1549901369-Elements_of_Real_Analysis__Denlinger_

7.6 The Fundamental Theorem of Calculus 407

Proof. Suppose f is integrable on a compact interval I , and a E J. Define

the function F on I by the formula F(x) = J: f. Let x 0 be a point of 1° at

which f is continuous. Then, '\Ix =/= x 0 in I,

F(x) - F(xo) = J: f - J:^0 f

= J; 0 f by Theorem 7.6.5.

Hence, F(x) -F(xo) ____ l _ 1x f.

x -xo x - xo xo

1

Moreover, f(xo) = --(x -xo)f(xo)

x-xo

= --^1 1x f(xo) (integral of constant).

X - Xo xo

Thus, F(x) -F(xo) - f(xo) = -^1 - [1x f -1x f(xo)]

x - xo x -xo xo x 0

= --^1 1x [f(t) -f(xo)] dt.

x - xo xo

Let i:: > 0. Since f is continuous at xo, 3 8 > 0 3

/t - xo/ < 8 * /f(t) - f(xo)/ < i::.

Case 1 (x > xo): Then, /x -xo/ < 8 * xo < x < xo + 8, so

11 : [f(t) - f(xo)] dtl ~ 1: /f(t) - f(xo)/ dt by Theorem 7.5.5

~ J; 0 E: by (23) and Corollary 7.5.2

= i::(x -x 0 ) (integral of a constant).

Case 2 (x < xo): Then, /x -xo/ < 8 * xo - 8 < x < xo, so

11: [f(t) - f(xo)] dtl = 11 xo [f(t) - f(xo)] dtl ~ i::(x 0 - x)

(following the same reasoning as in Case 1).

(22)

(23)

In either case, /x - xo/ < 8 * IJ: 0 [f(t) - f(xo)] dtl ~ i::/x - xo/- (24)

Combining (22) and (24), we have 0 < /x - xo/ < 8 *

I

F(x) - F(xo) - f(xo)I = 1 llx [f(t) - f(xo)] dtl

x -xo /x - Xo I xo

1

~I

1

-i::/x-xo/=i::.

X -Xo