Chapter 6.3, Problem 18E

To determine

To Find:Theaverage value of the function on the given interval without integrating.

Answer to Problem 18E

Theaverage value of the function on the given interval without integrating is 0.

Explanation of Solution

Giveninformation :

  $f(\theta )=\tan \theta ,\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$

Calculation :

If $f$ is integrated on $[a,b],$ , it's average (mean) value on $[a,b]$ is

  $av(f)=\frac{1}{b-a}{{\displaystyle \int }}_a^{b}f(x)dx$

So,

  $\begin{array}{cc}av(f)& =\frac{1}{\frac{\pi }{4}-{\left(-\frac{\pi }{4}\right)}^{\frac{\pi }{4}}\int \frac{\pi }{4}}\tan \theta d\theta \\ & =\frac{2}{\pi }{{\displaystyle \int }}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\tan \theta d\theta \end{array}$

Now,

For any integrated function,

  ${\displaystyle {\int }_a^{b}f}(x)dx=($ area above the $x$ -axis) $-($ area below the $x$ -axis)

  

So, the area above the x-axis and area below the x-axis are equal.

  $\begin{array}{cc}{{\displaystyle \int }}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\tan \theta d\theta & =\text{area above the}x\text{-axis}-\text{area below the}x\text{-axis}\\ & =0\end{array}$

Hence,

  $\begin{array}{cc}av(f)& =\frac{2}{\pi }{{\displaystyle \int }}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\tan \theta d\theta \\ & =\frac{2}{\pi }\times 0\\ & =0\end{array}$