Calculus: Analytic Geometry and Calculus, with Vectors

226 Integrals

bounded and hence that there is a constant positive M for which -M <_

f (x) S M or I f (x) I<= M. Therefore,

1F(x + Ax) - F(x) I <

I fxz+AxM dt M

The sandwich theorem then implies that

(4.354) lim F(x + Ax) = F(x)

Ax- o

and hence that F is continuous at x. It can be observed that we have

proved more than was promised; the function F must have bounded

difference quotients. To prove (4.352), let x be a point at which f is

continuous. From the two formulas

F(x + Ax) - F(x) 1 rx+-'x

Ax

_

Ax

f (t) dt, Ax) =

Ax 1

x+Axf(x)

dt

we obtain

F(x + Ax) - F(x) -

(4.355)

Ax Ax) Yx x - f(x)]

dt.

Let e > 0. Choose a number 3 > 0 such that

If(t) - f(x) I < E/2 (it - xI < s).

Then when JAxI < S, we can use Theorems 4.343 and 4.341 to obtain

F(x + Ax) - F(x)- AX)

I

<

Ax I

1 x+Az

x x

If(t) - f(x) I dt

1 fx+AxE

2 dt = 2 < E.

Ax x

A

Therefore,

(4.356) lim

F(x + Ax) - F(x) = f(x)

AX-0 Ax

and (4.352) follows from the definition of F'(x). This completes the proof

of Theorem 4.35.

Supposing now that f is continuous over a < x < b, we proceed to show

how Theorem 4.35 can be used to obtain the promised method for evalu-

ating Riemann integrals. Putting x = a in (4.351) shows thatF(a) = 0.

Putting x = b in (4.351) and then changing the dummy variable of inte-

gration from t to x gives

F(b) = f a'f(x) dx.

Therefore,

(4.36) fab f(x) dx= F(b) - F(a).