Calculus: Analytic Geometry and Calculus, with Vectors

350 Functions, graphs, and numbers

represents an Archimedes integral, then the integral exists and has the value 1,

but that if the symbol represents a Riemann integral, then the integral does not

exist.

5 For each n = 3,4, 5, the broken line joining in order the points

(0,0) (1/n,n), (2/n,0), (1,0) is the graph of a function fn defined over the interval

0 5 x 5 1. Prove that

(1) lim

roi

dx = 1, Jot [ lim f (x)] dx = 0.

n-- ro n-+ m

Hint: Observe that f,(0) = 0 for each n > 3 and hence lim 0. If 0 <

n--

x 5 1, then fn(x) = 0 when 2/n < x, and hence when n > 2/x, so again lim f (x)

n-,

= 0. Remark: Persons who push very far into the theory of Fourier series learn

that if

(2) Fn(x) _? (sin nx12

na\sinxI

(n=1,2,3, )

then

(3) lim /2 F(x) dx= dx =

w

the in

and us to a

and the formulas can bevalid.

6 This problem invites investment of time in a speculative venture. It was

proved in Section 4.3 that the formula

(1) f ab F(x) dx=F(x),a= F(b) - F(a)

is valid whenever F has a continuous derivative over the interval a 5 x < b.

It was proved in Problem 18 of Problems 5.59 that (1) is valid whenever F'

exists and is Riemann integrable over the interval a _<- x <- b. Even though

nobody requires us to learn everything, we may sometime be benefited by knowl-

edge that there is a function F for which (i) F'(x) exists when -1 <= x 5 1 and

(ii) F' is not continuous but is Riemann integrable over -1 5 x 5 1. Let F

be defined by the formulas F(0) = 0 and

(2) F(x) = -x2 cos 1 + I:2t cos^1 dt (x 0).

x o t

That F'(0) = 0 can be proved by using the inequality

(3)

IF(x)-0(0)ICIxcoszI+I

xJo

I2t(dtI <_21xl

and the sandwich theorem. When x 74 0, differentiating (2) gives

(4) F'(x) = sin 1 (x 0 0).