There is no need to keep
integration constant C because if we consider F(x) + C instead of F(x), we get
( )
[F ( )
C]
[F( )
C]– [F( )
C]
F( ) – F( )
b
ab
a f x dx
x
b
a
b
a
=
+
=
+
+
=
∫ Thus, the arbitrary constant disappears in evaluating the value of the definite
integral (ii)
Evaluate F(b) – F(a) = [F ( )]b
xa
, which is the value of  
( )
b
∫a f x dx We now consider some examples
Example 27 Evaluate the following integrals:
(i)
3
2
∫2 x dx
(ii)
9
3
4
22
(30–
)
x
dx
x
∫
(iii)
2
1 (
1) (
2)
x dx
x
x
+
+
∫
(iv)   
3
4
0 sin 2 cos 2
t
t dt
π
∫
Solution
(i)
Let 
3
2
2
I
= ∫x dx