Calculus: Analytic Geometry and Calculus, with Vectors

220 Integrals

10 In a campaign to obtain good ideas about Riemann sums and integrals,

we can use the discontinuous function 4, defined over the interval 0 i x 5 1,

for which

(1)

O(x) = 0 when x s 1 1 1 1 1

1 when x =

1

and m is a positive integer.

M

Sketch a figure which shows the nature of the graph of (P. Then mark the end

points xa, xi, - , x,,and the intermediate points x1 , x2 ,. ., xn of a partition

P of the interval 0 < x <= 1 for which 1PI < 0.1; to mark the end points of a

partition for which JPJ < 0.0001 would be a tedious operation requiring sharper

pencils and better microscopes than we normally carry around. Try to see

reasons why the Riemann sum

(2) (xk) Axk

k=1

must be near 0 whenever JP1 is near 0 and hence that

(3) fo1 0(x) dx= 0.

Then start cultivating the art of understanding and originating thoughts about

Riemann sums more or less like the following. Let e be a given positive number

for which 0 < e < 1. Let h = e/10 and suppose at first that IPI 5 h. The

terms of the Riemann sum (2) are all nonnegative. Those terms for which xk

can be a point of the interval 0 S x <= h contribute at most 2h to the sum.

Those terms for which xk > h will be 0 unless xk = 1/m, where in is an integer

for which 1/m > h or in < 1/h. Thus there are less than 1/h nonzero terms for

which xk > Jr and O(xk) 5,6 0. Since each one of these terms can contribute at

most JPI to the sum of these terms, it follows that the sum of all of these terms

cannot exceed (1/h)jP1. Therefore,

(4) 0 S 1 0(xk) Lixk S 2h + h JPJ < 0.2e +

101Pl-

k=1

If we let S = 2E2/25, we will have

(5) 0 < LI O(xk)Oxk < E

k=1

whenever JPJ < 3. This implies the first and hence the second of the formulas

(6) Iimo 4(xk) .xk = 0, f olo(x) dx = 0.

k-1

11 Supposing that g is the corn-popper function of Problem 16 of Problems

3.49, determine the value (if any) of r0 1 g(x) dx.