
Successive Partial Integration 16-11
= +
∫
= +
7
2
7
2 2
1
2
2
1
2
y dy
y
y
= ⋅ +
− ⋅ +
=
7
2
2
2
2
7
2
1
1
2
5
2 2
(2)
From Eqs. (1) and (2), we obtain
f x y dx dy f x y dy dx( , ) ( , ) ,
3
4
1
2
1
2
3
4
∫
∫
=
∫
∫
ii.e. I J= .
Example 16.11 Verify that
( ) ( )xy e dx dy xy e dy dx
y y
+
∫
∫
= +
∫
3
4
1
2
1
2
3
4
∫∫
.
Solution:
( )xy e dx dy
x
y xe dy
y y
x
x
+
∫
∫
= +
=
=
3
4
1
2
2
3
4
1
2
2
∫∫
= ⋅ +
− ⋅ +
1
2
4 4
1
2
3 3
2 2
1
2
y e y e dy
y y
∫∫
= +
∫
= ⋅ +
7
2
7
2 2
1
2
2
1
2
y e dy
y
e
y
y
= ⋅ +
− ⋅ +
= + −
7
2
2
2
7
2
1
2
21
4
2
2
2
2
e e e e. (1)
( )xy