676 Appendix C 11 Answers & Hints for Selected Exercises
= l n-1 "' [ 9i^2 + Qi ] = l [ 9 n-1 "' i2 + 2. n-1 "' i ] = 27 (n-l)n(2(n-l)+l) + 18 (n-l)n
n~ rt2 n n rl:i~ n ~ Tt3 6 rl:i 2
i=l i=l i=l
= ~ (1 - ~) (2 - ~) + 9 (1 - ~) ~ 18.By Thm. 7.2.12 ( c), f is integrable on [2, 5] and J: f = 18.
(b) Let Pn = {1, 1+~' 1 + ~, · · · , 1 +^5 : }. For each i, Xi = 1 + ~ and
6i = ~· Since f is increasing on [1, 6], mi = f(xi-1) and M i = f(xi )· Thus,
= 25 + 725 (1 + ~) + 1~5 (1 + ~) (2 + ~) ~ 6~5.Q_(f, Pn) = i~ mi6i = ~ i~ [ ( 1 + 5(i~l)) 2 + ( 1 + 5(i~l)) + 3]
[n n-1 n-1 ]
= ~ i~ 5 + ~ i~ i + ~ i~ i2= 25 +^7 { (1 - ~) +^1 ~^5 (1 - ~) (2 - ~) ~ 6~^5.
By Thm. 7.2.12 (c), f is integrable on [1, 6] and J 16 f =^6 ~^5 = 104. 16.
(c) Let P n = { (^0) 'n'n' 1 i · · · 'n^2 n }. For each i ' x i · = ~ n and 6 · i = 1. n Since f
is increasing on [O, 2], mi = f (xi_i) and M i = f (xi)· Thus,
n n 3 n 2 2 2
S(j ' P n ) = U "'Mi · 6 t · = (^1) n "' L...t (~) n = 12. n4 "'L...t i^3 = 12. n4 n (n+l) 4 = 4 (1 + .!.) n ~ 4· '
i=l i=l i=l
S(j p ) = "'m·6· n = 1"' n ( 2(i-l). )^3 = 16 n"'i3 -1 = 12. (n-1)^2 n^2 ~
- ' n L.,; i i n L.,; n n4 L-1 n4 4 4.
i=l i=l i=l
By Thm. 7.2.12 (c), f is integrable on [O, 2] and J~ f = 4.
(d) Let P n = {-1 ' -1 + l n' -l + 2. n' · · · ' -1 +^3 n n } · For , each i ' x t · = -1 + ~ n
and 6i = ~· Since f is decreasing on [-1, 2], mi= f(xi ) and Mi= f(xi_i).