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FDWK_3ed_Ch05_pp262-319

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Section 5.4 Fundamental Theorem of Calculus 295<br />

We can therefore continue our proof, letting 1h xh<br />

x<br />

f t dt f c,<br />

d F<br />

lim 1 dx<br />

h→0<br />

xh<br />

h f t dt<br />

x<br />

lim f c, where c lies between x and x h.<br />

h→0<br />

What happens to c as h goes to zero As x h gets closer to x, it carries c along with it like<br />

a bead on a wire, forcing c to approach x. Since f is continuous, this means that f c<br />

approaches f x:<br />

lim f c f x.<br />

h→0<br />

Putting it all together,<br />

d F<br />

lim Fx h Fx<br />

Definition of derivatives<br />

dx<br />

h→0 h<br />

lim<br />

h→0<br />

xh<br />

x<br />

f t dt<br />

h<br />

Rules for integrals<br />

lim<br />

h→0<br />

f c for some c between x and x h.<br />

f x.<br />

Because f is continuous<br />

This concludes the proof.<br />

It is difficult to overestimate the power of the equation<br />

■<br />

x<br />

d<br />

d<br />

x<br />

a<br />

f t dt f x. (1)<br />

It says that every continuous function f is the derivative of some other function, namely<br />

a<br />

x<br />

f t dt. It says that every continuous function has an antiderivative. And it says that<br />

the processes of integration and differentiation are inverses of one another. If any equation<br />

deserves to be called the Fundamental Theorem of Calculus, this equation is surely<br />

the one.<br />

EXAMPLE 1<br />

Find<br />

Applying the Fundamental Theorem<br />

d<br />

d<br />

x<br />

x<br />

<br />

by using the Fundamental Theorem.<br />

d<br />

cos tdt and d x<br />

x<br />

0<br />

1<br />

dt 1 <br />

t 2<br />

SOLUTION<br />

x<br />

x<br />

d<br />

d<br />

x<br />

d<br />

d<br />

x<br />

<br />

0<br />

cos tdt cos x<br />

Eq. 1 with ft cos t<br />

1 1<br />

1<br />

dt . Eq. 1 with ft <br />

1 1 1 <br />

t 2<br />

x 2<br />

t 2<br />

Now try Exercise 3.

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