SOLUTION OF HW3
MINGFENG ZHAO
October 01, 2011
1. [Page 64, Problem 4a] Prove [x + n] = [x] + n for all n ∈ Z, and all x ∈ R.
Proof. For any x ∈ R, by the Exercise 5, in Page 28,we know that there exists a unique m ∈ Z such
that
m ≤ x < m + 1.
By the definition of [x], we have
[x] = m.
And also for all n ∈ Z, we have
m + n ≤ x + n < m + n + 1.
So we get [x + n] = m + n = [x] + n.
2. [Page 70, Problem 2] Give an example of a step function s, defined on the closed interval [0, 5],
which has the following properties:
Z
2
Z
s(x) dx = 5,
and
0
5
s(x) dx = 2.
0
Proof. Let s be the function such that:
s(t) =





5
2,
if 0 ≤ t ≤ 2



 −1, if 2 < t ≤ 5.
Then we know
Z
2
s(x) dx =
0
5
· [2 − 0] = 5.
2
1
2
MINGFENG ZHAO
And
Z
5
s(x) dx
=
5
· [2 − 0] + (−1) · [5 − 2]
2
=
5−3
=
2.
0
3. [Page 70, Problem 11] If instead of defining integrals of step function by using formula (1.3), in
Page 65, we used the definition
b
Z
s(x) dx =
a
n
X
s3k · [xk − xk−1 ],
k=1
a new and different theory of integration would result which of the following properties would
remain valid in this new theory?
a.
Rb
b.
Rb
c.
Rb
d.
R b+c
a
a
a
s(x) dx +
Rc
b
s(x) dx =
Rb
[s(x) + t(x)] dx =
Rb
c · s(x) dx = c ·
a+c
s(x) dx =
a
a
Rb
Rc
a
s(x) dx.
s(x) dx +
Rb
a
t(x) dx.
s(x) dx.
s(x + c) dx.
a
e. If s(x) < t(x) for each x in [a, b], then
Proof. a. We have
Rb
a
s(x) dx +
Rc
b
s(x) dx =
Rb
a
Rc
a
s(x) dx <
Rb
a
t(x) dx.
s(x) dx. In fact, without loss of generality, we can
assume
a = x0 < x1 < · · · < xn = b < xn+1 < · · · < xn+m = c.
And s(x) = sk for all x ∈ (xk−1 , xk ). Then we have
Z
b
Z
s(x) dx +
a
c
s(x) dx
=
b
n
X
s3k · [xk − xk−1 ] +
n
X
k=1
s3n+j · [xn+j − xn+j−1 ]
j=1
k=1
=
m
X
s3k · [xk − xk−1 ] +
n+m
X
k=n+1
s3k · [xk − xk−1 ]
SOLUTION OF HW3
=
n+m
X
s3k · [xk − xk−1 ]
k=1
Z
c
s(x) dx.
=
a
b. In general,
Rb
a
Rb
[s(x) + t(x)] dx 6=
s(x) = 1,
Rb
s(x) dx +
a
a
t(x) dx. For example, [a, b] = [0, 1], and
for all 0 ≤ x ≤ 1.
and t(x) = 1,
Then s(x) + t(x) = 2 for all x ∈ [0, 1]. So we get
Z
1
1
Z
t(x) dx = 13 · [1 − 0] + 13 · [1 − 0] = 2.
s(x) dx +
0
0
And
Z
1
[s(x) + t(x)] dx = 23 · [1 − 0] = 8 6= 2.
0
c. In general,
Rb
a
c · s(x) dx 6= c ·
Rb
a
s(x) dx. For example, [a, b] = [0, 1], and c = 2, and
s(x) = 1,
for all 0 ≤ x ≤ 1.
Then c · s(x) = 2 for all 0 ≤ x ≤ 1. So we have
Z
c·
1
s(x) dx = 2 · 13 · [1 − 0] = 2.
0
But
1
Z
c · s(x) dx = 23 · [1 − 0] = 8 6= 2.
0
d. We have
R b+c
a+c
s(x) dx =
Rb
a
s(x + c) dx. In fact, without loss of generality, we can assume
a + c = x0 < x1 < · · · < xn = b + c.
And s(x) = sk for all x ∈ (xk−1 , xk ). Then we have
Z
b+c
s(x) dx =
a+c
n
X
k=1
s3k · [xk − xk−1 ].
3
4
MINGFENG ZHAO
Since s(x) = sk for all x ∈ (xk−1 , xk ), then we have s(x + c) = sk for all x ∈ (xk−1 − c, xk − c).
Hence
Z
b
s(x + c) dx =
a
n
X
s3k · [xk − c − (xk1 − c)] =
k=1
e. If s(x) < t(x) for each x in [a, b], then we have
n
X
s3k · [xk − xk−1 ] =
a
s(x) dx <
b+c
s(x) dx.
a+c
k=1
Rb
Z
Rb
a
t(x) dx. Because f (x) = x3 is
strictly increasing on R.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu