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’2 1 2 64
(’1)
= +3’ ’ ’3’ = .
3 5 3 5 15
21/4
21/4 2x 5
Area = [’x + 2] ’ [x ’ 2] dx = 4x ’
4 4
(e)
5
’21/4 ’21/4

2(21/4 )5 2(’21/4 )5
= 4·2 ’ ’ 4 · (’2 )’
1/4 1/4
5 5
32 · 21/4
= .
5
1
1 x3
Area = [’x 2 + 3] ’ 2x dx = ’ + 3x ’ x 2
(f)
3
’3 ’3
’27
1
= ’ + 3 · 1 ’ 12 ’ ’ + 3 · (’3) ’ (’3)2
3 3
32
= .
3


Chapter 5
limx’0 (cos x ’ 1) = 0 and limx’0 x 2 ’ x 3 = 0 so l™Hôpital™s
1. (a)
Rule applies. Thus

cos x ’ 1 ’ sin x
= lim
lim .
x’0 x 2 ’ x 3 x’0 2x ’ 3x 2
294 Solutions to Exercises

Now l™Hôpital™s Rule applies again to yield
’ cos x 1
= lim =’ .
x’0 2 ’ 6x 2
limx’0 e2x ’ 1 ’ 2x = 0 and limx’0 x 2 + x 4 = 0 so l™Hôpital™s
(b)
Rule applies. Thus
e2x ’ 1 ’ 2x 2e2x ’ 2
= lim
lim .
x2 + x4 x’0 2x + 4x 3
x’0
l™Hôpital™s Rule applies again to yield
4e2x
= lim = 2.
x’0 2 + 12x 2
limx’0 cos x = 0, so l™Hôpital™s Rule does not apply. In fact the
(c)
limit does not exist.
limx’1 [ln x]2 = 0 and limx’1 (x ’ 1) = 0 so l™Hôpital™s Rule
(d)
applies. Thus
[ln x]2 [2 ln x]/x
= lim = 0.
lim
x’1 (x ’ 1) 1
x’1

limx’2 (x ’ 2)3 = 0 and limx’2 sin(x ’ 2) ’ (x ’ 2) = 0 so
(e)
l™Hôpital™s Rule applies. Thus
(x ’ 2)3 3(x ’ 2)2
= lim
lim .
x’2 sin(x ’ 2) ’ (x ’ 2) x’2 cos(x ’ 2) ’ 1

Now l™Hôpital™s Rule applies again to yield
6(x ’ 2)
= lim .
x’2 ’ sin(x ’ 2)

We apply l™Hôpital™s Rule one last time to obtain
6
= lim = ’6.
x’2 ’ cos(x ’ 2)

limx’1 (ex ’ 1) = 0 and limx’1 (x ’ 1) = 0 so l™Hôpital™s Rule
(f)
applies. Thus
ex ’ 1 ex
= lim = e.
lim
x’1 x ’ 1 x’1 1

limx’+∞ x 3 = limx’+∞ (ex ’ x 2 ) = +∞ so l™Hôpital™s Rule
2. (a)
applies. Thus
x3 3x 2
= lim x
lim .
x’+∞ e ’ 2x
x’+∞ ex ’ x 2
295
Chapter 5

l™Hôpital™s Rule applies again to yield
6x
= lim .
x’+∞ ex ’ 2

l™Hôpital™s Rule applies one more time to ¬nally yield
6
= 0.
lim
x’+∞ ex

limx’+∞ ln x = limx’+∞ x = +∞ so l™Hôpital™s Rule applies.
(b)
Thus
ln x 1/x
= lim = 0.
lim
x’+∞ 1
x’+∞ x

limx’+∞ e’x = limx’+∞ ln[x/(x + 1)] = 0 so l™Hôpital™s Rule
(c)
applies. Thus
e’x ’e’x
= lim
lim .
x’+∞ ln[x/(x + 1)] x’+∞ 1/x ’ 1/[x + 1]

It is convenient to rewrite this expression as
x2 + x
lim .
x’+∞ ’ex

Now l™Hôpital™s Rule applies once more to yield
2x + 1
lim .
x’+∞ ’ex

We apply l™Hôpital™s Rule a last time to obtain
2
= lim = 0.
x’+∞ ’ex

(d) Since limx’+∞ sin x does not exist, l™Hôpital™s Rule does not apply.
In fact the requested limit does not exist.
(e) It is convenient to rewrite this limit as
x
lim ’x .
x’’∞ e

Since limx’’∞ x = limx’’∞ e’x = ±∞, l™Hôpital™s Rule
applies. Thus
1
x
= lim = 0.
lim
x’’∞ e’x x’’∞ ’e’x
296 Solutions to Exercises

Since limx’’∞ ln |x| = limx’’∞ e’x = +∞, l™Hôpital™s Rule
(f)
applies. Thus
ln |x| 1/x
= lim = 0.
lim
x’’∞ e’x x’’∞ ’e’x

x3
We write the limit as limx’+∞ x . Since limx’+∞ x 3 = limx’+∞
3. (a)
e
ex = +∞, l™Hôpital™s Rule applies. Thus
x3 3x 2
3 ’x
= lim x = lim
lim x e .
x’+∞ ex
x’+∞ e
x’+∞

We apply l™Hôpital™s Rule again to obtain
6x
= lim .
x’+∞ ex

Applying l™Hôpital™s Rule one last time yields
6
= lim = 0.
x’+∞ ex

sin(1/x)
(b) We write the limit as limx’+∞ . Since limx’+∞ sin(1/x)
1/x
= limx’+∞ 1/x = 0, l™Hôpital™s Rule applies. Hence
sin(1/x)
lim x · sin[1/x] = lim
1/x
x’+∞ x’+∞

[cos(1/x)] · [’1/x 2 ]
= lim
’1/x 2
x’+∞
cos(1/x)
= lim = 1.
1
x’+∞

ln[x/(x + 1)]
(c) We rewrite the limit as limx’+∞ . Since limx’+∞
1/(x + 1)
ln[x/(x + 1)] = limx’+∞ 1/(x + 1) = 0, l™Hôpital™s Rule applies.
Thus
ln[x/(x + 1)]
lim ln[x/(x + 1)] · (x + 1) = lim
x’+∞ 1/(x + 1)
x’+∞

[(x + 1)/x] · [1/(x + 1)2 ]
= lim
’1/(x + 1)2
x’+∞
’(x + 1)
= lim .
x
x’+∞
297
Chapter 5

Now l™Hôpital™s Rule applies again and we obtain
’1
= lim
= ’1.
x’+∞ 1

[ln x]
We rewrite the limit as limx’+∞ x . Since limx’+∞ ln x =
(d)
e
limx’+∞ ex = +∞, l™Hôpital™s Rule applies. Thus
ln x 1/x
lim ln x · e’x = lim = lim = 0.
x’+∞ ex x’+∞ ex
x’+∞

x2
. Since limx’’∞ lim x 2 =
(e) We write the limit as limx’’∞
e’2x
limx’’∞ e’2x = 0, l™Hôpital™s Rule applies. Thus
x2
2x
· x = lim = lim
2x 2
lim e .
x’’∞ e’2x x’’∞ ’2e’2x
x’’∞

l™Hôpital™s Rule applies one more time to yield
2
= lim = 0.
4e’2x
x’’∞

e1/x
. Since limx’0 e1/x =
(f) We rewrite the limit as limx’0
[1/x]
limx’0 1/x = +∞, l™Hôpital™s Rule applies. Thus
e1/x · [’1/x 2 ]
e1/x e1/x
= lim = lim = lim = +∞ .
1/x
lim x·e
’1/x 2
x’0 1/x x’0 1
x’0 x’0

4. We do (a), (b), (c), (d).
1
1 1 x 1/4
’3/4 ’3/4
dx = lim dx = lim
(a) x x
’0+ ’0+ 1/4
0

11/4 1/4
= lim ’ = 4.
’0+ 1/4 1/4
3 3’
’4/3
(x ’ 3)’4/3 dx
(x ’ 3) dx = lim
(b)
’0+ 1
1
3’
(x ’ 3)’1/3 ’ ’1/3 ’2’1/3
= lim = lim ’ . But
’1/3 ’1/3 ’1/3
’0+ ’0+
1
the limit does not exist; so the integral does not converge.
’1’
2 1 1
dx = lim
(c) dx
(x + 1)1/3 (x + 1)1/3
’0+
’2 ’2
298 Solutions to Exercises

2 1
+ lim dx
(x + 1)1/3
’0+ ’1+
’1’ 2
(x + 1)2/3 (x + 1)2/3
= lim + lim
’0+ ’0+
2/3 2/3
’2 ’1+

(’ )2/3 (’1)2/3 32/3 ( )2/3
= lim ’ + lim ’
’0+ ’0+
2/3 2/3 2/3 2/3
3
= · 32/3 ’ 1 .
2
’2’
6 x x
dx = lim
(d) dx
(x ’ 1)(x + 2) (x ’ 1)(x + 2)
’0+
’4 ’4
0 1’
x x
+ lim dx + lim dx
(x ’1)(x +2) (x ’1)(x +2)
’0+ ’2+ ’0+ 0




Y
6 x
+ lim dx. Now
FL
(x ’ 1)(x + 2)
’0+ 1+

1/3 2/3
x
= + .
AM

(x ’ 1)(x + 2) x’1 x+2
Therefore
6 x
TE



dx
’4 (x ’1)(x +2)
’2’ 1/3 0
2/3 1/3 2/3
= lim + dx + lim + dx
x ’1 x +2 x ’1 x +2
’0+ ’4 ’0+ ’2+
1’ 6
1/3 2/3 1/3 2/3
+ lim + dx + lim + dx
x ’1 x +2 x ’1 x +2
’0+ 0 ’0+ 1+
’2’
1 2
= lim ln |x ’1|+ ln |x +2|
’0+ 3 3 ’4
0
1 2
+ lim ln |x ’1|+ ln |x +2|
’0+ 3 3 ’2+
1’
1 2
+ lim ln |x ’1|+ ln |x +2|
’0+ 3 3 0
6
1 2
+ lim ln |x ’1|+ ln |x +2| .
’0+ 3 3 1+
299
Chapter 5

Now this equals

1 2 1 2
· ln | ’ 3 ’ | + ln ’ · ln 5 + ln 2 + etc.
lim
’0+ 3 3 3 3

The second limit does not exist, so the original integral does not
converge.
5. We do (a), (b), (c), (d).
N
∞ e’3x
’3x ’3x
dx = dx =
(a) lim e lim
e
’3
N ’+∞ N ’+∞
1 1
e’3N e’3 e’3
= ’ =
lim .
’3 ’3
N ’+∞ 3
∞ N
2 ’x
x 2 e’x dx
dx =
(b) lim
xe
N ’+∞ 2
2

’e’x x 2 ’ 2xe’x ’ 2e’x
N
= lim 2
N ’+∞

’e’N N 2 ’ 2N e’N ’ 2e’N
= lim
N ’+∞

’ ’e’2 22 ’ 2 · 2 · e’2 ’ 2e’2
= e’2 22 + 4e’2 + 2e’2 .
∞ 1 N
x ln x dx = lim x ln x dx + lim
(c) x ln x dx
’0+ N’+∞ 1
0
= lim [x ln x ’ x]1 + lim [x ln x ’ x]N
1
’+ N’+∞
= lim [(1 · ln 1 ’ 1) ’ ( · ln ’ )]
’0+
+ lim [(N · ln N ’ N ) ’ (1 ln 1 ’ 1)]
N ’+∞
= lim [’1 + ] + lim [N ln N ’ N + 1]
’0+ N’+∞
= lim [N ln N ’ N]. This last limit diverges, so
N ’+∞
the integral diverges.
∞ N
dx dx
= lim = lim [arctan x]N
(d)
1 + x 2 N ’+∞ 1 1+x
1
2 N’+∞
1
π π π
= lim (arctan N ’ arctan 1) = ’ = .
N ’+∞ 2 4 4
300 Solutions to Exercises

Chapter 6
2 ln a ’ 3 ln b ’ 4 ln c ’ ln d
1. (a)
ln 2
(b)
ln 3
(c) 3x + 4 ln z ’ 3 ln w
1
(d) 2w +
2
2. We do (a) and (b).
(a)
3x · 5’x = 2x · e3
x ln 3 ’ x ln 5 = x ln 2 + 3
x · [ln 3 ’ ln 5 ’ ln 2] = 3
3
x= .
ln 3 ’ ln 5 ’ ln 2
(b)
3x
= 10x · 102
’x · 42x
5
x log10 3 + x log10 5 ’ 2x log10 4 = x + 2
x[log10 3 + log10 5 ’ 2 log10 4 ’ 1] = 2
2
x= .
log10 3 + log10 5 ’ 2 log10 4 ’ 1
2x · cos(x 2 )
3. (a)
sin(x 2 )
2 1

(b)
x’1
x
sin(ex )
· cos(ex ) · ex
(c) e
1
cos(ln x) ·
(d)
x
’e x ’ e’x · 3x 2 ’ e’x · 6x ’ 6e’x + C
’x 3
4. (a)
x3 2 2 x3
23
ln x ’ x ln x + +C
(b)
3 9 39
e
ln2 x 1
=
(c)
2 2
1
2
ln(e + 1) = ln(e2 + 1) ’ ln(e + 1)
x
(d) 1
301
Chapter 6

5. We do (a) and (b).
x2 + 1
Let A = ·3
x3
(a) . Then
x ’x
ln A = 3 ln x + ln(x 2 + 1) ’ ln(x 3 ’ x)
hence
3x 2 ’ 1
3 2x
dA/dx d
= ln A = + 2 ’3 .
x +1 x ’x
A dx x
Multiplying through by A gives

3 x +1 3x 2 ’ 1
2 3 2x
dA
= x·3 · +2 ’3 .
x ’x x +1 x ’x
dx x
sin x · (x 3 + x)
Let A =
(b) . Then
x 2 (x + 1)
ln A = ln sin x + ln(x 3 + x) ’ ln x 2 ’ ln(x + 1)
hence
3x 2 + 1 2x
cos x 1
dA/dx d
= ln A = +3 ’ 2’ .
x+1
x +x
sin x
A dx x
Multiplying through by A gives
sin x · (x 3 + x) 3x 2 + 1 2x
cos x 1
dA
= · +3 ’ 2’ .
x+1
x 2 (x + 1) x +x
sin x
dx x
6. Let R(t) denote the amount of substance present at time t. Let noon on
January 10 correspond to t = 0 and noon on February 10 correspond to
t = 1. Then R(0) = 5 and R(1) = 3. We know that
R(t) = P · eKt .
Since
5 = R(0) = P · eK·0 ,
we see that P = 5. Since
3 = R(1) = 5 · eK·1 ,
we ¬nd that K = ln 3/5. Thus
t
3
R(t) = 5 · e =5·
t ln(3/5)
.
5
302 Solutions to Exercises

Taking March 10 to be about t = 2, we ¬nd that the amount of radioactive
material present on March 10 is
2
3 9
R(2) = 5 · =.
5 5
7. Let the amount of bacteria present at time t be
B(t) = P · eKt .
Let t = 0 be 10:00 a.m. We know that B(0) = 10000 and B(3) = 15000.
Thus
10000 = B(0) = P · eK·0
so P = 10000. Also
15000 = B(3) = 10000 · eK·3
hence
1
K= · ln(3/2).
3
As a result,
B(t) = 10000 · et·[1/3] ln(3/2)
or
t/3
3
B(t) = 10000 · .
2
We ¬nd that, at 2:00 p.m., the number of bacteria is
4/3
3
B(4) = 10000 · .
2
8. If M(t) is the amount of money in the account at time t then we know that
M(t) = 1000 · e6t/100 .
Here t = 0 corresponds to January 1, 2005. Then, on January 1, 2009, the
amount of money present is
M(4) = 1000 · e6·4/100 ≈ 1271.25.
1
· 1 · ex + x · ex
9. (a)
1 ’ (x · ex )2
1 1
·
(b)
1 + (x/[x + 1])2 (x + 1)2
303
Chapter 7

2x + 1
1
·2
(c)
1 + [ln(x 2 + x)]2 x + x
1
· sec2 x
(d)
| tan x| [tan x]2 ’ 1
Tan’1 x 2 + C
10. (a)
Sin’1 x 3 + C
(b)
π
π/2
’1
= Sin’1 1 ’ Sin’1 0 = .
2
(c) Sin (sin x)
2
0

1 1 2x
dx
= √ · Tan’1 √ +C

(d)
1 + [ 2/5x]2
5 5
10

Chapter 7
1. We do (a), (b), (c), (d).
Let u = log2 x and dv = 1 dx. Then
(a)
1
log2 x dx = log2 x · x ’ x · 2 log x · dx
x
= x log2 x ’ 2 log x dx.

Now let u = log x and dv = 1 dx. Then
1
log2 x dx = x log2 x ’ 2 log x · x ’ x· dx
x
= x log2 x ’ 2x log x + 2x + C.
Let u = x and dv = e3x dx. Then
(b)
e3x e3x
x · e x dx = x · ’ · 1 dx
3
3 3
e3x e3x
=x· ’ + C.
3 9
Let u = x 2 and dv = cos x dx. Then
(c)

x 2 cos x dx = x 2 · sin x ’ sin x · 2x dx.

Now let u = 2x and dv = sin x dx. Then

x 2 cos x dx = x 2 · sin x ’ 2x · (’ cos x) ’ (’ cos x) · 2 dx

= x 2 sin x + 2x cos x ’ 2 sin x + C.
304 Solutions to Exercises

Notice that t sin 3t cos 3t dt = 1 t sin 6t dt. Now let u = t and
(d) 2
dv = sin 6t dt. Then
1 1 1 1
t sin 6t dt = t · ’ cos 6t ’ ’ cos 6t · 1 dt
2 2 6 6
1
t
= ’ cos 6t + sin 6t + C.
12 72
2. We do (a), (b), (c), (d).
’1/7
1 1/7
= +
(a) hence
(x + 2)(x ’ 5) x+2 x’5
’1/7 1/7
dx
= +
(x + 2)(x ’ 5) x+2 x’5
’1 1
= ln |x + 2| + ln |x ’ 5| + C.
7 7
’x/2 + 1/2
1 1/2
= +
(b) hence
x+1
(x + 1)(x 2 + 1) x2 + 1
’x/2
1/2 1/2
dx
= dx + dx + dx
x +1
(x +1)(x 2 +1) x 2 +1 x 2 +1
1 1 1
= ln |x +1|’ ln |x 2 +1|+ Tan’1 x +C.
2 4 2
Now x 3 ’ 2x 2 ’ 5x + 6 = (x ’ 3)(x + 2)(x ’ 1). Then
(c)
’1/6
1 1/10 1/15
= + + .
x’3 x+2 x’1
x 3 ’ 2x 2 ’ 5x + 6
As a result,
’1/6
1/10 1/15
dx
= dx + dx + dx
x ’3 x +2 x ’1
x 3 ’2x 2 ’5x +6
1 1 1
= ln |x ’3|+ ln |x +2|’ ln |x ’1|+C.
10 15 6
Now x 4 ’ 1 = (x 2 ’ 1)(x 2 + 1) = (x ’ 1)(x + 1)(x 2 + 1).
(d)
Hence
’x/2
1/4 1/4
x
= + +2 .
x’1 x+1 x +1
x4 ’ 1
We conclude that
1 1 1
x dx
= ln |x ’ 1| + ln |x + 1| ’ ln |x 2 + 1| + C.
x4 + 1 4 4 4
305
Chapter 7

3. We do (a), (b), (c), (d).
Let u = sin x, du = cos x dx. Then the integral becomes
(a)
(1 + u2 )3
(1 + u ) 2u du = + C.
22
3
Resubstituting x, we obtain the ¬nal answer
(1 + sin2 x)3
(1 + sin x) 2 sin x cos x dx = + C.
2 2
3
√ √
Let u = x, du = 1/[2 x] dx. Then the integral becomes
(b)

2 sin u du = ’2 cos u + C.

Resubstituting x, we obtain the ¬nal answer


sin x
dx = ’2 cos x + C.

x
Let u = ln x, du = [1/x] dx. Then the integral becomes
(c)
1 1
cos u sin u du = sin 2u du = ’ cos 2u + C.
2 4
Resubstituting x, we obtain the ¬nal answer
cos(ln x) sin(ln x) 1
dx = ’ cos(2 ln x) + C.
4
x
Let u = tan x, du = sec2 x dx. Then the integral becomes
(d)

eu du = eu + C.

Resubstituting x, we obtain the ¬nal answer

etan x sec2 x dx = etan x + C.

4. We do (a), (b), (c), (d).
Let u = cos x, du = ’ sin x dx. Then the integral becomes
(a)
u3
’ u du = ’ + C.
2
3
Resubstituting x, we obtain the ¬nal answer
cos3 x
sin x cos x dx = ’ + C.
2
3
306 Solutions to Exercises

(b) Write

sin3 x cos2 x dx = sin x(1 ’ cos2 x) cos2 x dx.

Let u = cos x, du = ’ sin x dx. Then the integral becomes
u3 u5
’ (1 ’ u )u du = ’ + + C.
2 2
3 5
Resubstituting x, we obtain the ¬nal answer
cos3 x cos5 x
sin x cos x dx = ’ + + C.
3 2
3 5
Let u = tan x, du = sec2 x dx. Then the integral becomes
(c)
u4
u du = + C.
3
4
Resubstituting x, we obtain the ¬nal answer
tan4 x
tan x sec x dx = + C.
3 2
4
Let u = sec x, du = sec x tan x. Then the integral becomes
(d)
u3
u2 du = + C.
3
Resubstituting x, we obtain the ¬nal answer
sec3 x
tan x sec x dx = + C.
3
3
5. We do (a), (b), (c), (d).
(a) Use integration by parts twice:
1
1 1
e sin x dx = sin x ·e ’
x x
ex cosx dx
0 0
0
1 1
= [e·sin 1’0]’ cosxe ’
x
ex (’sin x)dx
0
0
1
= e·sin 1’e·cos1+1’ ex sin x dx.
0
307
Chapter 7

We may now solve for the desired integral:
1 1
ex sin x dx = [e · sin 1 ’ e · cos 1] .
2
0

Integrate by parts with u = ln x, dv = x 2 dx. Thus
(b)
e
x3 x3 1
e e
x ln x dx = ln x · ’ · dx
2
3 3x
1 1
1
e
e3 13 x 3
=1· ’0· ’
3 3 6 1
e3 e3 13
= ’ + .
3 9 9
(c) We write
2x + 1 ’1
1 1
= + 2+ .
x+1
x 2 (x + 1) x x
Thus
(2x + 1) dx ’1
4 4 4 4
1 1
= dx + dx + dx
x+1
x3 + x2 x2
x
2 2 2 2

’1 ’1
= [ln 4 ’ ln 2] + ’ + [ln 3 ’ ln 5]
4 2
61
= ln +.
54
(d) We write
π π
1
sin x cos x dx =
2 2
sin2 2x dx
4
0 0
1 ’ cos 4x
π
1
= dx
4 2
0
π
1 sin 4x
= x’
8 4 0
1
= [(π ’ 0) ’ (0 ’ 0)]
8
π
=.
8
308 Solutions to Exercises

Chapter 8

1. At position x in the base circle, the y-coordinate is 1 ’ x 2 . Therefore the

half-disk slice has radius 1 ’ x 2 and area π(1 ’ x 2 )/2. The volume of
the solid is then
π(1 ’ x 2 )
1
V= dx
2
’1
1
x3
π
= x’
2 3 ’1

’1
1
π
= 1’ ’ (’1) ’
2 3 3

=
.
3



Y
2. We calculate the volume of half the solid, and then double the answer.

For 0√ x ¤ 1/ 2, at position x in the base square, the y-coordinate
¤ √
FL
is 1/ 2 ’ x. Thus the disk slice has radius (1/ 2 ’ x) and area

π(1/ 2 ’ x)2 . Thus the volume of the solid is

AM


1/ 2
V =2 π(1/ 2 ’ x)2 dx
0

1/ 2
3
TE



2π 1
=’ √ ’x
3 2 0
3
2π 3 1
=’ 0’ √
3 2
π
= √.
32
5
π [x 2 ]2 dx
3. (a)
2
3
π [y 2 ]2 dy
(b)
1
2
π [x 3/2 + 1]2 dx
(c)
0
2
π [5 ’ (x + 3)]2 dx
(d)
’1
309
Chapter 8

6
π [y 2 + 2]2 dy
(e)
2
π/2
π[sin x]2 dx
(f )
0
4
1 + [(2/3)x ’1/3 ]2 dx
2π · x 2/3 ·
4. (a)
0

3
2π · y 2 · 1 + [2y]2 dy
(b)
0

3
2π · [x 2 ’ 2] · 1 + [2x]2 dx
(c)
0
π
2π · sin x · 1 + [cos x]2 dx
(d)
0
2
2π · [y 2 + 2] · 1 + [2y]2 dy
(e)
1
1
2π · x 3 · 1 + [3x 2 ]2 dx
(f )
0
5. The depth of points in the window ranges from 6 to 10 feet. At depth x

in this range, the window has chord of length 2 16x ’ x 2 ’ 60. Thus the
total pressure on the lower half of the window is
10
P= 62.4 · x · 2 16x ’ x 2 ’ 60 dx .
8

6. At depth x, the corresponding subtriangle has side-length 2(5 ’ x/ 3).
Therefore the total pressure on one end of the pool is


53
P= 62.4 · x · 2(5 ’ x/ 3) dx.
0
7. Let t = 0 be the moment when the climb begins. The weight of the sack at
time t is then 100 ’ t pounds. Then the work performed during the climb is
8
W= (100 ’ t) · 5 dt.
0
Thus
8
5t 2 320 0
W = 500t ’ = 4000 ’ ’ 100 · 0 ’ = 3840 ft lbs.
2 2 2
0
310 Solutions to Exercises

8. The work performed is
100
W= [3x 2 + 4x + 6] dx
3
100
= x 3 + 2x 2 + 6x
3
= (1000000 + 20000 + 600) ’ (27 + 18 + 18)
= 1020547 ft lbs.
π
1 + [cos x]2 dx
9. (a)
0
8
1 + [(2/3)x ’1/3 ]2 dx
(b)
1
π/2
1 + [’ sin y]2 dy
(c)
0
4
1 + [2x]2 dx
(d)
1
5
1
sin2 x dx
10. (a)
3 2
π/4
1
(b) tan x dx
π/4 0
2
1 x
(c) dx
x+1
4 ’2

1 sin x
(d) dx
2 + cos x
3π ’π
6
2
e’(2j/3) ·
2
11. (a)
3
j =1
10
2
sin(e’2+2j/5 ) ·
(b)
5
j =1
5
2
cos(’2 + 2j/5)2 ·
(c)
5
j =1
12
ej/3 1
·
(d)
2 + sin(j/3) 3
j =1
311
Chapter 8

12. We do (a) and (b).
2/3 ’02
+ 2 · e’(2/3) + 2 · e’(4/3) + 2 · e’(6/3)
2 2 2
(a) e
2
+2 · e’(8/3) + 2 · e’(10/3) + e’(12/3)
2 2 2


2/5
sin(e’10/5 ) + 2 · sin(e’8/5 ) + 2 · sin(e’6/5 )
(b)
2
+ 2 · sin(e’4/5 ) + 2 · sin(e’2/5 ) + 2 · sin(e0/5 )
+ 2 · sin(e2/5 ) + 2 · sin(e4/5 ) + 2 · sin(e6/5 )
+ 2 · sin(e8/5 ) + · sin(e10/5 )
13. We do (a) and (b)
2/3 ’02
+ 4e’(2/3) + 2 · e’(4/3) + 4 · e’(6/3)
2 2 2
(a) e
3
+ 2 · e’(8/3) + 4 · e’(10/3) + e’(12/3)
2 2 2


2/5
sin(e’10/5 ) + 4 · sin(e’8/5 ) + 2 · sin(e’6/5 )
(b)
3
+ 4 · sin(e’4/5 ) + 2 · sin(e’2/5 ) + 4 · sin(e0/5 )
+ 2 · sin(e2/5 ) + 4 · sin(e4/5 ) + 2 · sin(e6/5 )
+ 4 · sin(e8/5 ) + sin(e10/5 )
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FINAL EXAM




1. The operations that preserve rational numbers are
(a) addition, multiplication, subtraction, division
(b) addition and multiplication
(c) multiplication and division
(d) square roots and logarithm
(e) sine and cosine
2. The number 3.157575757 . . . , expressed as a rational fraction, is
3
(a)
2
2976
(b)
355
1563
(c)
495
2
(d)
3
111
(e)
222
√ √2
The number ( 3 + 2) is
3.
(a) rational
(b) irrational

313
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
314 Final Exam

(c) transcendental
(d) indeterminate
(e) the quotient of rational numbers
4. The decimal expansion of 4/7 is
(a) 0.213535353 . . .
(b) 0.141414114 . . .
(c) 0.1357357357 . . .
(d) 0.7981818181 . . .
(e) 0.571428571428 . . .
√ √
The number 3 ’ 2
5.
(a) lies between 1 and 2
(b) is rational
(c) is a perfect square
lies between ’1 and 0
(d)
(e) lies between 0 and 1
The set {x : 3 ¤ x < 7} is
6.
(a) a closed interval
(b) an open interval
(c) a discrete set
(d) a half-open interval
(e) a half-line
The set [2, 5) © [4, 8] is
7.
{x : 4 < x < 8}
(a)
{t : 4 ¤ t < 5}
(b)
{s : 2 ¤ s ¤ 4}
(c)
{w : 2 < w < 8}
(d)
{u : 4 ¤ u ¤ 5}
(e)
The set Q © (’3, 2) is
8.
(a) in¬nite
(b) ¬nite
(c) discrete
(d) unbounded
(e) arbitrary
The set {(x, y) : x = y 2 } has graph that is
9.
(a) a line
315
Final Exam

(b) a circle
(c) a parabola
(d) a hyperbola
(e) a directrix
10. The line that passes through the point (’4, 5) and has slope 3 has
equation
x + 3y = 2
(a)
x ’ 3y = ’4
(b)
’4x + 5y = 3
(c)
3x ’ y = ’17
(d)
3x ’ 5y = 4
(e)
The line 2x + 5y = 10 has slope
11.
(a) 3
(b) 1
(c) 1/5
’1/5
(d)
’2/5
(e)
The equation 2x 2 + 2y 2 = 4 describes
12.
(a) A circle with center (0, 0) and radius 2

(b) A circle with center (0, 0) and radius 2
(c) A circle with center (2, 2) and radius 2
(d) A circle with center (4, 4) and radius 4
(e) A circle with center (2, 4) and radius 1
The equation x + x 2 + y = 0 has graph that is
13.
(a) a circle
(b) a line
(c) a parabola
(d) two crossed lines
(e) a hyperbola
14. The sine, cosine, and tangent of the angle 5π/3 (measured in radians)
are
√ √
(a) 1/2, 3/2, 3
√ √
(b) √3/2, 1/2, 1/ 3

(c) 2/2, 2/2, 1 √

(d) ’ 3/2, 1/2, ’ 3
(e) 1, 0, unde¬ned
316 Final Exam

15. The tangent, cotangent, and secant of the angle 3π/4 (measured in
radians) are

(a) ’ 3/2, ’1/2, 1
√ √
(b) 1/ 2, 1/ 2, ’1
√ √
2, ’ 2, 2
(c)

(d) 1, ’1, 3 √
(e) ’1, ’1, ’ 2

The domain and range of the function g(x) = 1 + 2x are
16.
{x : x ≥ ’1/2} and {x √0 ¤ x < ∞}
:
(a)
{x : x ≥ 1/2} and {x : 2 ¤ x ¤ 2}
(b)
{x : x ¤ ’1/2} and {y : ’2 ¤ y < ∞}
(c)
{s : 1 ¤ s ¤ 2} and {t : 2 ¤ t ¤ 4}
(d)
{x : 0 ¤ x ¤ 2} and {x : 1 ¤ x ¤ 4}
(e)
The graph of the function y = 1/|x| is
17.
(a) Entirely in the second and third quadrants
(b) Entirely in the ¬rst and fourth quadrants
(c) Entirely above the x-axis
(d) Increasing as x moves from left to right
(e) Decreasing as x moves from left to right
The graph of y = 2x/(1 + x 2 ) includes the points
18.
(a) (0, 1), (2, 4), (3, 3)
(b) (1, 1), (2, 2), (4, 4)
(’1, 1), (1, ’1), (3, 6)
(c)
(1, 1), (2, 4/5), (’2, ’4/5)
(d)
(e) (0, 0), (’4, 3), (4, 5)
Let f (x) = x 2 + x and g(x) = x 3 ’ x. Then
19.
f —¦ g(x) = (x 2 + x)x and g —¦ f (x) = (x 2 ’ x)2x
(a)
f —¦ g(x) = (x 2 + x)3 + x, g —¦ f (x) = (x 3 ’ x)2 + x
(b)
f —¦ g(x) = (x 3 ’x)2 +(x 3 ’x) and g —¦ f (x) = (x 2 +x)3 ’(x 2 +x)
(c)
f —¦ g(x) = (x 2 + x) · (x 3 ’ x) and g —¦ f (x) = (x 2 + x)/(x 3 ’ x)
(d)
f —¦ g(x) = (x 2 + x) + (x 3 ’ x) and g —¦ f (x) = (x 3 ’ x)x +x
2
(e)

Let f (x) = 3 x + 1. Then
20.
f ’1 (x) = x 3 ’ 1
(a) √
f ’1 (x) = 3 x ’ 1
(b)
317
Final Exam

f ’1 (x) = x 3 ’ x
(c)
f ’1 (x) = x/(x + 1)
(d)
f ’1 (x) = x 3 ’ 1
(e)
a 3 · b’2
21. The expression ln 4 ’3 simpli¬es to
c /d
3 ln a ’ 2 ln b ’ 4 ln c + 3 ln d
(a)
3 ln a + 2 ln b + 4 ln c ’ 3 ln d
(b)
4 ln a ’ 3 ln b + 2 ln c ’ 4 ln d
(c)
3 ln a ’ 4 ln b + 3 ln c ’ 2 ln d
(d)
4 ln a ’ 2 ln b + 2 ln c + 2 ln d
(e)
2 ’ln b3
The expression eln a
22. simpli¬es to
2a · 3b
(a)
2a
(b)
3b
a 2 · b3
(c)
a2
(d)
b3
6a 2 b3
(e)
x2 if x < 1
The function f (x) =
23. has limits
if x ≥ 1
x
2 at c = 1 and ’1 at c = 0
(a)
1 at c = 1 and 4 at c = ’2
(b)
0 at c = 0 and 3 at c = 5
(c)
’3 at c = ’3 and 2 at c = 1
(d)
1 at c = 0 and 2 at c = 2
(e)
x
The function f (x) = 2
24. has limits
x ’1
(a) 3 at c = 1 and 2 at c = ’1
(b) ∞ at c = 1 and 0 at c = ’1
(c) 0 at c = 0 and nonexistent at c = ±1
(d) 2 at c = ’2 and ’2 at c = 2
(e) ’∞ at c = 1 and +∞ at c = ’1
x3 if x < 2
The function f (x) = √
25. is continuous at
if x ≥ 2
x
x = 2 and x = 3
(a)
318 Final Exam

= 2 and x = ’2
(b) x

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