Calculus: Analytic Geometry and Calculus, with Vectors

4.4 Areas and integrals 235

When (2) holds and u is a differentiable function of x, we can use the chain rule

to obtain

(3) dxfouf(t) dt=[dufou f(t) dt]dx

= f(u)du

Use these ideas to show that (1) is 2xe -'.
19 Letting
2T
F(x) = o e-1 dt,

use the ideas ofthe preceding problem to obtain a simple formula for F'(x).

Then find F(x), that is, find a simpler expression for F(x), and differentiate it to

obtain F'(x). Make the results agree.

20 Prove that if f is continuous and u and v are differentiable, then

TX!Haf(t) dt = f(u)

du

• f(v) dx

Hint: Use the formula

fu f(t) dt=fuf(t) dt-f"f(t) dt

and the ideas of Problem 18:

21 Supposing that A is a positive constant, x > 0, and

F(x) _

f

az 1

dt,

x t

show that F(x) = 0 without use of the formulaf dt = log t + c.

4.4 Areas and integrals We all know what is meant by a rectangular

region R having base length b and height h. When the x axis of a rec-

tangular coordinate system is parallel to the base, R is the set of points

(x,y) for which xo < x < xo + b and

yo < y < yo + h as in Figure 4.41.

We are all familiar with the idea that yo+h

R has an area and that this area is

bh, the product of the base length

and the height. There is an old- p xo xo+b x

fashioned view that this matter is

quite simple, but modern mathema-

Figure 4.41

ticians, like modern atomic physicists, find that there is much to be

learned about things that our ancestors thought were simple. It is

quite absurd to presume that it is easy to prove that the area of R is bh;

in fact it is quite absurd to presume that it is possible to prove that there

is a number (bh or not) which is the area of R unless we have some defini-

tions or postulates or something upon which proofs can be based. We