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
=
+
+…+
= ∑ (3)
Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles
raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively In view of the inequality (1) for an arbitrary subinterval [xr–1, xr], we have
sn < area of the region PRSQP < Sn