7.3 The Integral as a Limit of Riemann Sums 381x
t Yk 1 - It Yk 1 Yk 1 +1 f krlt Yk 2 Yk 2 +1
Yk,-2 x, Yk 2 -2 x2Figure 7.6Putting (7) and (8) together, we havem161 + m262:::; (m1 + m2 + · · · + mk 2 ) 6 + (2 + 3)B6.Continuing in this way, we see thatm161 +m262+· ··+mn6n :S:: (m1 + ··· +mkn)6+(2+3+···+(n+l))B 6
_ _ _ - n(n + 1) -
=(m1+m2+···+mkn)6+
2
B6.So, $(!, P) < $(!, Qm) + n(n 2+ 1) B6.
n(n+l)B/\ n(n+l)B(b-a) B
Now, u =. ut m >n(n + l)B(b - a)
2 2m 2c ' so
n(n --'-------+ l)B(b - a) < c. Th us,
2m
$(!, P) < $(!, Qm) + c
i.e., $(!, Qm) > $(!, P) - c.
Similarly (Exercise 1 5) we can show that S(f, Qm) < S(f, P) + c. •
Corollary 7.3.12 If f is defined and bounded on [a, b], where a < b, then
(a) l:l =sup{$_(!, Qm): Qm is a regular partition of [a, b]};
(b) l:l = inf{S(f, Qm): Qm is a regular partition of[a,b]}.