Chapter 5.2, Problem 56E

To determine

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

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{1}{x}$ if $0<x\le 1$ and $f\left(0\right)=0$ .

Proof:

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

  $\Delta x=\frac{b-a}{n}$

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

  $\begin{array}{c}\Delta x=\frac{1-0}{n}\\ =\frac{1}{n}\end{array}$

If left endpoints is used then the height of the first rectangle be 0 because $f\left(0\right)=0$ .Now, the second rectangle made arbitrarily large value. Consider a sample point that is closer to 0 for the first rectangle, such as $x_1^{*}=\frac{1}{n^2}$ .

The area of the first rectangle is as follows:

  $\begin{array}{c}f\left(x_1^{*}\right)\cdot \Delta x=\frac{1}{\frac{1}{n^2}}\cdot \frac{1}{n}\\ =\frac{n^2}{n}\\ =n\end{array}$

The function $f\left(x\right)$ 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,

  ${\displaystyle \sum _{i=1}^{n}f\left(x_1^{*}\right)\cdot \Delta x>n}$

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

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_1^{*}\right)\cdot \Delta x>\underset{n\to \infty }{\lim }n}=\infty$

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

Hence, it is proved that the function is not integrableon $\left[0,1\right]$ .