Therefore
( )
b
∫a f x dx
 =
(
– )
a
b f a
b
t dt
−
+
∫
=
(
– )
b
a f a
b
t dt
+
∫
 (by P1)
=
(
– )
b
a f a
b
x
+
∫
dx by P0
Proof of P4 Put t = a – x Then dt = – dx When x = 0, t = a and when x = a, t = 0 Now
proceed as in P3