Calculus: Analytic Geometry and Calculus, with Vectors

5.7 Darboux sums and Riemann integrals 349

whenever xk-1 < xk < xk for each k. Defining Mk and mk by (5.712), we

conclude that

n n

(5.783) k1 Mk Oxk C ISI + E', (^1) I Mk Oxk > ISI -. E'.
k
Let Rk and Rk be, for each k, the rectangles (meaning rectangular regions)
consisting of points (x,y) for which xk-1 < X < Xk, 0 < y < Mk and
Xk-1< x < xk, Mk < y < M. The two formulas (5.783) then give
n n)t
(5.784) k11 IRkI < ISI + e, kI IRkI < IRI - ISI +
If P(x,y) lies in S, then there is at least one k for which xk_1 5 x S Xk and
hence 0 <= y < f (x) < Mk, so P is a point of at least one rectangle Rk.
Similarly, if P(x,y) is a point of the set S' consisting of the points in R but
not in S, then there is at least one k for which xk_1 < x 5 xk and hence
mk < f(x) S M, soP is a point of at least one rectangle R. It is there-
fore a consequence of Definition 4.4 that the set S does possess an area
and that its area is ISI. This completes the proof of Theorem 5.78.

Problems 5.79

1 Sketch a dozen graphs that look like graphs of functions f that are bounded

and piecewise monotone over the interval 0 =< x < 1. Be sure to include graphs

of some discontinuous functions and of some nonmonotone functions.

2 Sketch a figure which is like Figure 5.721 except that the partition P con-

tains 10 or 20 partition points that are roughly equally spaced. Then look at

your figure and see how LDS(P) and UDS(P) are related.

3 Sketch a figure which shows the geometric meanings in the statement and

proof of Theorem 5.76.

4 As was remarked, Archimedes (287-212 s.c.) knew about some special

Riemann sums, and this matter may be worthy of brief consideration here.

When f is defined over rational values of x in the interval 0 < x < 1, we can

make a partition of the interval 0 <= x =< 1 into n equal subintervals of length

1/n by partition points xk for which xk = k/n and form the special Riemann sum

`4n - f

k-1 n n

which we can call an Archimedes sum. Without implying that Archimedes used

modern terminology involving sums, limits, and integrals, we can recognize that

there is historical evidence that we are merely putting ideas of Archimedes into

modern terminology when we say that f is Archimedes integrable over the interval

0 5 x < 1 and that f has the Archimedes integral I if A --- I as n ---+ oo. Now

comes the problem. Supposing that f(x) = 0 when x is irrational and f(x) = 1

when x is rational, show that if the symbol

foi f(x) dx