·
b
n
+
2
b
n
·
b
n
+
3
b
n
·
b
n
+
· · ·
+
nb
n
·
b
n
=
b
2
n
2
(1 + 2 + 3 +
· · ·
+
n
)
=
b
2
n
2
·
n
(
n
+ 1)
2
→
1
2
as
n
→ ∞
Therefore,
f
(
x
) =
x
is integrable and
integraldisplay
b
0
x dx
=
1
2
.
2
8. Set
f
(
x
) =
1
,
x
rational

1
,
x
irrational
x
∈
[

1
,
1]
Then,
f
2
(
x
)
≡
1
is integrable,
integraldisplay
1

1
f
2
(
x
)
dx
=
integraldisplay
1

1
1
dx
= 2.
f
is not integrable since for any partition
P
,
L
(
f,
P
) =

2
and
U
(
f,
P
) = 2.
12. Suppose that
f
is bounded and
f
(
x
)
≥
0 for all
x
∈
[
a, b
]. Then, for any partition
P
=
{
a
=
x
0
, x
1
, x
2
,
· · ·
, x
n
=
b
}
,
m
i
= inf
f
(
x
)
, x
∈
[
x
i

1
,x
i
]
≥
0.
Therrefore,
L
(
f,
P
)
≥
0. Since
L
(
f
)
≥ L
(
f,
P
), it follows that
L
(
f
)
≥
0.
13.
f
is continuous on [
a, b
],
f
(
x
)
≥
0 for all
x
, and
L
(
f
) =
integraltext
b
a
f
= 0. Prove
f
(
x
) = 0
for all
x
.
Suppose there is a point
c
∈
[
a, b
]
such that
f
(
c
)
>
0,
and suppose, first, that
c
∈
(
a, b
). By Exercise 21.13 there is a neighborhood
N
(
c
) (i.e., an interval (
c

δ, c
+
δ
)
)
and a number
α >
0
such that
f
(
x
)
≥
α
on
N
(
c
). Let
P
be the partition
{
a, c

δ, c
+
δ, b
}
. Then
L
(
{
,
P
) = inf
{·
(
floorright
δ
turnstileright
)+inf
{·
(
floorright
+
δ

(
floorright
δ
))+inf
{·
(
floorleftfloorright
+
δ
)
≥
+
α
·∈
δ
+
prime·
(
floorleftfloorright
+
δ
)
≥ ∈
δα >
prime
This is a contradiction since
L
(
{
,
P
)
≤ L
(
{
) =
prime
.
Additional Problems
1. Suppose that
f
: (

1
,
1)
→
R
is infinitely differentiable, and that
vextendsingle
vextendsingle
f
(
k
)
(
x
)
vextendsingle
vextendsingle
≤
3
on
(

1
,
1) for
all
k
. Let
p
n
be the Taylor polynomial of degree
n
for
f
in powers
of
x
.
(a) Prove that
f
is uniformly continuous on
(

1
,
1).
Proof:
Let
epsilon1 >
0.
Set
δ
=
epsilon1/
3.
Choose any points
x
1
, x
2
∈
I
(Assume
x
1
< x
2
). By the Mean Value Theorem, there is a number
c
∈
(
x
1
, x
2
)
such
that
f
(
x
2
)

f
(
x
1
) =
f
prime
(
c
)(
x
2

x
1
)
.
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 Fall '08
 Staff
 Mean Value Theorem, Taylor Series, Trigraph, Taylor Polynomial, B2 Spirit