Integration 11-21
Example 11.11 Given that f x dx g x dx( ) ( ) ,=
∫
=
∫
13 37
2
5
2
5
and then evaluate f x g x dx( ) ( ) .+
( )
∫
2
5
Solution: We have by property 5 that
f x g x dx f x dx g x dx
a
b
a
b
a
b
( ) ( ) ( ) ( ) .+
[ ]
∫
=
∫
+
∫
∴ +
( )
∫
=
∫
+
∫
= + =
f x g x dx f x dx g x dx( ) ( ) ( ) ( )
.
2
5
2
5
2
5
13 37 50
Example 11.12 Assume that
f x dx( ) =
∫
−
33
3
30
and
g x dx( ) ,=−
∫
−
3
30
3
then evaluate
( ( ) ( )) .3 5
30
3
f x g x dx−
∫
−
Solution: Consider
3 5 3 5
30
3
30
3
30
3
f x g x dx f x dx g x dx( ) ( ) ( ) ( )−
( )
∫
= −
∫ ∫
− − −
(by property 5)
(by property 4)= −
∫ ∫
= − −
− −
−
3 5
3
30
3
30
3
3
3
f x dx g x dx
f x dx
( ) ( )
( )
00
30
3
5
3 33 5 3 84
∫ ∫
= − − − = −
−
g x dx( )
( ) ( ) .
(by property 2)
Example 11.13 Assume that
f x dx f x dx f x dx( ) , ( ) ( ) ,=
∫
=
∫
=
∫
− − −
16 24 4
15
10
100
10
100
5
and