# Analysis Homework #6 Solutions 1. 2. ```Analysis Homework #6
Solutions
1. Suppose
∫ 1 f is non-negative and integrable on [0, 1] with f (x) = 0 for all x ∈ Q. Show
that 0 f (x) dx = 0.
• Given a partition P = {x0 , x1 , . . . , xn } of [0, 1], one clearly has
inf
[xk ,xk+1 ]
f (x) = 0 for each 0 ≤ k ≤ n − 1
because f is non-negative and every subinterval contains a rational. Thus,
−
S (f, P ) =
n−1
∑
k=0
inf
[xk ,xk+1 ]
f (x) &middot; (xk+1 − xk ) = 0
and we may take the supremum of both sides to conclude that
∫ 1
f (x) dx = sup S − (f, P ) = 0.
0
2. Suppose f is increasing on [0, 1]. Show that f is integrable on [0, 1]. Hint: let P be the
partition of [0, 1] into n subintervals of the same length and show that
S + (f, P ) − S − (f, P ) =
f (1) − f (0)
.
n
• Let P = {x0 , x1 , . . . , xn } be as in the hint. Since f is increasing, we then have
inf
[xk ,xk+1 ]
f (x) = f (xk ) for each 0 ≤ k ≤ n − 1
and this implies that
−
S (f, P ) =
n−1
∑
k=0
inf
[xk ,xk+1 ]
f (x) &middot; (xk+1 − xk )
)
1(
f (x0 ) + f (x1 ) + . . . + f (xn−1 ) .
n
Using the exact same reasoning to compute the upper Darboux sum, we now get
=
+
S (f, P ) =
n−1
∑
sup f (x) &middot; (xk+1 − xk )
k=0 [xk ,xk+1 ]
)
1(
=
f (x1 ) + f (x2 ) + . . . + f (xn ) ,
n
so we may combine the last two equations to arrive at
f (1) − f (0)
f (xn ) − f (x0 )
=
.
n
n
Letting n be suﬃciently large, we can certainly make this expression as small as we
wish, hence f is integrable by Theorem 4.9 in your notes.
S + (f, P ) − S − (f, P ) =
3. Show that there exists a unique function f , defined for all x ∈ R, such that
f ′ (x) =
• Since
1
1+x4
1
,
1 + x4
f (0) = 0.
is continuous, the fundamental theorem of calculus ensures that
∫
f (x) =
0
x
dt
1 + t4
has the desired properties. If g(x) also does, then
g ′ (x) =
1
= f ′ (x)
1 + x4
=⇒
g(x) = f (x) + C
and the fact that g(0) = 0 = f (0) implies that C = 0, hence g(x) = f (x).
4. Suppose f is continuous on [a, b]. Show that there exists some c ∈ (a, b) such that
∫
b
f (t) dt = (b − a) &middot; f (c).
a
Hint: apply the mean value theorem to the function F (x) =
∫x
a
f (t) dt.
• According to the mean value theorem, there exists some c ∈ (a, b) such that
F (b) − F (a)
= F ′ (c).
b−a
Moreover, F ′ (x) = f (x) by the fundamental theorem of calculus, while
∫
F (a) =
∫
a
f (t) dt = 0,
F (b) =
a
b
f (t) dt.
a
Once we now combine all these facts, we may conclude that
∫ b
′
F (b) − F (a) = (b − a) &middot; F (c) =⇒
f (t) dt = (b − a) &middot; f (c).
a
```