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.