BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5.2, Problem 56E
To determine

To show:The function f is not integrable on [0,1] .

Expert Solution

Explanation of Solution

Given information:The function is f(x)=1x if 0<x1 and f(0)=0 .

Proof:

It is known that a function f for the interval [a,b] should be divided into n subintervals of equal width:

  Δx=ban

Consider that the number of rectanglesof subintervals for [0,1] be n . Substitute 0 for a and 1 for b in the above formula to find the width of the rectangle.

  Δx=10n=1n

If left endpoints is used then the height of the first rectangle be 0 because f(0)=0 .Now, the second rectangle made arbitrarily large value. Consider a sample point that is closer to 0 for the first rectangle, such as x1*=1n2 .

The area of the first rectangle is as follows:

  f(x1*)Δx=11n21n=n2n=n

The function f(x) is positive on the interval. If n is a finite number, then the whole sum is greater than the area of the first rectangle. So,

  i=1nf(x1*)Δx>n

Now, increase the value of n toward infinity to get:

  limni=1nf(x1*)Δx>limnn=

The first term and the entire sum tendtowards infinity. A function is not integrable on a certain interval if the sum tend towards .

Hence, it is proved that the function is not integrableon [0,1] .

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