Calculus Exercises: Changing the Order of Integration, Exercises of Differential and Integral Calculus

In order to integrate with respect to x , we can't have x's in the limits. So, to reverse the order, it is best to first sketch the region. Notice first that ...

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Section 5.4 - Changing the Order of Integration
Problem 1. Evaluate the integral by first reversing the order of integration,
x=3
Z
x=0
y=9
Z
y=x2
x3ey3dy dx.
Solution. Even if we tried to integrate with respect to yfirst, we cannot do it. We can’t just
switch either. In order to integrate with respect to x, we can’t have x’s in the limits. So,
to reverse the order, it is best to first sketch the region. Notice first that our region has two
properties:
0x3x2y9.
We then can draw the region:
Since we want to integrate with respect to xfirst, we will need limits for xas functions of y
and we need constant bounds for y. Looking at the picture, we get
0y9 0 xy.
With this information, we can now set up our new integral and hopefully be able to solve it!
x=3
Z
x=0
y=9
Z
y=x2
x3ey3dy dx =
y=9
Z
y=0
x=y
Z
x=0
x3ey3dx dy
=
y=9
Z
y=0
1
4x4ey3
x=y
x=0
dy
=
y=9
Z
y=0
1
4y2ey3dy
=1
12ey3
y=9
y=0
=1
12 e729 1.
1
pf3

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Section 5.4 - Changing the Order of Integration

Problem 1. Evaluate the integral by first reversing the order of integration,

∫^ x=

x=

∫^ y=

y=x^2

x

3 e

y^3 dy dx.

Solution. Even if we tried to integrate with respect to y first, we cannot do it. We can’t just

switch either. In order to integrate with respect to x , we can’t have x’s in the limits. So,

to reverse the order, it is best to first sketch the region. Notice first that our region has two

properties:

0 ≤ x ≤ 3 x

2 ≤ y ≤ 9.

We then can draw the region:

Since we want to integrate with respect to x first, we will need limits for x as functions of y

and we need constant bounds for y. Looking at the picture, we get

0 ≤ y ≤ 9 0 ≤ x ≤

y.

With this information, we can now set up our new integral and hopefully be able to solve it!

∫^ x=

x=

∫^ y=

y=x^2

x

3 e

y^3 dy dx =

∫^ y=

y=

x=

√ y ∫

x=

x

3 e

y^3 dx dy

∫^ y=

y=

x

4 e

y^3

x=

√ y

x=

dy

∫^ y=

y=

y

2 e

y^3 dy

e

y^3

y=

y=

e

729 − 1

Problem 2. Evaluate the integral by first reversing the order of integration,

∫^ y=

y=

∫^ x=

x= 3

√ y

x^4 + 1 dx dy.

Solution. The region is described by the following two inequalities:

0 ≤ y ≤ 8 3

y ≤ x ≤ 2.

We sketch the region and get the picture

Now we can find our new inequalities and we get that

0 ≤ x ≤ 2 0 ≤ y ≤ x

3 .

With this information, we can now set up our new integral and evaluate:

∫^ y=

y=

x∫=

x= 3

√ y

x^4 + 1 dx dy =

∫^ x=

x=

y ∫=x^3

y=

(x

4

1 (^2) dy dx

∫^ x=

x=

y(x

4

y=x^3

y=

dx

∫^ x=

x=

x

3 (x

4

1 (^2) dx

3 (^2) − 1