Chapter 6.3, Problem 34E

To determine

To find:The average value of the function on the interval, using antiderivative to compute the integral.

  $y=\frac{1}{1+x^2},[0,]$

Answer to Problem 34E

  $\frac{\pi }{4}$

Explanation of Solution

Given information: $y=\frac{1}{1+x^2},[0,]$

Calculation:

  $\begin{array}{l}av(f)=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}\\ =\frac{1}{1-0}{\displaystyle {\int }_0^1\frac{1}{1+x^2}dx}\\ =\frac{1}{1}{\left[{\tan }^{-1}x\right]}_0^1\\ =1\left[{\tan }^{-1}(1)-{\tan }^{-1}(0)\right]\\ =1\left[\frac{\pi }{4}-0\right]\\ =\frac{\pi }{4}\end{array}$

The average value of $y=\frac{1}{1+x^2}$ over the interval $[0,]$ is $\frac{\pi }{4}$ .