A Guide to Changing the Order of an Iterated Integral
It often occurs that a double integral is easily evaluated if iterated in one direction and very
difficult, or impossible, if iterated in the other direction. Knowing how to reverse the order of
integration is, therefore, an important skills to learn.
Example 1 Consider
∫
2
∫
ln x
I=
f (x, y) dy dx.
1
0
The outer limits give a lower and an upper bound for the variable x, 1 ≤ x ≤ 2, while the inner
limits bound the variable y, 0 ≤ y ≤ ln x. Note that the inner limits are functions of x whereas
the outer limits are constants. We use this information to sketch the region R of integration:
Reversing the order of integration means finding new outer and inner limits to enable us to
write
∫ ∫
I=
f (x, y) dx dy.
The outer limits will bound the variable y while the inner limits, which will be functions of y,
bound the variable x. The above sketch is used to find these new limits. The outer limits are
the maximum and minimum values of y in the region, i.e. 0 ≤ y ≤ ln 2. The inner limits are
found by drawing a line across the region parallel to the x-axis. This enters the region on the
curve y = ln x ⇔ x = ey and leaves the region along the line x = 2. Thus ey ≤ x ≤ 2 and it
follows that
∫ ln 2 ∫ 2
f (x, y) dx dy.
I=
0
ey
1
Example 2
∫
1
∫
1
x2 sin(xy) dx dy.
I=
0
y
In this example the outer limits bound y, 0 ≤ y ≤ 1, and the inner limits bound x, y ≤ x ≤ 1:
Once we have reversed the order of integration the outer limits will bound x while the inner
limits will bound y. The outer limits are the greatest and least values of x in the region R:
0 ≤ x ≤ 1. The inner limits are obtained by drawing a line across the region parallel to the
y-axis. This line enters the region across the x-axis, i.e. y = 0, and leaves the region on the
line y = x. Thus 0 ≤ y ≤ x and we have
∫ 1∫ x
I=
x2 sin(xy) dy dx.
0
0
2
Example 3 Consider
∫ ∫
I=
(2xy + 1) dx dy
A
where A is the finite region in the first quadrant enclosed by the parabola y = x2 and the line
y = 2x. The curves y = x2 and y = 2x intersect when x2 = 2x ⇔ x = 0 or x = 2.
From the above sketch we see that the outer limits are 0 ≤ y ≤ 4 and the inner limits are
√
y
≤ x ≤ y. The latter are obtained by drawing a line across the region parallel to the x-axis.
2
Therefore
∫ 4 ∫ √y
I=
(2xy + 1) dx dy.
0
y
2
Let us now reverse the order of integration. The outer limits are the maximum and minimum
values of x: 0 ≤ x ≤ 2. By drawing a line across the region parallel to the y-axis we find that
the inner limits are x2 ≤ y ≤ 2x. Hence
∫ 2 ∫ 2x
I=
(2xy + 1) dy dx.
0
x2
3
Example 4
∫
4
∫
2
x
dy dx
5
0
x 1+y
√
The limits of the region of integration are 0 ≤ x ≤ 4 and x ≤ y ≤ 2.
I=
√
Using the above sketch we reverse the order of integration as follows:
∫
2
∫
I=
0
0
y2
x
dx dy.
1 + y5
Exercise Reverse the order of integration in the following integrals.
∫ 1∫ 1
1. I =
f (x, y) dy dx
0
∫
x
2
∫
1
x2
∫ √
2. I =
4
f (x, y) dy dx
0
∫
x
3. I =
f (x, y) dy dx
0
∫
x
2
∫
4. I =
2y
f (x, y) dx dy
1
2
4