Then
dt = – dx When x = a, t = a and when x = 2a, t = 0 Also x = 2a – t Therefore, the second integral becomes
2
( )
a
a
f x dx
∫
 =
–0
(2 – )
a f
a
t dt
∫
 = 
0
(2 – )
a f
a
t dt
∫
 = 
0
(2 – )
a f
a
x dx
∫
Hence
2
0
( )
∫a f x dx
 =
0
0
( )
(2
)
a
a
f x dx
f
a
x dx
+
−
∫
∫
Proof of P6 Using P5, we have 
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫