   Chapter 4.3, Problem 74E

Chapter
Section
Textbook Problem

Finding a Definite Integral The function f ( x ) = { 0 ,         x = 0 1 x ,       0 < x ≤ 1 Is defined on [0, 1], as shown in the figure. Show that ∫ 0 1 f ( x )   d x Does not exist. Does this contradict Theorem 4.4? Why or why not? To determine

To prove: The integral 01f(x)dx does not exist.

Explanation

Given:

The function

f(x)={0, x=01x, 0 < x1

is defined on [0,1].

The graph of the function is given below.

Formula Used:

If a function f is continuous on the closed interval [a,b], then f is integrable on [a,b].

But the converse of this theorem is that if a function f is integrable on [a,b], then it is not necessary to be continuous on that interval

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