Fig. 8.23

SOLUTION

Following the way that we usually do things in calculus, we break the problem

up into pieces. In moving the object from position x to position x + x, the

234 CHAPTER 8 Applications of the Integral

distance moved is x feet and the force applied is about F (x) = 3x 2 ’ x + 10.

See Fig. 8.23. Thus work performed in that little bit of the move is w(x) =

(3x 2 ’ x + 10) · x. The aggregate of the work is obtained by summation. In

this instance, the integral is the appropriate device:

10

10 x2

W= 3x ’ x + 10 dx = x ’ + 10x = 1050 foot-pounds.

2 3

2

0 0

EXAMPLE 8.14

A man is carrying a 100 lb sack of sand up a 20-foot ladder at the rate of

5 feet per minute. The sack has a hole in it and sand leaks out continuously

at a rate of 4 lb per minute. How much work does the man do in carrying

the sack?

Fig. 8.24

SOLUTION

It takes four minutes for the man to climb the ladder. At time t, the sack has

100 ’ 4t pounds of sand in it. From time t to time t + t, the man moves

5 · t feet up the ladder. He therefore performs about w(t) = (100 ’ 4t) · 5 t

foot-pounds of work. See Fig. 8.24. The total work is then the integral

4

4

W= (100 ’ 4t) 5 dt = 500t ’ 10t = 1840 foot-pounds.

2

0 0

235

CHAPTER 8 Applications of the Integral

You Try It: A man drags a 100 pound weight from x = 0 to x = 300. He resists

a wind which at position x applies a force of magnitude F (x) = x 3 + x + 40. How

much work does he perform?

EXAMPLE 8.15

According to Hooke™s Law, the amount of force exerted by a spring is propor-

tional to the distance of its displacement from the rest position.The constant

of proportionality is called the Hooke™s constant. A certain spring exerts a

force of 10 pounds when stretched 1/2 foot beyond its rest state. What is

the work performed in stretching the spring from rest to 1/3 foot beyond its

rest length?

Fig. 8.25

SOLUTION

Let the x-variable denote the position of the right end of the spring (Fig. 8.25),

with x = 0 the rest position. The left end of the spring is pinned down. Imagine

that the spring is being stretched to the right. We know that the force exerted

by the spring has the form

F (x) = kx,

with k a negative constant (since the spring will pull to the left). Also F (0.5) =

’10. It follows that k = ’20, so that

F (x) = ’20x.

Now the work done in moving the spring from position x to position x + x

will be about (20x) · x (the sign is + since we will pull the spring to the

right”against the pull of the spring). Thus the total work done in stretching

the right end of the spring from x = 0 to x = 1/3 is

1/3

1/3 10

W= (20x) dx = 10x =

2

foot-pounds.

9

0 0

236 CHAPTER 8 Applications of the Integral

EXAMPLE 8.16

Imagine that a water tank in the shape of a hemisphere of radius 10 feet

is being pumped out (Fig. 8.26). Find the work done in lowering the water

level from 1 foot from the top of the tank to 3 feet from the top of the tank.

Radius at depth x

equals √100 _ x 2

Fig. 8.26

SOLUTION

A glance at Fig. 8.27 shows that the horizontal planar slice of the tank, at the

√

level x feet below the top, is a disk of radius 100 ’ x 2 . This disk therefore

has area A(x) = π · (100 ’ x 2 ). Thus a slice at that level of thickness x will

have volume

V (x) = π · (100 ’ x 2 ) · x

100

x

√100 _ x 2

Fig. 8.27

and (assuming that water weights 62.4 pounds per cubic foot) weight equal to

w(x) = 62.4π · (100 ’ x 2 ) · x.

Thus the work in raising this slice to the top of the tank (where it can then

be dumped) is

W (x) = 62.4π · (100 ’ x 2 ) · x · x foot-pounds.

237

CHAPTER 8 Applications of the Integral

We calculate the total work by adding all these elements together using an

integral. The result is

3

W= 62.4π · (100 ’ x 2 ) · x dx

1

3

= 62.4π · 100x ’ x 3 dx

1

3

x4

= 62.4π 50x ’ 2

41

81 1

= 62.4π 450 ’ ’ 50 ’

4 4

= 23,712π foot-pounds.

You Try It: A spring has Hooke™s constant 5. How much work is performed in

stretching the spring half a foot from its rest position?

8.4 Averages

In ordinary conversation, when we average a collection p1 , . . . , pk of k numbers,

we add them together and divide by the number of items:

p1 + · · · + pk

σ = Average = .

k

The signi¬cance of the number σ is that if we wanted all the k numbers to be equal,

but for the total to be the same, then that common value would have to be σ .

Now suppose that we want to average a continuous function f over an interval

[a, b] of its domain. We can partition the interval,

P = {x0 , x1 , . . . , xk },

with x0 = a and xk = b as usual. We assume that this is a uniform partition, with

xj ’ xj ’1 = x = (b ’ a)/k for all j . Then an “approximate average” of f would

be given by

f (x1 ) + f (x2 ) + · · · + f (xk )

σapp = .

k

It is convenient to write this expression as

k k

b’a

1 1

= f (xj ) · = f (xj ) ·

σapp x.

b’a b’a

k

j =1 j =1

238 CHAPTER 8 Applications of the Integral

b

This last is a Riemann sum for the integral (1/[b ’ a]) · Thus, letting

a f (x) dx.

the mesh of the partition go to zero, we declare

b

1

average of f = σ = f (x) dx.

b’a a

EXAMPLE 8.17

In a tropical rain forest, the rainfall at time t is given by •(t) = 0.1 ’ 0.1t +

0.05t 2 inches per hour, 0 ¤ t ¤ 10. What is the average rainfall for times

0 ¤ t ¤ 6?

SOLUTION

We need only average the function •:

6

1

average rainfall = σ = •(t) dt

6’0 0

16

= 0.1 ’ 0.1t + 0.05t 2 dt

Y

60

FL

6

1 0.05 3

= 0.1t ’ 0.05t 2 + t

6 3 0

AM

= 0.1 ’ 0.3 + 0.6

= 0.4 inches per hour.

TE

EXAMPLE 8.18

Let f (x) = x/2 ’ sin x on the interval [’2, 5]. Compare the average value

of this function on the interval with its minimum and maximum.

SOLUTION

Observe that

1

f (x) = ’ cos x.

2

Thus the critical points occur when cos x = 1/2, or at ’π/3, π/3. We also

must consider the endpoints ’2, 5. The values at these points are

f (’2) = ’1 + sin 2 ≈ ’0.0907026

√

3

π

f (’π/3) = ’ + ≈ 0.3424266

6 √2

3

π

f (π/3) = ’ ≈ ’0.3424266

6 2

5

f (5) = ’ sin 5 ≈ 3.458924.

2

239

CHAPTER 8 Applications of the Integral

Plainly, the maximum value is f (5) = 5/2 ’ sin 5 ≈ 3.458924. The minimum

value is f (π/3) ≈ ’0.3424266.

The average value of our function is

5

1 x

σ= ’ sin x dx

5 ’ (’2) 2

’2

5

1 x2

= + cos x

74 ’2

1 25 4

= + cos 5 ’ + cos 2

7 4 4

1 21

= + cos 5 ’ cos 2

74

≈ 0.84997.

You can see that the average value lies between the maximum and the minimum,

as it should. This is an instance of a general phenomenon.

You Try It: On a certain tree line, the height of trees at position x is about 100 ’

3x + sin 5x. What is the average height of trees from x = 2 to x = 200?

EXAMPLE 8.19

What is the average value of the function g(x) = sin x over the interval

[0, 2π ]?

SOLUTION

We calculate that

2π

2π

1 1 1

σ= sin x dx = [’ cos x] = [’1 ’ (’1)] = 0.

2π ’ 0 2π 2π

0 0

We see that this answer is consistent with our intuition: the function g(x) =

sin x takes positive values and negative values with equal weight over the

interval [0, 2π]. The average is intuitively equal to zero. And that is the actual

computed value.

You Try It: Give an example of a function on the real line whose average over

every interval of length 4 is 0.

240 CHAPTER 8 Applications of the Integral

8.5 Arc Length and Surface Area

Just as the integral may be used to calculate planar area and spatial volume, so this

tool may also be used to calculate the arc length of a curve and surface area. The

basic idea is to approximate the length of a curve by the length of its piecewise

linear approximation. A similar comment applies to the surface area. We begin by

describing the basic rubric.

8.5.1 ARC LENGTH

Suppose that f (x) is a function on the interval [a, b]. Let us see how to calculate

the length of the curve consisting of the graph of f over this interval (Fig. 8.28).

We partition the interval:

a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xk’1 ¤ xk = b.

Look at Fig. 8.29. Corresponding to each pair of points xj ’1 , xj in the partition

is a segment connecting two points on the curve; the segment has endpoints

(xj ’1 , f (xj ’1 )) and (xj , f (xj )). The length j of this segment is given by the

y = f (x)

a b

Fig. 8.28

y = f (x)

(xj , f (xj))

(xj _ 1, f (xj _ 1))

xj “1 xj

Fig. 8.29

241

CHAPTER 8 Applications of the Integral

usual planar distance formula:

1/2

= [xj ’ xj ’1 ]2 + [f (xj ) ’ f (xj ’1 )]2 .

j

We denote the quantity xj ’ xj ’1 by x and apply the de¬nition of the derivative

to obtain

f (xj ) ’ f (xj ’1 )

≈ f (xj ).

x

Now we may rewrite the formula for as

j

≈ ([ x]2 + [f (xj ) x]2 )1/2

j

= (1 + [f (xj )]2 )1/2 x.

Summing up the lengths (Fig. 8.30) gives an approximate length for the curve:

j

k k

length of curve ≈ = (1 + [f (xj )]2 )1/2 x.

j

j =1 j =1

Fig. 8.30

But this last is a Riemann sum for the integral

b

= (1 + [f (x)]2 )1/2 dx. ()

a

As the mesh of the partition becomes ¬ner, the approximating sum is ever closer to

what we think of as the length of the curve, and it also converges to this integral.

Thus the integral represents the length of the curve.

EXAMPLE 8.20

Let us calculate the arc length of the graph of f (x) = 4x 3/2 over the interval

[0, 3].

242 CHAPTER 8 Applications of the Integral

SOLUTION

The length is

3 3

(1 + [f (x)] ) dx = (1 + [6x 1/2 ]2 )1/2 dx

2 1/2

0 0

3

= (1 + 36x)1/2 dx

0

3

1

= · (1 + 36x) 3/2

54 0

1

= [1093/2 ’ 13/2 ]

54

(109)3/2 ’ 1

= .

54

EXAMPLE 8.21

Let us calculate the length of the graph of the function f (x) = (1/2)—

(ex + e’x ) over the interval [1, ln 8].

SOLUTION

We calculate that

f (x) = (1/2)(ex ’ e’x ).

Therefore the length of the curve is

ln 8

1 + [(1/2)(ex ’ e’x )]2

1/2

dx

1

1/2