FDWK_3ed_Ch05_pp262-319
FDWK_3ed_Ch05_pp262-319
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.