Calculus: Analytic Geometry and Calculus, with Vectors

348 Functions, graphs, and numbers

Similarly, we apply the fundamental Theorem 5.731 to prove Riemann

integrability of continuous functions and piecewise continuous functions.

Theorem 5.77 If f is continuous over a < x 5 b, then the Riemann

b

integral f f(x) dx exists.

To prove this, let e > 0. Theorem 5.58 then enables us to choosea

positive number 3 such that If(X2) - f(xi)I < c./(b - a) whenever

a<x1<b,a=< x2__<b,andJx2-x11 <3. LetPbeapartitionofthe

interval a < x < b for which IPI < S. Then, with the notation of

(5.712) and (5.72), we have Mk - mk 5 e/(b - a) and hence

n

(5.771) UDS(P) - LDS(P) < 1 h Oxk = e.

k=1

The required conclusion then follows from Theorem 5.751.

A function f is said to be piecewise continuous over the closed interval

a < x < b if it is defined over a < x < b and if there is a partition P of

the interval a < x < b such that, whenever xk-1 and xk are two con-

secutive partition points, f is continuous over the open interval xk_1 <

x < xk and, in addition, the unilateral limits

lim AX), lim f (x)

X Xk-

both exist. On account of the fact that functions that are piecewise

continuous over a closed interval must be bounded, it is not difficult to

use Theorem 5.77 to prove the following more general theorem.

Theorem 5.772 If f is piecewise continuous over a < x < b, then the

Riemann integral exists.

Finally, we use ideas and notation of this section to prove the following

theorem.

Theorem 5.78 If f is Riemann integrable over the interval a < x <_ b

and f(x) >= 0 when a < x 5 b, then the set S of points (x,y) for which

a <- x <- b, 0 <- y <_ f(x) possesses an area

M b

S, M,,

S

MA

Rk

b

ISI and 1,31 =

J.

f(x) dx.

The proof depends upon the funda-

mental definition of area given in Defini-

tion 4.44. Choose a constant M such

that f(x) <--_ M - 1 and observe that S is

a subset of the large rectangle R of Figure

5.781. Let e > 0 and let 0 < E < e. Let

a Xk_1 Xk

Figure 5.781

X

the number ISI be defined by the formula ISI = f s f(x) dx.a To prove

the theorem, we shall show that ISI is in fact the area of S. Let P be a

partition for which

n nn+

(5.782)

lf(k) Axk < ISI + W,

La f(4) Lxk > S - E

k kffi1