18.022 Recitation Handout (with solutions)
3 November 2014
y2
Z 1Z
1. Evaluate
tion.
0
0
x 2 y d x d y and sketch the region of integration in R2 indicated by the limits of integra-
Solution. The domain of integration is shown below.
1
1
Evaluating the integral, we find
y2
Z 1Z
0
0
Z 1·
2
x y dx dy =
0
1
Z
=
0
x3
y
3
¸y 2
dy
0
y7
dy
3
= 1/24 .
2. Evaluate
R π R π sin x
0
y
x
d x d y.
Solution. We reverse the order of integration. The region of integration is a triangle with vertices at (0, 0),
(π, 0), and (π, π). So the integral is equal to
Z πZ x
Z π
Z π
£ ¤x
sin x
sin x d x = 2 .
dy dx =
y 0 sin(x)/x d x =
x
0 0
0
0
aZ b
Z
3. (Putnam exam ’89) Evaluate
ted.
0
e max{b
2 2
x ,a 2 y 2 }
d y d x where a and b are positive. Solution. Omit-
0
4. (Fun/Challenge, based on 5.2.29 in Colley) Define a function f (x, y) on [0, 1] × [0, 2] by

 1
0
f (x, y) =

2
if x is rational
if x is irrational and y ≤ 1
if x is irrational and y > 1.
R1R2
Show that the iterated Riemann integral 0 0 f (x, y) d y d x exists, and find its value. Show that the iterR2R1
ated Riemann integral 0 0 f (x, y) d x d y does not exist.
R1R2
R2
Solution. To show that 0 0 f (x, y) d y d x exists, we first calculate 0 f (x, y) d y. If x is rational, then this
R2
R2
integral is 0 1 d y = 2. If x is irrational, then this integral reduces to 1 2 d y = 2. Therefore, regardless of
R2
R1R2
R1
the value of x, the inner integral 0 f (x, y) d y is equal to 2. Therefore, 0 0 f (x, y) d y d x = 0 2 d x = 2.
R1
On the other hand, there is no value of y for which the integral 0 f (x, y) d x exists. To see this, note that
(for y ≤ 1, say) the upper and lower Riemann sums are equal to 1 and 0, respectively. For if 0 = x 0 < x 1 <
. . . < x n = 1 is some partition of [0, 1], then the sum
n−1
X
k=0
f (x k∗ )(x k+1 − x k )
can be made as large as 1 by choosing each x k∗ ∈ (x k+1 − x k ) to be rational and as small as 0 by choosing
each x k∗ ∈ (x k+1 − x k ) to be irrational. Since the inner integral doesn’t exist, the iterated integral doesn’t
exist either.