Additional and Advanced Exercises

Chapter 5 Additional and Advanced Exercises

Theory and Examples

solves the initial value problem

dx

2 dy

1. a. If 7ƒsxd dx = 7, does

ƒsxd dx = 1 ?

2 +a y= ƒsxd,

and y = 0 when x = 0.

L 0 L 0 dx

b. If ƒsxd dx = 4 and ƒsxd Ú 0, does (Hint: sin sax - atd = sin ax cos at - cos ax sin at . ) L 0

4. Proportionality Suppose that x and y are related by the equation 2ƒsxd dx = 24 = 2?

0 x=

1 dt .

Give reasons for your answers. 0 21 + 4t 2

2 5 5 Show that d 2 y /dx 2 is proportional to y and find the constant of 2. Suppose

L proportionality.

ƒsxd dx = 4,

ƒsxd dx = 3,

5. Find ƒ(4) if

Which, if any, of the following statements are true?

x 2 2 5 ƒsxd

a. b. a. ƒstd dt = x cos px b. t 2 ƒsxd dx = - 3 px sƒsxd + g sxdd = 9 dt = x cos . L

ƒsxd … g sxd on the interval -2…x…5

Find ƒsp/2d

from the following information.

i. ƒ

is positive and continuous.

3. Initial value problem

Show that

ii. The area under the curve y= ƒsxd from x= 0 to x=a is

y= 1 ƒstd sin asx - td dt

2 aL p 0 a + a 2 2 sin a+ 2 cos a .

Chapter 5: Integration

7. The area of the region in the xy-plane enclosed by the x-axis, the

curve y= ƒsxd, ƒsxd Ú 0 , and the lines x= 1 and x=b is equal to 2b 2 + 1 - 22 for all b7 1. Find ƒ(x).

8. Prove that

xu

3 2 y ƒstd dt

L ƒsudsx - ud du .

0 a L 0 b du = L 0

(Hint: Express the integral on the right-hand side as the difference of two integrals. Then show that both sides of the equation have

the same derivative with respect to x.) 9. Finding a curve Find the equation for the curve in the xy-plane

x that passes through the point s1, - 1d if its slope at x is always

10. Shoveling dirt You sling a shovelful of dirt up from the bottom of a hole with an initial velocity of 32 ft/sec. The dirt must rise 17

FIGURE 5.34 Piecewise continuous functions ft above the release point to clear the edge of the hole. Is that

like this are integrated piece by piece. enough speed to get the dirt out, or had you better duck?

Piecewise Continuous Functions

x 2 Although we are mainly interested in continuous functions, many >3 ,

11. ƒsxd = e

-8 … x 6 0

functions in applications are piecewise continuous. A function ƒ(x) is

0…x…3

piecewise continuous on a closed interval I if ƒ has only finitely

2-x,

12. ƒsxd = e

-4…x60

many discontinuities in I, the limits

x 2 - 4,

0…x…3

lim - ƒsxd and lim + ƒsxd

exist and are finite at every interior point of I, and the appropriate one- sided limits exist and are finite at the endpoints of I. All piecewise

21 - z,

14. hszd = e -1 >3

0…z61

continuous functions are integrable. The points of discontinuity subdi-

1…z…2 vide I into open and half-open subintervals on which ƒ is continuous,

s7z - 6d ,

- 2 … x 6 -1 and the limit criteria above guarantee that ƒ has a continuous exten-

15. ƒsxd = • 1-x 2 ,

-1…x61

sion to the closure of each subinterval. To integrate a piecewise con-

1…x…2

tinuous function, we integrate the individual extensions and add the results. The integral of

-1 … r 6 0

16. hsrd = • 1-r 2 ,

ƒsxd = x ,

0…x62

17. Find the average value of the function graphed in the accompany- - 1,

2…x…3

ing figure.

(Figure 5.34) over [ - 1, 3] is

L -1 L -1

ƒsxd dx = s1 - xd dx +

Find the average value of the function graphed in the accompany- ing figure.

The Fundamental Theorem applies to piecewise continuous func- x

tions with the restriction that sd >dxd1 a ƒstd dt is expected to equal

ƒ (x) only at values of x at which ƒ is continuous. There is a similar re-

striction on Leibniz’s Rule below. Graph the functions in Exercises 11–16 and integrate them over

x their domains.

393 Leibniz’s Rule

Chapter 5 Additional and Advanced Exercises

dy >dx at which the carpet is being unrolled. That is, A(x) is being in- In applications, we sometimes encounter functions like

creased at the rate

and sin

x 2 21x

dy

ƒsxd = s1 + td dt

g sxd =

t 2 dt ,

ƒsysxdd dx .

L sin x L 1x

defined by integrals that have variable upper limits of integration and At the same time, A is being decreased at the rate variable lower limits of integration at the same time. The first integral

can be evaluated directly, but the second cannot. We may find the de- du ƒsusxdd

dx ,

rivative of either integral, however, by a formula called Leibniz’s Rule. the width at the end that is being rolled up times the rate du >dx . The net rate of change in A is

dx , ferentiable functions of x whose values lie in [a, b],

dA dy = ƒsysxdd du - ƒsusxdd If ƒ is continuous on [a, b] and if u(x) and y(x) are dif-

Leibniz’s Rule

dx

dx

which is precisely Leibniz’s Rule.

then

To prove the rule, let F be an antiderivative of ƒ on [a, b]. Then

d sxd dy

du

y sxd ƒstd dt = ƒsysxdd - ƒsusxdd .

dx L usxd

dx

dx

L ƒstd dt = Fsysxdd - Fsusxdd .

usxd

Differentiating both sides of this equation with respect to x gives the equation we want:

Figure 5.35 gives a geometric interpretation of Leibniz’s Rule. It y shows a carpet of variable width ƒ(t) that is being rolled up at the left sxd d

dx cFsysxdd - Fsusxdd d tation, time is x, not t.) At time x, the floor is covered from u(x) to y(x).

ƒstd dt = d

at the same time x as it is being unrolled at the right. (In this interpre-

dx L usxd

du The rate du >dx at which the carpet is being rolled up need not be the = F¿sysxdd - F¿susxdd dx dx Chain Rule same as the rate dy >dx at which the carpet is being laid down. At any given time x, the area covered by carpet is

dy

dx . Asxd =

dy - ƒsusxdd = ƒsysxdd du

dx sxd

L ƒstd dt .

usxd

Use Leibniz’s Rule to find the derivatives of the functions in Ex- ercises 19–21.

sin 1 x y

19. ƒsxd =

t dt 20. ƒsxd = 1 dt

f (u(x))

y Covering

22. Use Leibniz’s Rule to find the value of x that maximizes the value

f (y(x))

of the integral

Problems like this arise in the mathematical theory of political L u (x)

(x)

elections. See “The Entry Problem in a Political Race,” by Steven J. Brams and Philip D. Straffin, Jr., in Political Equilibrium, Peter

FIGURE 5.35 Rolling and unrolling a carpet: a geometric Ordeshook and Kenneth Shepfle, Editors, Kluwer-Nijhoff, interpretation of Leibniz’s Rule:

Boston, 1982, pp. 181–195.

dA dy = ƒsysxdd du - ƒsusxdd dx

dx .

dx

Approximating Finite Sums with Integrals

In many applications of calculus, integrals are used to approximate fi- At what rate is the covered area changing? At the instant x, A(x) is in-

nite sums—the reverse of the usual procedure of using finite sums to creasing by the width ƒ(y(x)) of the unrolling carpet times the rate

approximate integrals.

394

Chapter 5: Integration

For example, let’s estimate the sum of the square roots of the first 25. Let ƒ(x) be a continuous function. Express n positive integers,

21 + 22 + Á + 2n. The integral

nb+ƒ nb+

nbd

as a definite integral.

is the limit of the upper sums 26. Use the result of Exercise 25 to evaluate 1 1 2 1 n 1 a. lim S 1

=A s2 + 4 + 6 + Á + 2nd ,

n+ A n n+ Á+ A n # n n:q n 2

21 + 22 + Á + 2n b. lim 1 s1 15 +2 15 +3 15 +Á+n = 15 d,

c. lim 1 n asin n + sin 2p n + sin 3p n+ Á + sin

nb. y What can be said about the following limits?

n:q

d. lim 1 17 s1 15 +2 15 +3 15 +Á+n 15 d

n:q n

+Á+n d

e. lim 1 s1 15 +2 15 +3 15 15

n:q 15

27. a. Show that the area A n of an n-sided regular polygon in a circle

of radius r is

2 sin 2 n n. n

A n = nr 2p

b. Find the limit of A n as n:q. Is this answer consistent with

Therefore, when n is large, S n will be close to 2 >3 and we will have

what you know about the area of a circle? 28. A differential equation Show that y= sin x+

Root sum = 21 + 22 + Á + 2n = S n # n 3 >2 L 2 n 3/2 .

3 1 x cos 2t dt + 1 satisfies both of the following conditions:

i. y– = - sin x+ 2 sin 2x

The following table shows how good the approximation can be. ii. y= 1 and when y¿ = - 2 x=p .

n Root sum

s2 >3dn 3 >2

Relative error

29. A function defined by an integral The graph of a function ƒ consists of a semicircle and two line segments as shown. Let

1 ƒstd dt . 50 239.04 235.70 1.4% 100

lim 1 +2 5 +3 5 +Á+n 5

n:q

1 3 by showing that the limit is

c. Find gs- 1d . and evaluating the integral.

L 0 a. Find g (1).

b. Find g (3).

d. Find all values of x on the open interval s - 3, 4d at which g 24. See Exercise 23. Evaluate

has a relative maximum. e. Write an equation for the line tangent to the graph of g at

+3 3

lim s1 3 +2 3 +Á+n 3 d. x=- 1.

n:q n 4

f. Find the x-coordinate of each point of inflection of the graph of g on the open interval s - 3, 4d .

g. Find the range of g.