Solution Set #5
Section 7.8
1.
2.
By Theorem 7.8A
m
m
3
0
3
(3 x 2 & 5) dx
0
3
m
3
23
3
) (x)
'
x % x 6 . Hence, f ) (2)
'
(by Theorem 7.4C)
&5 dx '
(by Theorem 7.4B)
'
(by Theorem 7.8E)
3
0
&5
x 2 dx
0
3(
m
%
3x 2 dx
f
m
3
1 dx
0
'
66 .
& 0) &5 (2 & 0)
4.
By Theorem 7.8A F ’ (x) = f (x) for x 0 [a,b]. Since f (x) > 0 for x 0 [a,b], we have
F ’ (x) > 0 for x 0 [a,b]. By Theorem 7.7B which is a corollary of the Mean Value
Theorem F is strictly increasing for x 0 [a,b].
5.
Let F(x)
'
m
x
a
f (t) dt for x 0 [a,b]. By Theroem 7.8A, F ’ (x) exists for x 0 [a,b] (and
equals f (x)) . Note that F (a) = 0. By the Mean Value Theorem there exists a c 0 (a,b)
F(b) & F(a)
such that
' F )(c) ' f (c) . Hence, b f (x) dx ' f (c) (b & a) .
a
b&a
m
Section 7.9
1a)
No.
m
(
m
(
4
1
xp
4 1
1
1b)
1c)
1d)
1e)
Yes.
No.
Yes.
No.
(
1
xp
1
2x
#
dx diverges if p # 1).
dx converges if p > 1).
x
for x $ 1 and
1%x
43 x 2
(
1 % 2 x 2 % 12 x 4
(
m
s
2
x cos(x) dx
1
m
4
1
1
xp
dx diverges if p # 1).
# 43 12 for x $ 1 and
12 x
m
4
1
1
dx converges if p > 1).
xp
' cos(s)%s(sin(s) & cos(1) & sin(1) ' F(s) and lim F(s) does not
s64
exist.)
1f)
Yes.
(
1
1
(1 % x 3) 2
# 13 for x $ 1 and
x2
m
4
1
1
dx converges if p > 1).
xp
1g)
2.
No.
1
2x
#
1
1
3 3
for x $ 1 and
(1 % x )
' 1
Let F(x)
1
1
&
m
4
1
2
and G(x)
1
xp
dx diverges if p # 1).
' &1
1
. Then, F )(x)
x
'
and
2 (1 % x)
1%x
2 1%x
(1 % x)3
1
1
x
1
s
. Hence,
G )(x) '
dx ' F(s) & F(0) ' F(s) % and
0 (1 % x)3
2 (1 % x)2
2
1 s
1
1
dx ' G(s) & G(0) ' G(s) % . Since lim F(s) ' lim G(s) ' 0 , we have
2
2 0 (1 % x)
2
s 64
s 64
4
x
1
' 1 4 1 2 dx.
dx '
3
0 (1 % x)
2
2 0 (1 % x)
m
m
m
4.
(
m
False. Let f (x) ' sin(2 n x) for x 0 [n,n+1], n 0 I. Then, f is continuous on [1,4). {f is
obviously piece-wise continuous on [1,4) and at each integer in [1,4) f has two one-sided
1
limits which are the same (namely 0)}. Let g > 0 and let N ' [ ] % 1 . Then, for s $ N
g
we have |
m
4
f (x) dx |
s
lim f (x)
x64
5.
#
1
(N%1)
<
1
1
< g . Hence, the integral converges, but
g
ú 0.
Proceeding by contradiction. Suppose L ú 0. WoLog L > 0. Since, lim f (x)
x64
there exists N such that for x > N we must have f (x) > L / 2. But then,
diverges which implies that
m
m
4
' L , then
f (x) dx
N
4
f (x) dx diverges which is a contradiction.
1
6.
Let f (x)
m
4
' |sin( x)| . Then, f (n) = 0 for all n 0 I and hence
f (x) dx diverges .
1
2 (x&n)%
7.
Let f (x)
1
n
x 0 [n &
' (& 2 (x&n)% 1
n
0
1
, n]
2n
x 0 [n, n %
1 , for n 0 I.
]
2n
otherwise
4
j f (n) converges. But,
n'1
Then, f(n)
' 1 and hence,
n
which converges.
4
j
n'1
f (n) diverges. But,
m
4
1
f (x) dx
4
4
j
2
' 1 % 1
n'1
1
n2