, h → 0 all the three areas shown in (1) become
nearly equal to each other Now we form the following sums sn = h [f(x0) + … + f (xn - 1)] = 
1
0
(
)
n
r
r
h
f x
−
=∑ (2)
and
 Sn =
1
2
1
[ ( )
(
)
(
)]
(
)
n
n
r
r
h f x
f x
f x
h
f x
=
+
+…+
= ∑