Chapter Review Exercises
665
CHAPTER REVIEW EXERCISES
In Exercises 1–4, refer to the function f .x/ whose graph is shown in Figure 1.
y
3
2
1
1
2
3
4
x
FIGURE 1
1. Estimate L4 and M4 on Œ0; 4!.
SOLUTION
With n D 4 and an interval of Œ0; 4!, "x D
4!0
4
D 1. Then,
L4 D "x.f .0/ C f .1/ C f .2/ C f .3// D 1
!
"
1
5
23
C1C C2 D
4
2
4
and
! ! "
! "
! "
! ""
!
"
1
3
5
7
1
9
9
M4 D "x f
Cf
Cf
Cf
D1
C2C C
D 7:
2
2
2
2
2
4
4
2. Estimate R4 , L4 , and M4 on Œ1; 3!.
1
With n D 4 and an interval of Œ1; 3!, "x D 3!1
4 D 2 . Then,
! ! "
! "
"
!
"
3
5
1
5
9
35
R4 D "x f
C f .2/ C f
C f .3/ D
2C C C2 D
I
2
2
2
2
4
8
!
! "
! ""
!
"
3
5
1
5
9
31
L4 D "x f .1/ C f
C f .2/ C f
D
1C2C C
D
I and
2
2
2
2
4
8
! ! "
! "
! "
! ""
!
"
5
7
9
11
1 3
9
5
17
67
M4 D "x f
Cf
Cf
Cf
D
C C C
D
:
4
4
4
4
2 2
4
2
8
16
Z b
3. Find an interval Œa; b! on which R4 is larger than
f .x/ dx. Do the same for L4 .
SOLUTION
a
SOLUTION
In general, RN is larger than
Rb
a
f .x/ dx on any interval Œa; b! over which f .x/ is increasing. Given the graph of
Rb
a f .x/ dx, f .x/ must be decreasing over the interval Œa; b!.
f .x/, we may take Œa; b! D Œ0; 2!. In order for L4 to be larger than
We may therefore take Œa; b! D Œ2; 3!.
Z 2
3
9
4. Justify !
f .x/ dx ! .
2
4
1
SOLUTION
Because f .x/ is increasing on Œ1; 2!, we know that
LN !
Z
2
f .x/ dx ! RN
1
for any N . Now,
L2 D
1
3
.1 C 2/ D
2
2
and
R2 D
so
3
!
2
Z
1
2
f .x/ dx !
9
:
4
!
"
1
5
9
2C
D ;
2
2
4
666
THE INTEGRAL
CHAPTER 5
In Exercises 5–8, let f .x/ D x 2 C 3x.
5. Calculate R6 , M6 , and L6 for f .x/ on the interval Œ2; 5!. Sketch the graph of f .x/ and the corresponding rectangles for each
approximation.
SOLUTION
Let f .x/ D x 2 C 3x. A uniform partition of Œ2; 5! with N D 6 subintervals has
"x D
5"2
1
D ;
6
2
xj D a C j"x D 2 C
j
;
2
and
!
"
1
7
j
xj" D a C j "
"x D C :
2
4
2
Now,
R6 D "x
D
1
2
6
X
j D1
!
f .xj / D
! ! "
! "
! "
"
1
5
7
9
f
C f .3/ C f
C f .4/ C f
C f .5/
2
2
2
2
"
55
91
135
625
C 18 C
C 28 C
C 40 D
:
4
4
4
8
The rectangles corresponding to this approximation are shown below.
y
35
30
25
20
15
10
x
2.0
2.5
3.0
3.5
4.0
4.5
Next,
M6 D "x
D
1
2
6
X
j D1
!
f .xj" / D
! ! "
! "
! "
! "
! "
! ""
1
9
11
13
15
17
19
f
Cf
Cf
Cf
Cf
Cf
2
4
4
4
4
4
4
189
253
325
405
493
589
C
C
C
C
C
16
16
16
16
16
16
"
D
2254
1127
D
:
32
16
The rectangles corresponding to this approximation are shown below.
y
35
30
25
20
15
10
x
2.0
2.5
3.0
3.5
4.0
4.5
Finally,
L6 D "x
D
5
X
j D0
!
f .xj / D
!
! "
! "
! ""
1
5
7
9
f .2/ C f
C f .3/ C f
C f .4/ C f
2
2
2
2
1
55
91
135
10 C
C 18 C
C 28 C
2
4
4
4
"
D
505
:
8
The rectangles corresponding to this approximation are shown below.
y
35
30
25
20
15
10
x
2.0
2.5
3.0
3.5
4.0
4.5
Chapter Review Exercises
6. Use FTC I to evaluate A.x/ D
SOLUTION
x
f .t/ dt.
!2
Let f .x/ D x 2 C 3x. Then
A.x/ D
Z
x
!2
"ˇ
!
"
1 3 3 2 ˇˇx
1
3
8
1
3
10
t C t ˇ D x3 C x2 " " C 6 D x3 C x2 " :
3
2
3
2
3
3
2
3
!2
Z 5
for f .x/ on Œ2; 5! and compute
f .x/ dx by taking the limit.
.t 2 C 3t/ dt D
7. Find a formula for RN
SOLUTION
Z
!
2
5"2
3
D
and a D 2. Hence,
N
N
!
!
"
!
"!
N
N
N
X
3 X
3j 2
3j
3 X
21j
9j 2
D "x
f .2 C j"x/ D
2C
C3 2C
D
10 C
C 2
N
N
N
N
N
N
Let f .x/ D x 2 C 3x on the interval Œ2; 5!. Then "x D
RN
j D1
D 30 C
j D1
N
N
63 X
27 X 2
j
C
j
N2
N3
j D1
!
N2
N
C
2
2
63
D 30 C 2
N
D
j D1
j D1
27
C 3
N
N3
N2
N
C
C
3
2
6
!
141
45
9
C
C
2
N
2N 2
and
!
8. Find a formula for LN
SOLUTION
"
141
45
9
141
lim RN D lim
C
C
D
:
2
N
2
2N 2
N !1
N !1
Z 2
for f .x/ on Œ0; 2! and compute
f .x/ dx by taking the limit.
0
Let f .x/ D x 2 C 3x and N be a positive integer. Then
"x D
2"0
2
D
N
N
and
xj D a C j"x D 0 C
for 0 ! j ! N . Thus,
LN D "x
D
N
!1
X
j D0
N !1
2 X 4j 2
6j
f .xj / D
C
N
N
N2
j D0
2j
2j
D
N
N
!
D
N !1
N !1
8 X 2
12 X
j
C
j
N3
N2
j D0
j D0
4.N " 1/.2N " 1/
6.N " 1/
26 10
4
C
D
"
C
:
N
3
N
3N 2
3N 2
Finally,
Z
2
0
f .x/ dx D lim
N !1
!
26 10
4
"
C
3
N
3N 2
9. Calculate R5 , M5 , and L5 for f .x/ D .x 2 C 1/!1 on the interval Œ0; 1!.
SOLUTION
"
D
26
:
3
Let f .x/ D .x 2 C 1/!1 . A uniform partition of Œ0; 1! with N D 5 subintervals has
"x D
1"0
1
D ;
5
5
xj D a C j"x D
j
;
5
and
!
"
2j " 1
1
xj" D a C j "
"x D
:
2
10
Now,
R5 D "x
5
X
j D1
f .xj / D
! ! "
! "
! "
! "
"
1
1
2
3
4
f
Cf
Cf
Cf
C f .1/
5
5
5
5
5
667
668
CHAPTER 5
THE INTEGRAL
1
5
D
!
25
25
25
25
1
C
C
C
C
26
29
34
41
2
"
# 0:733732:
Next,
M5 D "x
1
D
5
5
X
j D1
!
f .xj" / D
! ! "
! "
! "
! "
! ""
1
1
3
1
7
9
f
Cf
Cf
Cf
Cf
5
10
10
2
10
10
100
100
4
100
100
C
C C
C
101
109
5
149
181
"
# 0:786231:
Finally,
L5 D "x
4
X
j D0
!
! "
! "
! "
! ""
1
1
2
3
4
f .0/ C f
Cf
Cf
Cf
5
5
5
5
5
f .xj / D
!
"
1
25
25
25
25
D
1C
C
C
C
# 0:833732:
5
26
29
34
41
10. Let RN be the N th right-endpoint approximation for f .x/ D x 3 on Œ0; 4! (Figure 2).
64.N C 1/2
(a) Prove that RN D
.
N2
(b) Prove that the area of the region within the right-endpoint rectangles above the graph is equal to
64.2N C 1/
N2
y
64
32
x
1
2
3
4
FIGURE 2 Approximation RN for f .x/ D x 3 on Œ0; 4!.
SOLUTION
(a) Let f .x/ D x 3 and N be a positive integer. Then
"x D
4"0
4
D
N
N
and
xj D a C j"x D 0 C
4j
4j
D
N
N
for 0 ! j ! N . Thus,
RN D "x
N
X
j D1
f .xj / D
(b) The area between the graph of y D
x3
N
N
4 X 64j 3
256 X 3
256 N 2 .N C 1/2
64.N C 1/2
D
j
D
D
:
N
4
N3
N4
N4
N2
j D1
j D1
and the x-axis over Œ0; 4! is
ˇ
Z 4
1 ˇ4
x 3 dx D x 4 ˇˇ D 64:
4
0
0
The area of the region below the right-endpoint rectangles and above the graph is therefore
64.N C 1/2
64.2N C 1/
" 64 D
:
N2
N2
11. Which approximation to the area is represented by the shaded rectangles in Figure 3? Compute R5 and L5 .
y
30
18
6
1
2
3
FIGURE 3
4
5
x
Chapter Review Exercises
669
SOLUTION There are five rectangles and the height of each is given by the function value at the right endpoint of the subinterval.
Thus, the area represented by the shaded rectangles is R5 .
From the figure, we see that "x D 1. Then
R5 D 1.30 C 18 C 6 C 6 C 30/ D 90
and
x2
L5 D 1.30 C 30 C 18 C 6 C 6/ D 90:
12. Calculate any two Riemann sums for f .x/ D
on the interval Œ2; 5!, but choose partitions with at least five subintervals of
unequal widths and intermediate points that are neither endpoints nor midpoints.
Let f .x/ D x 2 . Riemann sums will, of course, vary. Here are two possibilities. Take N D 5,
SOLUTION
P D fx0 D 2; x1 D 2:7; x2 D 3:1; x3 D 3:6; x4 D 4:2; x5 D 5g
and
C D fc1 D 2:5; c2 D 3; c3 D 3:5; c4 D 4; c5 D 4:5g:
Then,
R.f; P; C / D
5
X
j D1
"xj f .cj / D 0:7.6:25/ C 0:4.9/ C 0:5.12:25/ C 0:6.16/ C 0:8.20:25/ D 39:9:
Alternately, take N D 6,
P D fx0 D 2; x1 D 2:5; x2 D 3:5; x3 D 4; x4 D 4:25; x5 D 4:75; x6 D 5g
and
C D fc1 D 2:1; c2 D 3; c3 D 3:7; c4 D 4:2; c5 D 4:5; c6 D 4:8g:
Then,
R.f; P; C / D
6
X
"xj f .cj /
j D1
D 0:5.4:41/ C 1.9/ C 0:5.13:69/ C 0:25.17:64/ C 0:5.20:25/ C 0:25.23:04/ D 38:345:
In Exercises 13–16, express the limit as an integral (or multiple of an integral) and evaluate.
13.
lim
N !1
!
"
N
# X
#
#j
sin
C
6N
3
6N
j D1
Let f .x/ D sin x and N be a positive integer. A uniform partition of the interval Œ#=3; #=2! with N subintervals has
SOLUTION
"x D
#
6N
and
xj D
#
#j
C
3
6N
for 0 ! j ! N . Then
!
"
N
N
X
# X
#
#j
sin
C
D "x
f .xj / D RN I
6N
3
6N
j D1
j D1
consequently,
ˇ!=2
!
" Z !=2
N
ˇ
# X
#
#j
1
1
sin
C
D
sin x dx D " cos x ˇˇ
D0C D :
3
6N
2
2
N !1 6N
!=3
!=3
lim
14.
lim
N !1
SOLUTION
3
N
N
!1 !
X
kD0
10 C
3k
N
"
j D1
Let f .x/ D x and N be a positive integer. A uniform partition of the interval Œ10; 13! with N subintervals has
"x D
3
N
and
xj D 10 C
3j
N
for 0 ! j ! N . Then
"
N !1 !
N
!1
X
3 X
3k
10 C
D "x
f .xj / D LN I
N
N
kD0
j D0
670
THE INTEGRAL
CHAPTER 5
consequently,
lim
N !1
ˇ
" Z 13
N !1 !
3 X
3k
1 ˇ13
10 C
D
x dx D x 2 ˇˇ
N
N
2
10
10
kD0
169 100
69
"
D
:
2
2
2
D
15.
lim
N !1
N
5 Xp
4 C 5j=N
N
j D1
p
Let f .x/ D
SOLUTION
x and N be a positive integer. A uniform partition of the interval Œ4; 9! with N subintervals has
"x D
5
N
and
5j
N
xj D 4 C
for 0 ! j ! N . Then
N
N
X
5 Xp
4 C 5j=N D "x
f .xj / D RN I
N
j D1
j D1
consequently,
ˇ9
Z 9
N
ˇ
p
5 Xp
2
54 16
38
4 C 5j=N D
x dx D x 3=2 ˇˇ D
"
D
:
3
3
3
3
N !1 N
4
4
lim
j D1
16.
lim
1k
N !1
SOLUTION
C 2k
C $$$ C Nk
N kC1
(k > 0)
Observe that
"!
1k C 2k C 3k C $ $ $ C N k
1
D
N
N kC1
1
N
"k
C
!
2
N
"k
C
!
3
N
"k
!
N
C $$$
N
"k #
D
"
N !
1 X j k
:
N
N
j D1
Now, let f .x/ D x k and N be a positive integer. A uniform partition of the interval Œ0; 1! with N subintervals has
"x D
1
N
and
xj D
j
N
for 0 ! j ! N . Then
"
N !
N
X
1 X j k
D "x
f .xj / D RN I
N
N
j D1
j D1
consequently,
ˇ1
"
Z 1
N !
ˇ
1 X j k
1
1
D
x k dx D
x kC1 ˇˇ D
:
N
N
k
C
1
k
C
1
N !1
0
0
lim
j D1
In Exercises 17–20, use the given substitution to evaluate the integral.
Z 2
dt
17.
, u D 4t C 12
0 4t C 12
SOLUTION
18.
Z
Let u D 4t C 12. Then du D 4dt, and the new limits of integration are u D 12 and u D 20. Thus,
.x 2 C 1/ dx
,
.x 3 C 3x/4
SOLUTION
Z
2
0
dt
1
D
4t C 12
4
u D x 3 C 3x
Z
20
12
ˇ20
ˇ
du
1
1
1 20
1 5
D ln uˇˇ D .ln 20 " ln 12/ D ln
D ln :
u
4
4
4
12
4 3
12
Let u D x 3 C 3x. Then du D .3x 2 C 3/ dx D 3.x 2 C 1/ dx and
Z
1
.x 2 C 1/ dx
D
4
3
3
.x C 3x/
Z
1
1
u!4 du D " u!3 C C D " .x 3 C 3x/!3 C C:
9
9
Chapter Review Exercises
19.
Z
!=6
sin x cos4 x dx,
u D cos x
0
SOLUTION
Let u D cos x. Then du D " sin x dx and the new limits of integration are u D 1 and u D
Z
!=6
sin x cos x dx D "
0
20.
Z
sec2 .2$/ tan.2$/ d$,
4
Z
p
3=2
u4 du
1
ˇp
1 5 ˇˇ 3=2
D " u ˇ
5
1
p !
1
9 3
D
1"
:
5
32
u D tan.2$/
Let u D tan.2$/. Then du D 2 sec2 .2$/ d$ and
Z
Z
1
1
1
sec2 .2$/ tan.2$/ d$ D
u du D u2 C C D tan2 .2$/ C C:
2
4
4
In Exercises 21–70, evaluate the integral.
Z
21. .20x 4 " 9x 3 " 2x/ dx
Z
9
SOLUTION
.20x 4 " 9x 3 " 2x/ dx D 4x 5 " x 4 " x 2 C C .
4
Z 2
22.
.12x 3 " 3x 2 / dx
SOLUTION
0
SOLUTION
23.
Z
Z
2
0
ˇ2
ˇ
.12x 3 " 3x 2 / dx D .3x 4 " x 3 /ˇˇ D .48 " 8/ " 0 D 40.
0
.2x 2 " 3x/2 dx
SOLUTION
24.
Z
1
0
.2x 2 " 3x/2 dx D
Z
4 5
x " 3x 4 C 3x 3 C C .
5
.4x 4 " 12x 3 C 9x 2 / dx D
.x 7=3 " 2x 1=4 / dx
SOLUTION
Z
Z
1
0
.x 7=3 " 2x 1=4 / dx D
!
"ˇ
3 10=3 8 5=4 ˇˇ1
3
8
13
x
" x
ˇ D 10 " 5 D " 10 .
10
5
0
C 3x 4
dx
x2
Z 5
Z
x C 3x 4
1
SOLUTION
dx
D
.x 3 C 3x 2 / dx D x 4 C x 3 C C .
4
x2
Z 3
26.
r !4 dr
25.
x5
Z
1
SOLUTION
27.
Z
3
!3
Z
3
r
1
!4
jx 2 " 4j dx
ˇ
!
"
1 !3 ˇˇ3
1 1
26
dr D " r ˇ D "
"1 D
.
3
3
27
81
1
SOLUTION
Z
3
!3
jx 2 " 4j dx D
Z
2
!3
.x 2 " 4/ dx C
Z
2
!2
.4 " x 2 / dx C
Z
3
2
.x 2 " 4/ dx
"ˇ!2 !
"ˇ3
"ˇ2
!
ˇ
ˇ
ˇ
1
1 3
1 3
x " 4x ˇˇ C 4x " x 3 ˇˇ C
x " 4x ˇˇ
3
3
3
!3
2
!2
!
" !
" !
"
16
16
16
16
D
"3 C
C
C "3 C
3
3
3
3
D
!
D
46
:
3
p
3=2. Thus,
671
672
28.
THE INTEGRAL
CHAPTER 5
Z
4
!2
j.x " 1/.x " 3/j dx
SOLUTION
Z
4
!2
29.
Z
j.x " 1/.x " 3/j dx D
Z
1
2
!2
D
!
D
62
:
3
.x " 4x C 3/ dx C
Z
3
1
2
."x C 4x " 3/ dx C
Z
4
3
.x 2 " 4x C 3/ dx
"ˇ1
!
"ˇ3 !
"ˇ4
ˇ
ˇ
ˇ
1 3
1
1 3
x " 2x 2 C 3x ˇˇ C " x 3 C 2x 2 " 3x ˇˇ C
x " 2x 2 C 3x ˇˇ
3
3
3
!2
1
3
!
"
! "
4
50
4
4
D " "
C0" "
C "0
3
3
3
3
3
Œt! dt
1
SOLUTION
30.
Z
2
0
Z
Z
3
1
Œt! dt D
2
1
Œt! dt C
Z
3
2
Œt! dt D
Z
2
1
dt C
Z
3
2
.t " Œt!/2 dt
ˇ2
ˇ3
ˇ
ˇ
2 dt D t ˇˇ C 2t ˇˇ D .2 " 1/ C .6 " 4/ D 3:
1
2
SOLUTION
Z
2
0
.t " Œt!/2 dt D
Z
D
31.
Z
D
Z
3p
33.
34.
2
1
.t " 1/2 dt
ˇ
ˇ2
ˇ
1 3 ˇˇ1
1
t ˇ C .t " 1/3 ˇˇ
3 0
3
1
1
1
2
C D :
3
3
3
Let u D 10t " 7. Then du D 10dt and
Z
Z
1
1 15
1
.10t " 7/14 dt D
u14 du D
u CC D
.10t " 7/15 C C:
10
150
150
Let u D 7y " 5. Then du D 7dy and when y D 2, u D 9 and when y D 3, u D 16. Finally,
Z
.2x 3 C 3x/ dx
.3x 4 C 9x 2 /5
SOLUTION
Z
0
Z
7y " 5 dy
2
SOLUTION
Z
t 2 dt C
.10t " 7/14 dt
SOLUTION
32.
1
!1
!3
SOLUTION
3p
2
7y " 5 dy D
1
7
Z
16
9
u1=2 du D
ˇ
1 2 3=2 ˇˇ16
2
74
$ u ˇ D
.64 " 27/ D
:
7 3
21
21
9
Let u D 3x 4 C 9x 2 . Then du D .12x 3 C 18x/ dx D 6.2x 3 C 3x/ dx and
Z
.2x 3 C 3x/ dx
1
D
6
.3x 4 C 9x 2 /5
Z
u!5 du D "
1 !4
1
u C C D " .3x 4 C 9x 2 /!4 C C:
24
24
x dx
.x 2 C 5/2
Let u D x 2 C 5. Then du D 2x dx and
Z
!1
!3
ˇ6
ˇ
1
u!2 du D " u!1 ˇˇ
2
14
14
!
"
1
1 1
1
"
D" :
D"
2 6 14
21
x dx
1
D
2
2
2
.x C 5/
Z
6
Chapter Review Exercises
35.
Z
5
0
p
15x x C 4 dx
Let u D x C 4. Then x D u " 4, du D dx and the new limits of integration are u D 4 and u D 9. Thus,
SOLUTION
Z
5
0
Z
p
15x x C 4 dx D
9
4
D 15
p
15.u " 4/ u du
Z
9
4
.u3=2 " 4u1=2 / du
"ˇ
2 5=2 8 3=2 ˇˇ9
u
" u
ˇ
5
3
4
!!
" !
""
486
64 64
D 15
" 72 "
"
5
5
3
D 15
36.
Z
t
p
2
!
D 506:
t C 8 dt
SOLUTION
Let u D t C 8. Then du D dt, t D u " 8, and
Z
Z
Z
p
p
t 2 t C 8 dt D .u " 8/2 u du D .u5=2 " 16u3=2 C 64u1=2 / du
2 7=2 32 5=2 128 3=2
u
" u
C
u
CC
7
5
3
2
32
128
D .t C 8/7=2 " .t C 8/5=2 C
.t C 8/3=2 C C:
7
5
3
D
37.
Z
1
cos
0
##
3
$
.t C 2/ dt
p
$
##
$ˇˇ1
3
3 3
.t C 2/ dt D
sin
.t C 2/ ˇˇ D "
.
3
#
3
2#
0
0
!
"
Z !
5$ " #
38.
sin
d$
6
!=2
SOLUTION
Z
SOLUTION
Let
1
cos
##
uD
5$ " #
6
so that
"
6
5
du D
5
d$:
6
Then
Z
!
!=2
39.
Z
Z
!
5$ " #
6
d$ D
Z
2!=3
sin u du
!=4
ˇ2!
ˇ
6
D " cos uˇˇ
3
5
!=4
p !
p
6
1
2
3
D"
" "
D .1 C 2/:
5
2
2
5
t 2 sec2 .9t 3 C 1/ dt
SOLUTION
40.
sin
Let u D 9t 3 C 1. Then du D 27t 2 dt and
Z
Z
1
1
1
t 2 sec2 .9t 3 C 1/ dt D
sec2 u du D
tan u C C D
tan.9t 3 C 1/ C C:
27
27
27
sin2 .3$/ cos.3$/ d$
SOLUTION
Let u D sin.3$/. Then du D 3 cos.3$/d$ and
Z
Z
1
1
1
sin2 .3$/ cos.3$/ d$ D
u2 du D u3 C C D sin3 .3$/ C C:
3
9
9
673
674
41.
THE INTEGRAL
CHAPTER 5
Z
csc2 .9 " 2$/ d$
Let u D 9 " 2$. Then du D "2 d$ and
Z
Z
1
1
1
csc2 .9 " 2$/ d$ D "
csc2 u du D cot u C C D cot.9 " 2$/ C C:
2
2
2
Z
p
42.
sin $ 4 " cos $ d$
SOLUTION
SOLUTION
43.
Z
!=3
0
Let u D 4 " cos $. Then du D sin $ d$ and
Z
Z
p
2
2
sin $ 4 " cos $ d$ D u1=2 du D u3=2 C C D .4 " cos $/3=2 C C:
3
3
sin $
d$
cos2=3 $
Let u D cos $. Then du D " sin $ d$ and when $ D 0, u D 1 and when $ D
SOLUTION
Z
!=3
0
44.
Z
sec2 t dt
.tan t " 1/2
Z
Z
Z
3
Z
Z
Z
x2e
49.
ln 3
Z
1
D "3.2
p
334
" 1/ D 3 "
:
2
sec2 t dt
D
.tan t " 1/2
Z
u!2 du D "u!1 C C D "
1
C C:
tan t " 1
x3
3
1
e 4x!3 dx D
dx
ˇ
1 4x!3 ˇˇ3
1 9
e
ˇ D 4 .e " e/.
4
1
x
e x!e dx
0
x
x
Note e x!e D e x e !e . Now, let u D e x . Then du D e x dx, and the new limits of integration are u D e 0 D 1 and
3. Thus,
ˇ3
Z ln 3
Z ln 3
Z 3
ˇ
x
x
e x!e dx D
e x e !e dx D
e !u du D "e !t ˇˇ D ".e !3 " e !1 / D e !1 " e !3 :
0
0
1
1
e x 10x dx
SOLUTION
50.
ˇ
!1=3
Let u D x 3 . Then du D 3x 2 dx, and
Z
Z
1
1
1 3
3
x 2 e x dx D
e u du D e u C C D e x C C:
3
3
3
SOLUTION
u D e ln 3 D
Z
du D "3u
ˇ1=2
ˇ
1=3 ˇ
e 4x!3 dx
1
SOLUTION
48.
u
1
!2=3
Let u D 9 " 2x. Then du D "2 dx, and
Z
Z
1
1
1
e 9!2x dx D "
e u du D " e u C C D " e 9!2x C C:
2
2
2
SOLUTION
47.
1=2
e 9!2x dx
SOLUTION
46.
Z
u D 12 . Finally,
Let u D tan t " 1. Then du D sec2 t dt and
SOLUTION
45.
sin $
d$ D "
cos2=3 $
!
3,
Z
e x 10x dx D
Z
.10e/x dx D
.10e/x
.10e/x
10x e x
CC D
CC D
C C.
ln.10e/
ln 10 C ln e
ln 10 C 1
e !2x sin.e !2x / dx
SOLUTION
Let u D e !2x . Then du D "2e !2x dx, and
Z
Z
#
$
#
$
cos u
1
1
e !2x sin e !2x dx D "
sin u du D
C C D cos e !2x C C:
2
2
2
Chapter Review Exercises
51.
Z
e !x dx
.e !x C 2/3
Let u D e !x C 2. Then du D "e !x dx and
SOLUTION
52.
Z
Z
2
sin $ cos $e cos
Z
!=6
54.
1
1
CC D
C C:
2
!x
2u
2.e C 2/2
d$
tan 2$ d$
0
SOLUTION
Z
u!3 du D
Let u D cos2 $ C 1. Then du D "2 sin $ cos $ d$ and
Z
Z
1
1
1
2
2
sin $ cos $e cos " C1 d$ D "
e u du D " e u C C D " e cos " C1 C C:
2
2
2
SOLUTION
53.
" C1
Z
e !x dx
D"
.e !x C 2/3
Z
!=6
tan 2$ d$ D
0
2!=3
cot
!=3
!
1
$
2
"
d$
ˇ!=6
ˇ
1
1
ln j sec 2$jˇˇ
D ln 2.
2
2
0
SOLUTION
Z
55.
Z
dt
t.1 C .ln t/2 /
SOLUTION
Let u D ln t. Then, du D
Z
2!=3
!=3
1
t
!
1
cot
$
2
Z
Z
e
dt
2
58.
1
Z
Z
du
D tan!1 u C C D tan!1 .ln t/ C C:
1 C u2
Let u D ln x. Then du D dx
x , and
Z
Z
cos.ln x/
dx D cos u du D sin u C C D sin.ln x/ C C:
x
Let u D ln x. Then du D
dx
x
and the new limits of integration are u D ln 1 D 0 and u D ln e D 1. Thus,
Z
e
1
ln x dx
D
x
dx
p
x ln x
SOLUTION
59.
D
ln x dx
x
SOLUTION
Z
!=3
#
#
#$
D 2 ln sin " ln sin
3
6
!
p
3
1
D 2 ln
" ln
D ln 3:
2
2
cos.ln x/ dx
x
SOLUTION
57.
ˇ
ˇ ˇ2!
ˇ $ ˇ ˇˇ
ˇ
d$ D 2 ln ˇsin ˇˇ ˇ 3
2 ˇ
dt and
t.1 C .ln t/ /
56.
"
Let u D ln x. Then du D
Z
dx
4x 2 C 9
1
x
Z
1
0
u du D
ˇ
1 2 ˇˇ1
1
u ˇ D :
2
2
0
dx, and
dx
p
D
x ln x
Z
p
p
u!1=2 du D 2 u C C D 2 ln x C C:
675
676
SOLUTION
60.
THE INTEGRAL
CHAPTER 5
Z
2x
3 .
Let u D
Z
0:8
p
Then x D 23 u, dx D
dx
D
4x 2 C 9
Z
3
2
3
2 du
9 2
4u C 9
4$
du, and
D
1
6
Z
du
1
1
D tan!1 u C C D tan!1
6
6
u2 C 1
!
2x
3
"
C C:
dx
1 " x2
ˇ0:8
Z 0:8
ˇ
dx
SOLUTION
p
D sin!1 x ˇˇ D sin!1 0:8 " sin!1 0 D sin!1 0:8.
0
1 " x2
0
Z 12
dx
61.
p
4
x x2 " 1
ˇ12
Z 12
ˇ
dx
SOLUTION
p
D sec!1 x ˇˇ D sec!1 12 " sec!1 4:
2
4
x x "1
4
Z 3
x dx
62.
2
0 x C9
0
SOLUTION
63.
Z
3
0
Z
dx
2
x C9
SOLUTION
64.
Let u D x 2 C 9. Then du D 2x dx, and the new limits of integration are u D 9 and u D 18. Thus,
x
3.
Let u D
Z
3
0
Z
p
dx
3
x dx
1
D
2
x2 C 9
0
Then du D
dx
1
D
3
x2 C 9
dx
3 ,
Z
1
0
e 2x " 1
SOLUTION
Z
18
9
ˇ18
ˇ
du
1
1
1 18
1
D ln uˇˇ D .ln 18 " ln 9/ D ln
D ln 2:
u
2
2
2
9
2
9
and the new limits of integration are u D 0 and u D 1. Thus,
ˇ
$
dt
1 !1 ˇˇ1
1
1 ##
#
!1
!1
D
tan
t
D
.tan
1
"
tan
0/
D
"
0
D
:
ˇ
3
3
3
4
12
t2 C 1
0
Let u D e x . Then
du D e x dx
)
du D u dx
)
u!1 du D dx
By substitution, we obtain
Z
65.
Z
Z
1
0
4
0
Z
du
p
u u2 " 1
D sec!1 u C C D sec!1 .e x / C C
p
p
Let u D x 2 . Then du D 2x dx, and 1 " x 4 D 1 " u2 . Thus,
Z
Z
x dx
1
du
1
1
p
D
p
D sin!1 u C C D sin!1 .x 2 / C C:
4
2
2
2
2
1"x
1"u
Let x D 5u. Then dx D 5 du, and the new limits of integration are u D 0 and u D
Z
Z
e 2x " 1
D
dx
25 " x 2
SOLUTION
67.
dx
x dx
p
1 " x4
SOLUTION
66.
p
dx
2x 2 C 1
1
0
Thus,
Z 1=5
5 du
5
du
D
25 0
1 " u2
1 " u2
0
ˇ1=5
!
"
ˇ
1
1
1
1
1
D tanh!1 uˇˇ
D
tanh!1 " tanh!1 0 D tanh!1 :
5
5
5
5
5
0
dx
1
D
25
25 " x 2
Z
1
5.
1=5
Chapter Review Exercises
p
Let u D
SOLUTION
Z
4
0
2x. Then du D
dx
D
2x 2 C 1
Z
p
p
2 dx, and the new limits of integration are u D 0 and u D 4 2. Thus,
p
1
4 2 p du
2
D
u2 C 1
0
ˇ4p2
ˇ
1
!1 ˇ
D p tan
2
68.
Z
8
5
dx
p
x x 2 " 16
Z
8
5
69.
Z
1
1
p
2
Z
p
4 2
du
C1
u2
0
$
p
p
1 #
1
D p tan!1 .4 2/ " tan!1 0 D p tan!1 .4 2/:
2
2
uˇ
0
Let x D 4u. Then dx D 4 du, and the new limits of integration are u D
SOLUTION
dx
1
p
D
4
x x 2 " 16
Z
2
5=4
.tan!1 x/3 dx
1 C x2
0
SOLUTION
677
5
4
and u D 2. Thus,
ˇ
!
"
!
"
du
1 # !1 $ ˇˇ2
1
5
1 #
5
p
D
sec u ˇ
D
sec!1 2 " sec!1
D
" sec!1
:
4
4
4
4 3
4
u u2 " 1
5=4
Let u D tan!1 x. Then
1
dx
1 C x2
du D
and
70.
Z
cos!1 t
p
SOLUTION
Z
1
0
Z
.tan!1 x/3 dx
D
1 C x2
!=4
0
ˇ
1 4 ˇˇ!=4
1 # # $4
#4
u ˇ
D
D
:
4
4 4
1024
0
u3 du D
dt
1 " t2
Let u D cos!1 t. Then du D " p
1
1!t 2
dt, and
Z
Z
cos!1 t
1
1
2
p
dt D " u du D " u2 C C D " .cos!1 t/ C C:
2
2
2
1"t
71. Combine to write as a single integral:
Z 8
Z 0
Z 6
f .x/ dx C
f .x/ dx C
f .x/ dx
0
SOLUTION
!2
8
First, rewrite
Z
8
0
f .x/ dx D
Z
6
0
f .x/ dx C
Z
8
f .x/ dx
6
and observe that
Z
Z
6
f .x/ dx D "
8
8
f .x/ dx:
6
Thus,
Z
8
0
f .x/ dx C
Z
8
6
f .x/ dx D
Z
6
f .x/ dx:
0
Finally,
Z
8
0
Rx
0
f .x/ dx C
Z
0
!2
f .x/ dx C
Z
6
8
f .x/ dx D
Z
0
6
f .x/ dx C
Z
0
!2
f .x/ dx D
Z
6
f .x/ dx:
!2
72. Let A.x/ D
f .x/ dx, where f .x/ is the function shown in Figure 4. Identify the location of the local minima, the local
maxima, and points of inflection of A.x/ on the interval Œ0; E!, as well as the intervals where A.x/ is increasing, decreasing,
concave up, or concave down. Where does the absolute max of A.x/ occur?
678
CHAPTER 5
THE INTEGRAL
y
x
A
B
C
D
E
FIGURE 4
SOLUTION
Let f .x/ be the function shown in Figure 4 and define
Z x
A.x/ D
f .x/ dx:
0
A0 .x/
A00 .x/
0 .x/.
Then
D f .x/ and
Df
Hence, A.x/ is increasing when f .x/ is positive, is decreasing when f .x/ is negative,
is concave up when f .x/ is increasing and is concave down when f .x/ is decreasing. Thus, A.x/ is increasing for 0 < x < B, is
decreasing for B < x < D and for D < x < E, has a local maximum at x D B and no local minima. Moreover, A.x/ is concave
up for 0 < x < A and for C < x < D, is concave down for A < x < C and for D < x < E, and has a point of inflection at
x D A, x D C and x D D. The absolute maximum value for A.x/ occurs at x D B.
Z x
t dt
73. Find the local minima, the local maxima, and the inflection points of A.x/ D
.
2
3 t C1
SOLUTION
Let
Z
A.x/ D
x
3
t dt
:
C1
t2
Then
A0 .x/ D
x
x2 C 1
and
A00 .x/ D
.x 2 C 1/.1/ " x.2x/
1 " x2
D 2
:
2
2
.x C 1/
.x C 1/2
Now, x D 0 is the only critical point of A; because A00 .0/ > 0, it follows that A has a local minimum at x D 0. There are no local
maxima. Moreover, A.x/ is concave down for jxj > 1 and concave up for jxj < 1. A.x/ therefore has inflection points at x D ˙1.
74. A particle starts at the origin at time t D 0 and moves with velocity v.t/ as shown in Figure 5.
(a) How many times does the particle return to the origin in the first 12 seconds?
(b) What is the particle’s maximum distance from the origin?
(c) What is particle’s maximum distance to the left of the origin?
v(t) m/s
4
2
5
10
−2
t (s)
−4
FIGURE 5
SOLUTION
Because the particle starts at the origin, the position of the particle is given by
s.t/ D
Z
t
v.%/ d %I
0
that is by the signed area between the graph of the velocity and the t-axis over the interval Œ0; t!. Using the geometry in Figure 5,
we see that s.t/ is increasing for 0 < t < 4 and for 8 < t < 10 and is decreasing for 4 < t < 8 and for 10 < t < 12. Furthermore,
s.0/ D 0 m;
s.4/ D 4 m;
s.8/ D "4 m;
s.10/ D "2 m;
and
s.12/ D "6 m:
(a) In the first 12 seconds, the particle returns to the origin once, sometime between t D 4 and t D 8 seconds.
(b) The particle’s maximum distance from the origin is 6 meters (to the left at t D 12 seconds).
(c) The particle’s distance to the left of the origin is 6 meters.
Chapter Review Exercises
679
75. On a typical day, a city consumes water at the rate of r.t/ D 100 C 72t " 3t 2 (in thousands of gallons per hour), where
t is the number of hours past midnight. What is the daily water consumption? How much water is consumed between 6 PM and
midnight?
SOLUTION
With a consumption rate of r.t/ D 100 C 72t " 3t 2 thousand gallons per hour, the daily consumption of water is
Z
24
0
ˇ24
ˇ
%
2
3 &ˇ
.100 C 72t " 3t / dt D 100t C 36t " t ˇ D 100.24/ C 36.24/2 " .24/3 D 9312;
2
0
or 9.312 million gallons. From 6 PM to midnight, the water consumption is
Z 24
#
$ˇ24
ˇ
.100 C 72t " 3t 2 / dt D 100t C 36t 2 " t 3 ˇ
18
18
%
&
D 100.24/ C 36.24/ " .24/3 " 100.18/ C 36.18/2 " .18/3
2
D 9312 " 7632 D 1680;
or 1.68 million gallons.
76. The learning curve in a certain bicycle factory is L.x/ D 12x !1=5 (in hours per bicycle), which means that it takes a bike
mechanic L.n/ hours to assemble the nth bicycle. If a mechanic has produced 24 bicycles, how long does it take her or him to
produce the second batch of 12?
SOLUTION
The second batch of 12 bicycles consists of bicycles 13 through 24. The time it takes to produce these bicycles is
Z
24
13
ˇ24
ˇ
%
&
12x !1=5 dx D 15x 4=5 ˇˇ D 15 244=5 " 134=5 # 73:91 hours:
13
77. Cost engineers at NASA have the task of projecting the cost P of major space projects. It has been found that the cost C of
developing a projection increases with P at the rate dC =dP # 21P !0:65 , where C is in thousands of dollars and P in millions of
dollars. What is the cost of developing a projection for a project whose cost turns out to be P D $35 million?
SOLUTION Assuming it costs nothing to develop a projection for a project with a cost of $0, the cost of developing a projection
for a project whose cost turns out to be $35 million is
ˇ35
Z 35
ˇ
21P !0:65 dP D 60P 0:35 ˇˇ D 60.35/0:35 # 208:245;
0
0
or $208,245.
78. An astronomer estimates that in a certain constellation, the number of stars per magnitude m, per degree-squared of sky, is
equal to A.m/ D 2:4 % 10!6 m7:4 (fainter stars have higher magnitudes). Determine the total number of stars of magnitude between
6 and 15 in a one-degree-squared region of sky.
SOLUTION
The total number of stars of magnitude between 6 and 15 in a one-degree-squared region of sky is
Z
15
A.m/ d m D
6
79. Evaluate
SOLUTION
Z
8
x 15 dx
!8
3 C cos2 x
Let f .x/ D
Z
15
6
2:4 % 10!6 m7:4 d m
ˇ15
ˇ
2
!6 8:4 ˇ
D % 10 m ˇ
7
6
# 2162
, using the properties of odd functions.
x 15
3Ccos2 x
and note that
f ."x/ D
."x/15
x 15
D " 2 D "f .x/:
2
3 C cos ."x/
cos x
Because f .x/ is an odd function and the interval "8 ! x ! 8 is symmetric about x D 0, it follows that
Z 8
x 15 dx
D 0:
2
!8 3 C cos x
R1
80. Evaluate 0 f .x/ dx, assuming that f .x/ is an even continuous function such that
Z
2
1
f .x/ dx D 5;
Z
1
!2
f .x/ dx D 8
680
CHAPTER 5
SOLUTION
THE INTEGRAL
Using the given information
Z
2
!2
f .x/ dx D
Z
1
!2
f .x/ dx C
Z
2
1
f .x/ dx D 13:
Because f .x/ is an even function, it follows that
Z
0
!2
so
f .x/ dx D
Z
Z
2
f .x/ dx;
0
2
0
f .x/ dx D
13
:
2
Finally,
Z
1
0
f .x/ dx D
Z
2
0
f .x/ dx "
Z
2
1
f .x/ dx D
13
3
"5D :
2
2
81.
Plot the graph of f .x/ D sin mx sin nx on Œ0; #! for the pairs .m; n/ D .2; 4/, .3; 5/ and in each case guess the value
R!
of I D 0 f .x/ dx. Experiment with a few more values (including two cases with m D n) and formulate a conjecture for when I
is zero.
SOLUTION The graphs of f .x/ D sin mx sin nx with .m; n/ D .2; 4/ and .m; n/ D .3; 5/ are shown below. It appears as if the
positive areas balance the negative areas, so we expect that
Z !
I D
f .x/ dx D 0
0
in these cases.
y
y
(3, 5)
(2, 4)
0.5
0.5
x
x
0.5
1
1.5
2
2.5
0.5
3
1
1.5
2
2.5
3
−0.5
−0.5
We arrive at the same conclusion for the cases .m; n/ D .4; 1/ and .m; n/ D .5; 2/.
y
y
(5, 2)
(4, 1)
0.5
0.5
x
x
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
−0.5
−0.5
However, when .m; n/ D .3; 3/ and when .m; n/ D .5; 5/, the value of
Z !
I D
f .x/ dx
0
is clearly not zero as there is no negative area.
y
y
0.5
0.5
x
x
0.5
1
1.5
2
2.5
3
−0.5
−0.5
(3, 3)
We therefore conjecture that I is zero whenever m ¤ n.
(5, 5)
Chapter Review Exercises
82. Show that
Z
x f .x/ dx D xF .x/ " G.x/
Z
where F 0 .x/ D f .x/ and G 0 .x/ D F .x/. Use this to evaluate
x cos x dx.
Suppose F 0 .x/ D f .x/ and G 0 .x/ D F .x/. Then
SOLUTION
d
.xF .x/ " G.x// D xF 0 .x/ C F .x/ " G 0 .x/ D xf .x/ C F .x/ " F .x/ D xf .x/:
dx
Therefore, xF .x/ " G.x/ is an antiderivative of xf .x/ and
Z
xf .x/ dx D xF .x/ " G.x/ C C:
To evaluate
83. Prove
R
x cos x dx, note that f .x/ D cos x. Thus, we may take F .x/ D sin x and G.x/ D " cos x. Finally,
Z
x cos x dx D x sin x C cos x C C:
Z
2!
SOLUTION
2
2x dx ! 4
1
Z
1
!
9
and
2
1
3!x dx !
1
3
The function f .x/ D 2x is increasing, so 1 ! x ! 2 implies that 2 D 21 ! 2x ! 22 D 4. Consequently,
Z
2D
2
1
2 dx !
Z
2
1
2x dx !
Z
2
4 dx D 4:
1
On the other hand, the function f .x/ D 3!x is decreasing, so 1 ! x ! 2 implies that
1
1
D 3!2 ! 3!x ! 3!1 D :
9
3
It then follows that
Z
1
D
9
84.
SOLUTION
2
1
1
dx !
9
Z
2
1
3!x dx !
Plot the graph of f .x/ D x !2 sin x, and show that 0:2 !
Z
Z
2
1
1
1
dx D :
3
3
2
1
f .x/ dx ! 0:9.
Let f .x/ D x !2 sin x. From the figure below, we see that
0:2 ! f .x/ ! 0:9
for 1 ! x ! 2. Therefore,
0:2 D
Z
1
2
0:2 dx !
Z
1
2
f .x/ dx !
Z
2
1
0:9 dx D 0:9:
y
1
0.8
0.6
x−2sin x
0.4
0.2
x
85. Find upper and lower bounds for
Z
0.5
1
1.5
1
0
f .x/ dx, for f .x/ in Figure 6.
2
681
682
THE INTEGRAL
CHAPTER 5
y
f (x)
y = x2 + 1
y = x1/2 + 1
2
1
x
1
FIGURE 6
From the figure, we see that the inequalities x 2 C 1 ! f .x/ !
SOLUTION
Z
1
0
.x 2 C 1/ dx D
!
and
Z
1
0
p
. x C 1/ dx D
!
it follows that
4
!
3
Z
"ˇ1
ˇ
1 3
4
x C x ˇˇ D
3
3
0
"ˇ1
ˇ
2 3=2
5
x
C x ˇˇ D ;
3
3
0
1
0
p
x C 1 hold for 0 ! x ! 1. Because
f .x/ dx !
5
:
3
In Exercises 86–91, find the derivative.
Z x
86. A0 .x/, where A.x/ D
sin.t 3 / dt
Let A.x/ D
SOLUTION
Z
Let A.x/ D
88.
d
dy
Z
sin.t 3 / dt. Then A0 .x/ D sin.x 3 /.
3
87. A0 .#/, where A.x/ D
SOLUTION
3
x
Z
Z
x
2
x
2
cos t
dt
1Ct
cos t
cos x
dt. Then A0 .x/ D
and
1Ct
1Cx
A0 .#/ D
y
!2
SOLUTION
3x dx
d
dy
Z
y
!2
3x dx D 3y :
89. G 0 .x/, where G.x/ D
SOLUTION
cos #
1
D"
:
1C#
1C#
Let G.x/ D
Z
Z
sin x
t 3 dt
!2
sin x
t 3 dt. Then
!2
G 0 .x/ D sin3 x
90.
G 0 .2/,
SOLUTION
where G.x/ D
Let G.x/ D
Z
Z
x3
p
0
x3
p
0
t C 1 dt
t C 1 dt. Then
G 0 .x/ D
p
and G 0 .2/ D 3.2/2 8 C 1 D 36.
d
sin x D sin3 x cos x:
dx
p
x3 C 1
p
d 3
x D 3x 2 x 3 C 1
dx
Chapter Review Exercises
91. H 0 .1/, where H.x/ D
SOLUTION
Let H.x/ D
Z
Z
9
4x 2
9
4x 2
683
1
dt
t
1
dt D "
t
Z
4x 2
9
1
dt . Then
t
H 0 .x/ D "
1 d
8x
2
4x 2 D " 2 D "
2
x
4x dx
4x
and H 0 .1/ D "2.
92.
Explain with a graph: If f .x/ is increasing and concave up on Œa; b!, then LN is more accurate than RN . Which is
more accurate if f .x/ is increasing and concave down?
SOLUTION
Consider the figure below, which displays a portion of the graph of an increasing, concave up function.
y
x
The shaded rectangles represent the differences between the right-endpoint approximation RN and the left-endpoint approximation
LN . In particular, the portion of each rectangle that lies below the graph of y D f .x/ is the amount by which LN underestimates
the area under the graph, whereas the portion of each rectangle that lies above the graph of y D f .x/ is the amount by which RN
overestimates the area. Because the graph of y D f .x/ is increasing and concave up, the lower portion of each shaded rectangle is
smaller than the upper portion. Therefore, LN is more accurate (introduces less error) than RN . By similar reasoning, if f .x/ is
increasing and concave down, then RN is more accurate than LN .
Z b
1
93.
Explain with a graph: If f .x/ is linear on Œa; b!, then the
f .x/ dx D .RN C LN / for all N .
2
a
SOLUTION
Consider the figure below, which displays a portion of the graph of a linear function.
y
x
The shaded rectangles represent the differences between the right-endpoint approximation RN and the left-endpoint approximation
LN . In particular, the portion of each rectangle that lies below the graph of y D f .x/ is the amount by which LN underestimates
the area under the graph, whereas the portion of each rectangle that lies above the graph of y D f .x/ is the amount by which RN
overestimates the area. Because the graph of y D f .x/ is a line, the lower portion of each shaded rectangle is exactly the same size
as the upper portion. Therefore, if we average LN and RN , the error in the two approximations will exactly cancel, leaving
Z b
1
.RN C LN / D
f .x/ dx:
2
a
94. In this exercise, we prove
x"
(a) Show that ln.1 C x/ D
Z
x
0
x2
! ln.1 C x/ ! x
2
.for x > 0/
dt
for x > 0.
1Ct
1
! 1 for all t > 0.
1Ct
(c) Use (b) to prove Eq. (1).
(d) Verify Eq. (1) for x D 0:5, 0.1, and 0.01.
(b) Verify that 1 " t !
SOLUTION
(a) Let x > 0. Then
Z
x
0
ˇx
ˇ
dt
D ln.1 C t/ˇˇ D ln.1 C x/ " ln 1 D ln.1 C x/:
1Ct
0
1
684
CHAPTER 5
THE INTEGRAL
(b) For t > 0, 1 C t > 1, so
Hence,
1
1Ct
< 1. Moreover, .1 " t/.1 C t/ D 1 " t 2 < 1. Because 1 C t > 0, it follows that 1 " t <
1"t !
1
1Ct .
1
! 1:
1Ct
(c) Integrating each expression in the result from part (b) from t D 0 to t D x yields
x"
x2
! ln.1 C x/ ! x:
2
(d) For x D 0:5, x D 0:1 and x D 0:01, we obtain the string of inequalities
0:375 ! 0:405465 ! 0:5
0:095 ! 0:095310 ! 0:1
0:00995 ! 0:00995033 ! 0:01;
respectively.
95. Let
Z
p
F .x/ D x x 2 " 1 " 2
p
x
1
t 2 " 1 dt
Prove that F .x/ and cosh!1 x differ by a constant by showing that they have the same derivative. Then prove they are equal by
evaluating both at x D 1.
SOLUTION
Let
Z
p
F .x/ D x x 2 " 1 " 2
p
x
1
Then
t 2 " 1 dt:
p
p
p
dF
x2
x2
1
D x2 " 1 C p
" 2 x2 " 1 D p
" x2 " 1 D p
:
dx
x2 " 1
x2 " 1
x2 " 1
d
Also, dx
.cosh!1 x/ D
by a constant:
p 1
; therefore,
x 2 !1
F .x/ and cosh!1 x have the same derivative. We conclude that F .x/ and cosh!1 x differ
F .x/ D cosh!1 x C C:
Now, let x D 1. Because F .1/ D 0 and cosh!1 1 D 0, it follows that C D 0. Therefore,
F .x/ D cosh!1 x:
96.
Let f .x/ be a positive increasing continuous function on Œa; b!, where 0 ! a < b as in Figure 7. Show that the shaded
region has area
Z
I D bf .b/ " af .a/ "
b
f .x/ dx
2
a
y
f (b)
y = f(x)
f (a)
a
b
x
FIGURE 7
SOLUTION We can construct the shaded region in Figure 7 by taking a rectangle of length b and height f .b/ and removing a
rectangle of length a and height f .a/ as well as the region between the graph of y D f .x/ and the x-axis over the interval Œa; b!.
The area of the resulting region is then the area of the large rectangle minus the area of the small rectangle and minus the area under
the curve y D f .x/; that is,
I D bf .b/ " af .a/ "
Z
b
a
f .x/ dx:
Chapter Review Exercises
97.
685
How can we interpret the quantity I in Eq. (2) if a < b ! 0? Explain with a graph.
We will consider each term on the right-hand side of (2) separately. For convenience, let I, II, III and IV denote the
area of the similarly labeled region in the diagram below.
SOLUTION
y
f (b)
II
I
III
f (a)
x
IV
a
b
Because b < 0, the expression bf .b/ is the opposite of the area of the rectangle along the right; that is,
bf .b/ D "II " IV:
Similarly,
"af .a/ D III C IV
and
"
Z
b
a
f .x/ dx D "I " III:
Therefore,
bf .b/ " af .a/ "
Z
b
a
f .x/ dx D "I " III
that is, the opposite of the area of the shaded region shown below.
y
f (b)
f (a)
x
a
b
98. The isotope thorium-234 has a half-life of 24:5 days.
(a) What is the differential equation satisfied by y.t/, the amount of thorium-234 in a sample at time t?
(b) At t D 0, a sample contains 2 kg of thorium-234. How much remains after 40 days?
SOLUTION
(a) By the equation for half-life,
24:5 D
ln 2
;
k
so k D
ln 2
# 0:028 days!1 :
24:5
Therefore, the differential equation for y.t/ is
y 0 D "0:028y:
(b) If there are 2 kg of thorium-234 at t D 0, then y.t/ D 2e !0:028t . After 40 days, the amount of thorium-234 is
y.40/ D 2e !0:028.40/ D 0:653 kg:
99. The Oldest Snack Food? In Bat Cave, New Mexico, archaeologists found ancient human remains, including cobs of popping corn whose C14 -to-C12 ratio was approximately 48% of that found in living matter. Estimate the age of the corn cobs.
SOLUTION
Let t be the age of the corn cobs. The C 14 to C 12 ratio decreased by a factor of e !0:000121t which is equal to 0:48.
That is:
e !0:000121t D 0:48;
so
"0:000121t D ln 0:48;
and
t D"
1
ln 0:48 # 6065:9:
0:000121
We conclude that the age of the corn cobs is approximately 6065:9 years.
686
CHAPTER 5
THE INTEGRAL
100. The C14 -to-C12 ratio of a sample is proportional to the disintegration rate (number of beta particles emitted per minute) that
is measured directly with a Geiger counter. The disintegration rate of carbon in a living organism is 15:3 beta particles per minute
per gram. Find the age of a sample that emits 9.5 beta particles per minute per gram.
SOLUTION Let t be the age of the sample in years. Because the disintegration rate for the sample has dropped from 15:3 beta
particles=min per gram to 9.5 beta particles=min per gram and the C 14 to C 12 ratio is proportional to the disintegration rate, it
follows that
e !0:000121t D
9:5
;
15:3
so
t D"
1
9:5
ln
# 3938:5:
0:000121 15:3
We conclude that the sample is approximately 3938.5 years old.
101. What is the interest rate if the PV of $50,000 to be delivered in 3 years is $43,000?
Let r denote the interest rate. The present value of $50,000 received in 3 years with an interest rate of r is 50;000e !3r .
Thus, we need to solve
SOLUTION
43;000 D 50;000e !3r
for r. This yields
1 43
r D " ln
D 0:0503:
3 50
Thus, the interest rate is 5.03%.
102. An equipment upgrade costing $1 million will save a company $320,000 per year for 4 years. Is this a good investment if the
interest rate is r D 5%? What is the largest interest rate that would make the investment worthwhile? Assume that the savings are
received as a lump sum at the end of each year.
SOLUTION
With an interest rate of r D 5%, the present value of the four payments is
%
&
$320;000 e !0:05 C e !0:1 C e !0:15 C e !0:2 D $1;131;361:78:
As this is greater than the $1 million cost of the upgrade, this is a good investment. To determine the largest interest rate that would
make the investment worthwhile, we must solve the equation
%
&
320;000 e !r C e !2r C e !3r C e !4r D 1;000;000
for r. Using a computer algebra system, we find r D 10:13%.
103. Find the PV of an income stream paying out continuously at a rate of 5000e !0:1t dollars per year for 5 years, assuming an
interest rate of r D 4%.
ˇ
Z 5
Z 5
5000 !0:14t ˇˇ5
!0:1t !0:04t
!0:14t
SOLUTION P V D
5000e
e
dt D
5000e
dt D
e
ˇ D $17; 979:10.
"0:14
0
0
0
104. Calculate the limit:
!
"
4 n
(a) lim 1 C
n!1
n
(b) lim
n!1
!
1C
1
n
SOLUTION
"!
!
"
" #4
4 n
1 n=4
(a) lim 1 C
D lim
1C
D e4.
n!1
n!1
n
n=4
!
"
'!
" (4
1 4n
1 n
(b) lim 1 C
D lim
1C
D e4.
n!1
n!1
n
n
"!
!
"
" #12
4 3n
1 n=4
(c) lim 1 C
D lim
1C
D e 12 .
n!1
n!1
n
n=4
"4n
(c) lim
n!1
!
1C
4
n
"3n