Problem 1
Show that the function f (x) = x2 is integrable on the interval [1, 3] and calculate
the integral
Z 3
x2 dx.
1
Let n > 0 be an integer, partition [1, 3] into n intervals of length
intervals Ij . Let x∗j ∈ Ij . As usual, let Sn be the quantity
2
n,
and call the
n
X
2
f (x∗j ).
n
j=1
This is equal to
n
2X ∗ 2
(x ) .
n j=1 j
Recall that the interval Ij = [1 + n2 (j − 1), 1 + n2 j]. So x∗j is always between
1 + n2 (j − 1) and 1 + n2 j. So we can use the lower bound for our sum:
2
n 2X
2
Sn ≥
1 + (j − 1)
n j=1
n
We can re-index the sum by taking k = j − 1, and noting that the k = 0 term
is zero (You don’t have to re-index the sum to evaluate it, but the re-indexing
makes it slightly easier):
2
n−1 2
2X
1+ k
=
n
n
=
=
2
n
2
n
k=1
n−1
X
k=1
n−1
X
k=1
1+
1+
4
4
k + 2 k2
n
n
n−1
n−1
8 X
8 X 2
k
+
k
n2
n3
k=1
k=1
8 n2 + n
8 2n3 + 4n2 + 4n + 1
2
+ 3
.
= (n − 1) + 2
n
n
2
n
6
In the limit, this expression approaches 2 + 4 + 38 = 6 + 83 = 26
3 .
The upper sum is almost the same, but we have the k = n terms as well,
8
, which goes to zero as n → ∞. (If you don’t
which only contribute n2 + n8 + 3n
believe this, you should try writing out the entire sum yourself!) So both the
2
upper and lower bounds for Sn approach 26
3 as n → ∞. This shows that x is
integrable on [1, 3] and that
Z
3
x2 dx =
1
1
26
.
3
Problem 2
Show that the following function is integrable on [−1, 1] and evaluate the integral:
(
1 if x = 0
f (x) =
.
0 otherwise
Let n > 0 and partition [−1, 1] into n intervals Ij of equal length. Select
x∗j ∈ Ij , and let
n
X
2
Sn =
f (x∗j ).
n
j=1
Then f (x∗j ) is zero for all except for at most 2 values of j. Furthermore, f (x∗j )
is at least 0 for every j. So we have that Sn ≥ 0 and that
Sn ≤
2
4
2
+ =
n n
n
which goes to zero as n → ∞.
2