–33, conjecture the values

for c and the mean as a →∞, for f (x) = ce

−ax

, a > 0.

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486 CHAPTER 7

Integration Techniques 7-66

APPLICATIONS

35. In one version of the game of keno, you choose 10 numbers

between 1 and 80. A random drawing selects 20 numbers be-

tween 1 and 80. Your payoff depends on how many of your

numbers are selected. Use the given probabilities (rounded to

4 digits) to ﬁnd the probability of each event indicated below.

(To win, at least 5 of your numbers must be selected. On a

$2 bet, you win $40 or more if 6 or more of your numbers are

selected.)

Number selected 0 1 2 3 4

Probability 0.0458 0.1796 0.2953 0.2674 0.1473

Number selected 5 6 7 8 9 10

Probability 0.0514 0.0115 0.0016 0.0001 0.0 0.0

(a) winning (at least 5 selected)

(b) losing (4 or fewer selected)

(d) 3 or 4 numbers selected

36. Suppose a basketball player makes 70% of her free throws.

If she shoots three free throws and the probability of making

each one is 0.7, the probabilities for the total number made

are as shown. Find the probability of each event indicated

below.

Number made 0 1 2 3

Probability 0.027 0.189 0.441 0.343

(a) She makes 2 or 3 (b) She makes at least 1

37. (a) Suppose that a game player has won m games out of n,

with a winning percentage of 100

< 75. The player then

wins severalgames in a row, so that the winning percentage

exceeds 75%. Show that at some point in this process the

player’s winning percentage is exactly 75%.

(b) Generalize to any winning percentage that can be written

as 100

k + 1

, for some integer k.

38. In example 8.5, we found the median (also called the sec-

ond quartile). Now ﬁnd the ﬁrst and third quartiles, the ages

such that the probability of being younger are 0.25 and 0.75,

respectively.

39. Thepdfinexample8.2is the pdf foranormally distributedran-

dom variable. The mean is easily read off from f (x); in exam-

ple 8.2, the mean is 68. The mean and a number called

the standard deviation characterize normal distributions. As

Figure7.31indicates,thegraphofthepdfhasamaximumatthe

mean and has two inﬂection points located on opposite sides

of the mean. The standard deviation equals the distance from

the mean to an inﬂection point. Find the standard deviation in

example 8.2.

40. In exercise 39, you found the standard deviation for the pdf

in example 8.2. Denoting the mean as μ and the standard

deviation as σ , ﬁnd the probability that a given height is

between μ − σ and μ + σ (that is, within one standard de-

viation of the mean). Find the probability that a given height

is within two standard deviations of the mean (μ − 2σ to

μ + 2σ ) and within three standard deviations of the mean.

These probabilities are the same for any normal distribution.

So, if you know the mean andstandard deviation of a normally

distributed random variable, you automatically know these

probabilities.

41. Iftheprobabilityofaneventis p, theprobabilitythatitwillhap-

pen m times in n tries is f (p) =

m!(n − m)!

(1 − p)

n−m

Find the value of p that maximizes f (p). This is called the

maximum likelihood estimator of p. Brieﬂy explain why your

answer makes sense.

42. The Buffon needle problem is one of the oldest and most

famous of probability problems. Suppose that a series of hor-

izontal lines are spaced one unit apart and a needle of length

one is placed randomly. What is the probability that the needle

intersects one of the horizontal lines?

In the ﬁgure, y is the distance from the center of the needle to

thenearestline and θ is thepositiveangle that theneedlemakes

with the horizontal. Show that the needle intersects the line if

and only if 0 ≤ y ≤

sinθ. Since 0 ≤ θ ≤ π and 0 ≤ y ≤

the desired probability is



sinθdθ



dθ

. Compute this.

43. The Maxwell-Boltzmann pdf for molecular speeds in a gas

at equilibrium is f (x) = ax

−b

, for positive parame-

ters a and b. Find the most common speed [i.e., ﬁnd x to

maximize f (x)].

44. The pdf for inter-spike intervals of neurons ﬁring in the

cochlear nucleus of a cat is f (t) = kt

−3/2

bt−a/t

, where a =

100, b = 0.38andt ismeasuredinmicroseconds.(See Mackey

and Glass, From Clocks to Chaos.) Use your CAS to ﬁnd the

value of k that makes f a pdf on the interval [0, 40]. Then ﬁnd

the probability that neurons ﬁre between 20 and 30 microsec-

onds apart.

45. Suppose that a soccer team has a probability p of scoring the

nextgoalin agame. The probability of a 2-goal game ending in

a 1-1 tie is 2p(1 − p), the probability of a 4-goal game ending

in a 2-2 tie is

4 · 3

2 · 1

(1 − p)

, the probability of a 6-goal game

ending in a 3-3 tie is

6 · 5 · 4

3 · 2 · 1

(1 − p)

and so on. Assume

that an even number of goals is scored. Show that the proba-

bility of a tie is a decreasing function of the number of goals

scored.

46. Two players toss a fair coin until either the sequence HTT or

HHT occurs. Player A wins if HTT occurs ﬁrst, and player B

wins if HHT occurs ﬁrst. Show that player B is twice as likely

to win.

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7-67 CHAPTER 7

Review Exercises 487

47. Let f be a function such that both f and g are pdf’s on [0, 1],

where g(x) = f (x

(a) Find such a function of the form f (x) = a + bx +cx

(b) Find the mean of any random variable with pdf g.

EXPLORATORY EXERCISES

1. The mathematical theory of chaos indicates that numbers gen-

erated by very simple algorithms can look random. Chaos

researchers look at a variety of graphs to try to distinguish

randomness from deterministicchaos. For example, iterate the

function f (x) = 4x(1 − x) starting at x = 0.1. That is, com-

pute f (0.1) = 0.36, f (0.36) = 0.9216, f (0.9216) ≈ 0.289

and so on. Iterate 50 times and record how many times each

ﬁrst digit occurs. (So far, we’ve got a 1, a 3, a 9 and a 2.) If

the process were truly random, the digits would occur about

the same number of times. Does this seem to be happening?

To unmask this process as nonrandom, you can draw a phase

portrait. To do this, take consecutive iterates as coordinates

of a point (x, y) and plot the points. The ﬁrst three points are

(0.1, 0.36), (0.36, 0.9216) and (0.9216, 0.289). Describe the

(nonrandom) pattern that appears,identifying it asprecisely as

possible.

2. Suppose that a spring is oscillating up and down with verti-

cal position given by u(t) = sint. If you pick a random time

and look at the position of the spring, would you be more

likely to ﬁnd the spring near an extreme (u = 1oru =−1) or

near the middle (u = 0)? The pdf is inversely proportional to

speed. (Why is this reasonable?) Show that speed is given by

|cost|=

√

1 − u

, so the pdf is f (u) = c/

√

1 − u

−1 ≤ u ≤ 1, for some constant c. Show that c = 1/π, then

graph f (x) and describe the positions in which the spring is

likely to be found. Use this result to explain the following. If

you are driving in a residential neighborhood, you are more

likely to meet a car coming the other way at an intersection

than in the middle of a block.

Review Exercises

WRITING EXERCISES

The following list includes terms that are deﬁned and theorems that

arestatedin this chapter.Foreach termor theorem, (1) give a precise

deﬁnition or statement, (2) state in general terms what it means and

(3) describe the types of problems with which it is associated.

Integration by parts Reduction formula

Partial fractions decomposition CAS

Improper integral Integral converges

Integral diverges Comparison Test

Probability density function

TRUE OR FALSE

State whether each statement is true or false, and brieﬂy explain

why. If the statement is false, try to “ﬁx it” by modifying the given

statement to a new statement that is true.

1. Integration by parts works only for integrals of the form



f (x)g(x)dx.

2. For an integral of the form



xf(x) dx, always use integration

by parts with u = x.

3. The trigonometric techniques in section 7.3 are all versions of

substitution.

4. If an integrand contains a factor of

√

1 − x

, you should sub-

stitute x = sinθ.

5. If p and q are polynomials, then any integral of the form



p(x)

q(x)

dx can be evaluated.

6. With an extensive integral table, you don’t need to know any

integration techniques.

7. If f (x) has a vertical asymptote at x = a, then



f (x) dx

diverges for any b.

8. If lim

x→∞

f (x) = L = 0, then



∞

f (x) dx diverges.

9. The mean of a random variable is always larger than the

median.

10. L’H ˆopital’s Rule states that the limit of the derivative equals

the limit of the function.

In exercises 1–44, evaluate the integral.



√

dx 2.



sin(1/x)



√

1 − x

dx 4.



√

9 − x



−3x

dx 6.



−x



1 + x

dx 8.



1 + x



4 + x

dx 10.



4 + x

11.



2lnx

dx 12.



cos4xdx

13.



x sin3xdx 14.



x sin4x

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488 CHAPTER 7

Integration Techniques 7-68

Review Exercises

15.



π/2

sin

xdx 16.



π/2

cos

xdx

17.



−1

x sinπ xdx 18.



cosπ xdx

19.



ln xdx 20.



π/4

sin x cos xdx

21.



cos x sin

xdx 22.



cos x sin

xdx

23.



cos

x sin

xdx 24.



cos

x sin

xdx

25.



tan

x sec

xdx 26.



tan

x sec

xdx

27.



√

sin x cos

xdx 28.



tan

x sec

xdx

29.



8 + 4x + x

dx 30.



√

−2x − x

31.



√

4 − x

dx 32.



√

9 − x

33.



√

9 − x

dx 34.



√

− 9

35.



√

+ 9

dx 36.



√

x + 9

37.



x + 4

+ 3x + 2

dx 38.



5x + 6

+ x − 12

39.



+ 6x − 12

− 4x

dx 40.



+ 2

+ x

41.



cos2xdx 42.



sin x

43.





+ 1dx 44.





1 − x

............................................................

In exercises 45–50, ﬁnd the partial fractions decomposition.

45.

− 3x − 4

46.

+ x − 6

47.

−6

+ x

− 2x

48.

− 2x − 2

+ x

49.

x − 2

+ 4x + 4

50.

− 2

+ 1)

............................................................

In exercises 51–60, use the Table of Integrals to ﬁnd the integral.

51.





4 + e

dx 52.





− 4dx

53.



sec

xdx 54.



tan

xdx

55.



x(3 − x)

dx 56.



cos x

sin

x(3 +4sin x)

57.



√

9 + 4x

dx 58.



√

4 − 9x

59.



√

4 − x

dx 60.



− 4)

3/2

............................................................

In exercises 61–68, determine whether the integral converges or

diverges. If it converges, ﬁnd the limit.

61.



− 1

dx 62.



√

x − 4

63.



∞

dx 64.



∞

−3x

65.



∞

4 + x

dx 66.



∞

−∞

−x

67.



−2

dx 68.



−2

1 − x

............................................................

In exercises 69–76, ﬁnd the limit.

69. lim

x→1

− 1

70. lim

x→0

sin x

+ 3x

71. lim

x→∞

+ 2

72. lim

x→∞

−3x

)

73. lim

x→2



x + 1

x − 2



√

−4

74. lim

x→∞

x ln(1 +1/x)

75. lim

x→0

(tan x ln x) 76. lim

x→0

tan

−1

sin

−1

............................................................

77. Show that f (x) = x +2x

is a pdf on the interval [0, 1].

78. Show that f (x) =

−2x

is a pdf on the interval [0, ln 2].

79. Find the value of c such that f (x) =

is a pdf onthe interval

[1, 2].

80. Find the value of c such that f (x) = ce

−2x

is a pdf on the

interval [0, 4].

81. The lifetime of a lightbulb has pdf f (x) = 4e

−4x

(x in years).

Find the probability that the lightbulb lasts (a) less than

6 months; (b) between 6 months and 1 year.

82. The lifetime of an organism has pdf f (x) = 9xe

−3x

(x in

years). Find the probability that the organism lasts (a) less

than 2 months; (b) between 3 months and 1 year.

83. Find the (a) mean and (b) median of a random variable with

pdf f (x) = x + 2x

on the interval [0, 1].

84. Find the (a) mean and (b) median of a random variable with

pdf f (x) =

−2x

on the interval [0, ln 2].

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7-69 CHAPTER 7

Review Exercises 489

Review Exercises

85. Cardiologists test heart efﬁciency by injecting a dye at a con-

stant rate R into a vein near the heart and measuring the con-

centration of the dye in the bloodstream over a period of T

seconds. If all of the dye is pumped through, the concentra-

tion is c(t) = R. Compute the total amount of dye



c(t)dt.

For a general concentration, the cardiac output is deﬁned by



c(t)dt

. Interpret this quantity. Compute the cardiac output

if c(t) = 3te

2Tt

86. For



ln(x + 1) dx, you can use integration by parts with

u = ln(x + 1) and dv = 1. Compare your answers using v = x

versus using v = x + 1.

87. Showthattheaveragevalueofln x ontheinterval(0, e

)equals

n − 1 for any positive integer n.

88. Manyprobability questions involveconditional probabilities.

Forexample,ifyouknowthatalightbulbhasalreadyburnedfor

30 hours, what is the probability that it will last at least 5 more

hours? This is the “probability that x > 35 given that x > 30”

and is written as P(x > 35|x > 30). In general, for events A

and B, P(A|B) =

P(A and B)

P(B)

. The failure rate function is

given as the limit of

P(x < t + t|x > t)

t

as t → 0. For the

pdf f (x) of the lifetime of a lightbulb, the numerator is the

probability that the bulb burns out between times t and t + t.

Use R(t) = P(x > t) to show that the failure rate function can

be written as

f (t)

R(t)

89. Show that the failure rate function (see exercise 88) of an ex-

ponential pdf f (x) = ce

−cx

is constant.

90. For the gamma distribution f (x) = xe

−x

, (a) use a CAS to

show that P(x > s + t|x > s) = e

−t

1 + s

−t

. (b) Show

that this is a decreasing function of s (for a ﬁxed t). (c) If this

is the pdf for annual rainfall amounts in a certain city, interpret

the result of part (b).

91. Scores on IQ tests are intended to follow the distribution

f (x) =

√

450π

−(x−100)

/450

. Based on this distribution, what

percentage of people are supposedto have IQs between 90 and

100?If the top1% of scoresare to be giventhe titleof “genius,”

how high do you have to score to get this title?

92. Deﬁne I(n) =



∞

(1+x

)

dx for positive integers n. Show

that I (1) =

. Using integration by parts with u =

(1+x

)

, show that I (n + 1) =

2n−1

I(n). Conclude that

I(n) =

1 · 3 · 5 ···(2n − 3)

2 · 4 · 6 ···(2n − 2)

EXPLORATORY EXERCISES

1. In this exercise, you will try to determine whether or

not



sin(1/x)dx converges. Since |sin(1/x)|≤1, the

integral does not diverge to ∞, but that does not necessar-

ily mean it converges. Explain why the integral



∞

sin xdx

diverges (not to ∞, but by oscillating indeﬁnitely). You

need to determine whether a similar oscillation occurs for



sin(1/x)dx. First, estimate



sin(1/x)dx numerically

for R = 1/π, 1/(2π), 1/(3π ) and so on. Note that once you

have



1/π

sin(1/x)dx, you can get



1/(2π)

sin(1/x)dx by

“adding”



1/π

1/(2π)

sin(1/x)dx. We put this in quotes because

this new integral is negative. Verify that the integrals



1/π

1/(2π)

sin(1/x)dx,



1/(2π)

1/(3π)

sin(1/x)dx and so on, are

alternately negative and positive, so that the sum



sin(1/x)dx does seem to converge as R → 0

. It turns

out that the limit does converge if the additional integrals



1/(nπ)

1/((n+1)π)

sin(1/x)dx tend to 0 as n →∞. Show that this is

true.

2. Supposethat f (x) is a function such that both



∞

−∞

f (x) dx and



∞

−∞

f (x − 1/x)dx converge. Start with



∞

−∞

f (x − 1/x)dx

and make the substitution u =−

. Show that



∞

−∞

f (x − 1/x)dx =



∞

−∞

f (u − 1/u)du. Then let

y = u − 1/u. Show that



∞

−∞

f (x) dx =



∞

−∞

f (x − 1/x)dx.

Use this result to evaluate



∞

−∞

+ (x

− 1)

dx and



∞

−∞

−x

+2−1/x

dx.

3. Evaluate



π/2

(a cos x + b sin x)

dx, by dividing all terms

by cos

x, using the substitution u = ab tan x and evaluating

the improper integral



∞

(u + a

)

dx.

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CHAPTER

First-Order

Differential Equations

Well-preserved fossils can provide paleontologists with priceless clues

about the early history of life on Earth. In 1993, an amateur fossil hunter

found the bones of a massive dinosaur in southern Argentina. The new

species of dinosaur, named Giganotosaurus, replaced Tyrannosaurus Rex

as the largest known carnivore. Measuring up to 45 feet in length and

standing 12 feet high at the hip bone, Giganotosaurus is estimated to

have weighed roughly 8 tons.

Paleontologists employ several techniques for dating fossils, in order

to place them in their correct historical perspective. The most well known

of these is radiocarbon dating using carbon-14, an unstable isotope of

carbon. In living plants and animals, the ratio of the amount of carbon-14 to the

total amount of carbon is constant. When a plant or animal dies, it stops taking in

carbon-14 and the existing carbon-14 begins to decay, at a constant (though nearly

imperceptible) rate. An accurate measurement of the proportion of carbon-14

remaining can then be converted into an estimate of the time of death. Estimates

from carbon-14 dating are considered to be reliable for fossils dating back tens of

thousands of years, due to its very slow rate of decay.

Dating using other radioisotopes with slower decay rates than that of

carbon-14 works on the same basic principle, but can be used to accurately date

very old rock or sediment that surrounds the fossils. Using such techniques, pale-

ontologists estimate that Giganotosaurus lived about 100 million years ago. This

is critical information to scientists studying life near the end of the Mesozoic era.

For example, based on this method of dating, it is apparent that Giganotosaurus

did not live at the same time and therefore did not compete with the smaller but

stronger Tyrannosaurus Rex.

The mathematics underlying carbon-14 and other radioisotope dating

techniques is developed in this chapter. The study of differential equations pro-

vides you with essential tools to analyze many important phenomena such as

radiocarbon dating and the population of a bacterial colony. In this chapter, we

introduce the basic theory and a few common applications of some elementary

differential equations. In Chapter 16, we return to the topic of differential equa-

tions and present additional examples. However, a more thorough examination of

this vast ﬁeld will need to wait for a course focused on this topic.

8.1 MODELING WITH DIFFERENTIAL EQUATIONS

Growth and Decay Problems

In this age, we are all keenly aware of how infection by microorganisms such as

Escherichia coli (E. coli) causes disease. Many organisms (such as E. coli) can

491

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492 CHAPTER 8

First-Order Differential Equations 8-2

reproduceinourbodiesatasurprisinglyfastrate,overwhelmingourbodies’naturaldefenses

with the sheer volume of toxin they are producing. The table shown in the margin indicates

thenumberofE. coli bacteria(inmillionsofbacteriaperml)inalaboratoryculturemeasured

at half-hour intervals during the course of an experiment. We have plotted the number

of bacteria per milliliter versus time in Figure 8.1. What would you say the graph most

resembles?Ifyousaid,“anexponential,”youguessedright.Carefulanalysisofexperimental

data has shownthat many populations grow at a rateproportional to their current level. This

is quite easily observed in bacterial cultures, where the bacteria reproduce by binary ﬁssion

(i.e., each cell reproduces by dividing into two cells). In this case, the rate at which the

bacterial culture grows is directly proportional to the current population (until such time as

resourcesbecome scarceorovercrowdingbecomesa limiting factor).If welet y(t)represent

the number of bacteria in a culture at time t, then the rate of change of the population with

respect to time is y



(t). Thus, since y



(t) is proportional to y(t), we have



(t) = ky(t), (1.1)

100

200

300

400

4321

Number of bacteria

(millions per ml)

Time (hours)

FIGURE 8.1

Growth of bacteria

Number of

Time Bacteria

(hours) (millions per ml)

0 1.2

0.5 2.5

1 5.1

1.5 11.0

2 23.0

2.5 45.0

3 91.0

3.5 180.0

4 350.0

for some constant of proportionality k (the growth constant). Since equation (1.1) involves

the derivative of an unknown function, we call it a differential equation. Our aim is to

solve the differential equation, that is, ﬁnd the function y(t). Assuming that y(t) > 0 (this

is a reasonable assumption, since y(t) represents a population), we have



(t)

y(t)

= k. (1.2)

Integrating both sides of equation (1.2) with respect to t, we obtain



(t)

y(t)

dt =



kdt. (1.3)

Substituting y = y(t) in the integral on the left-hand side, we have dy = y



(t) dt and so,

(1.3) becomes



dy =



kdt.

Evaluating these integrals, we obtain

ln|y|+c

= kt + c

where c

and c

are constants of integration. Subtracting c

from both sides yields

ln|y|=kt + (c

− c

) = kt + c,

for some constant c. Since y(t) > 0, we have

ln y(t) = kt + c

and taking exponentials of both sides, we get

y(t) = e

ln y(t)

= e

kt+c

= e

Since c is an arbitrary constant, we write A = e

and get

y(t) = Ae

. (1.4)

We refer to (1.4) as the general solution of the differential equation (1.1). For k > 0,

equation (1.4) is called an exponential growth law and for k < 0, it is an exponential

decay law. (Think about the distinction.)

In example 1.1, we examine how an exponential growth law predicts the number of

cells in a bacterial culture.

EXAMPLE 1.1 Exponential Growth of a Bacterial Colony

A freshly inoculated bacterial culture of Streptococcus A (a common group of

microorganisms that cause strep throat) contains 100 cells. When the culture is checked

60 minutes later, it is determined that there are 450 cells present. Assuming exponential