When x = – a, t = a and when
x = 0, t = 0 Also x = – t INTEGRALS         343
Therefore
( )
a
a f x dx
−∫
 =
0
0
–
(– )
( )
a
a f
t dt
f x dx
+
∫
∫
=
0
0
(– )
( )
a
a
f
x dx
f x dx
+
∫
∫
       (by P0) (1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
0
0
0
( )
( )
( )
2
( )
a
a
a
a
a f x dx
f x dx
f x dx
f x dx
−
=
+
=
∫
∫
∫
∫
(ii) If f is an odd function, then f (–x) = – f(x) and so (1) becomes
0
0
( )
( )
( )
0
a
a
a
a f x dx
f x dx
f x dx
−
= −
+
=
∫
∫
∫
Example 30 Evaluate 
2
3
1
x–
x dx
−∫
Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that
x3 – x ≥ 0 on [1, 2]