Math 3210-3
HW 22
Solutions
Properties of the Riemann Integral
1. ♣A function fon [a, b] is called a step function if there exists a partition P={a=u0< u1<··· <
um=b}of [a, b] such that fis constant on each interval (uj−1,uj), say f(x) = cjfor x∈(uj−1, uj).
(a) Show that a step function fis integrable and evaluate Rb
af.
Proof: Let P={x0, x1,...,xn}be a partition of [a, b] such that fis constant on (xi−1, xi) for
each i= 1,...,n. Then we can say f(x) = cifor x∈(xi−1, xi).
I claim that fis integrable on [xi−1, xi] for each i= 1,...,n. To prove this claim, notice that
if g(x) = cifor all x∈[xi−1, xi], then f(x) = g(x) except possibly at two points. Also gis
monotonic, so by Theorem 98 gis integrable on [xi−1, xi]. By problem 6 on homework 21, we
can conclude that fis integrable on [xi−1, xi] for each i= 1,...,n. Thus by Theorem 100 fis
integrable on [a, b] and Zb
a
f=
n
X
i=1 Zxi
xi−1
f=
n
X
i=1
ci∆xi.
˜
(b) Given P(x) = 15 if 0 ≤x < 1
15 + 13nif n≤x < n + 1 , evaluate R4
0P(x)dx. Note: P(x) is called the
postage-stamp function. Do you see why?
Proof: By part (a), we see Z4
0
P(x)dx =Z1
0
15 dx +Z2
1
(15 + 13) dx +Z3
2
(15 + 13(2)) dx +
Z4
3
(15 + 13(3)) dx = 15 + 28 + 41 + 54 = 138.
˜
2. ♣Prove that if fis integrable on [a, b] then so is f2. (Hint: If |f(x)| ≤ Mfor all x∈[a, b], then show
|f2(x)−f2(y)| ≤ 2M|f(x)−f(y)|for all x, y ∈[a, b]. Then use this to estimate U(f2, P )−L(f2, P )
in terms of U(f, P )−L(f, P ) for a given partition P.)
Proof: Let ǫ > 0. Since fit is bounded. Let M∈Rsuch that |f(x)|< M for all x∈[a, b]. From
Theorem 97, there is a partition Pof [a, b] such that U(f, P )−L(f , P )<ǫ
2M. I claim that with this
partition we have U(f2, P )−L(f2, P )< ǫ, so by Theorem 97, f2will be integrable. To this end, notice
that |f2(x)−f2(y)|=|f(x) + f(y)||f(x)−f(y)| ≤ (|f(x)|+|f(y)|)|f(x)−f(y)|<2M|f(x)−f(y)|
for all x, y ∈[a, b].
Thus we have
U(f2, P )−L(f2, P ) = X|f2(xi)−f2(yi)||xi−yi|
≤X2M|f(xi)−f(yi)||xi−yi|
= 2M(U(f, P )−L(f , P ))
<2Mǫ
2M
=ǫ
Therefore f2is integrable.
˜
3. Prove that if fand gare integrable on [a, b], then so is f g. (Hint: Use the previous problem to write
fg as the sum of two functions which you know are integrable.)
Proof: Notice that fand gare integrable, so f+g,f2,g2, and (f+g)2are integrable by the previous
problem and Theorem 100. Thus 1
2[(f+g)2−f2−g2] = fg is integrable.
