Chapter 5.2, Problem 52E

To determine

To express the value of the integral.

Answer to Problem 52E

The value of the integral is $\frac{2}{5}\le {\int }_0^2\frac{1}{1+x^2}dx\le 2$ .

Explanation of Solution

Given information:

The given equation is

  ${\int }_0^2\frac{1}{1+x^2}dx$

Formula used:

Used the comparison property

  $\begin{array}{l}\text{if }m\le f\left(x\right)\le M\text{ for }a\le x\le b,\text{then}\\ m\left(b-a\right)\le {\int }_a^{b}f\left(x\right)dx\le M\left(b-a\right)\end{array}$

The value of the integral can be express as

Find a values of m and M

The values of $a=0,b=2\text{ and }f\left(x\right)=\frac{1}{1+x^2}$

  $\begin{array}{l}b-a\\ =2-0\\ =2\end{array}$

Put the value of $x=0\text{ and }x=2\text{ in }f\left(x\right)=\frac{1}{1+x^2}$

  $\begin{array}{l}f\left(0\right)=\frac{1}{1+0^2}\\ \text{ =}\frac{1}{1}\\ \text{ =1}\\ \text{and}\\ f\left(2\right)=\frac{1}{1+2^2}\\ \text{ =}\frac{1}{1+4}\\ \text{ =}\frac{1}{5}\end{array}$

We have restricted function fand obtained value for

  $m=\frac{1}{5}\text{ and }M=1$

Therefore,

According to comparison properties of integrals, we have

  $\begin{array}{l}\frac{1}{5}\left(2-0\right)\le {\int }_0^2\frac{1}{1+x^2}dx\le 1\left(2-0\right)\\ =\frac{1}{5}\left(2\right)\le {\int }_0^2\frac{1}{1+x^2}dx\le 1\left(2\right)\\ =\frac{2}{5}\le {\int }_0^2\frac{1}{1+x^2}dx\le 2\end{array}$

Hence,

The value of the integral is $\frac{2}{5}\le {\int }_0^2\frac{1}{1+x^2}dx\le 2$ .