680 Appendix C • Answers & Hints for Selected Exercises
n n
(a) For xi = Xi-1, R(f, Q~) = ~ I: l(xi_i) = ~ I: 1 = ; = 1.
i=l i=l
n n
(b) For xi =Xi, R(f, Q~) = ~ I: l(xi) = ~ I: 1 = ; = 1.
i=l i=l
(c) For xi = Xi-~+xi' R(f, Q~) = ~ i~ I ( Xi-~+xi) = ~ i~ 1 = ; = 1.(d) 2 ; [l(O) + 2l(x1) + 2l(x2) + · · · + 2l(Xn-i) + l(xn)] = ~~ = 1.
(e) 3 (~n) [f(O) + 4l(xi) + 2l(x2) + · · · + 4l(x2n-i) + l(x2n)]
= 6 ; [2(1) +^2 ;1(4) + (2; - 1) (2)) = 6 ;[2 + 4n + 2n - 2] = 1.
EXERCISE SET 7.4- Modify the proof of (a) as appropriate.
- Suppose I is integrable on [a, b], and [c, d] s;;; [a, b]. If a = c < d < b, or if
a < c < d = b, then I is integrable on [c, d] by Thm. 7.4.2. If a < c < d < b,
the desired conclusion follows from Cor. 7.4.3. - Use mathematical induction. The case n = 1 is trivial; the case n = 2 is
Thm. 7.4.5. Suppose the theorem is true when n = k. Suppose I is integrable
on each subinterval created by P = {x1, x2, · · · , Xk+1}. By Thm. 7.4.5, I is
integrable on [xo, xk ] and [xk, Xk+1L and
Since P' = { x1, x2, · · · , xk} is a partition of [xo, xk], our induction hypoth-
esis says that I is integrable over [xo, x1], [xo, x2], · · · , [xk-i, xk] and
I:k I = i~ (I:i~l I). (2)
b k+l ( x · )
Putting (1) and (2) together, fa I= i~ fx;_ 1 I ·- (i) By Thm. 7.4.2, I is integrable on [a, b - h] and [b - h, b], and J: I =
J:-h I+ JLh f. Thus, by Exercise 6, II: I -J:-h II = IJLh II :S Mh.
Let c > 0. Choose o = c/M. Then 0 < h < 5 ~ II: I - J:-h II < M(j = c,
so h->O+ lim ta -h I= t a f.
(ii) For any c E (a, b), h->O+ lim t+-hh a I = h->O+ lim [r+h a I+ t-h c 1] = fac I+
t I = J: I by earlier parts of the proof.