Calculus: Analytic Geometry and Calculus, with Vectors

4.3 Properties of integrals 231

and we look at an example. Let A be a positive number and start with the fact

that

(1)

when 0 < x <= A. Replace x by t in (1) and integrate over the interval 0 <--_t 5

x to obtain,with the aid of Theorem 4.34,

(2) x < es - 1 S eAx

1 S ez <eA

(05x511).

Replace x by t in (2) and integrate over the interval 0 5 t < x to obtain

(3) 22 5 et- (1 + x)^5 e4 (05x51).

Continue the process to obtain

(4)

(5)

3!<e -+x+2!)5e3!

'x4 <ex_ (1 +x+'x2+x3) eAx'

¢! = 2! 3! 4!

when 0 <= x 511. Remark: Continuation of the process (with the aid of mathe-

matical induction) shows that, for each positive integer n,

(6) x-I 2 i

While we now have so many other things to do that we shall not look at the details,

we can observe that (6) provides a straightforward and foolproof way to obtain

decimal approximations to e',e, e2, etcetera, correct to 4 or 40 decimal places.

We can discard much of the information in (6) by observing that x"/n! approaches

0 as n -> oo and hence that

(7) es = lim (1 -{-x -{-x2 x3 x"l

' !-{- -}- i

The formula (7) is the spectacular one, but (6) is often much more useful. We

shall learn more about these things later.

9 Applying the idea of the preceding problem to the inequality

show that

-1 < sin x 5 1,

-x_-<1-cosx__<x,

• x2S x2
  2

x - sinx<=

2

when x > 0. Remark: More extensive information will appear in a problem of

Section 8.2.

10 The Bernoulli functions Bo(x), B1(x), B2(x), satisfy the conditions

(1) Bo(x) = 1

(2) B'n(x) = B,, i(x) (n = 1,2,3, ...)

(3) 101 dx = 0 (n = 1,2,3, ...)