Chapter 6.5, Problem 6E

To determine

To find the average value of the given function on the given interval.

Answer to Problem 6E

The average of a given function is:

Hence, the average of a given function is: $h_{average}=0.175$

Explanation of Solution

Given: The given function is: $h(u)={(3-2u)}^{-1},\left[-1,1\right]$

Calculation:

Let, $\left[a,b\right]=\left[-1,1\right]$

The average of a given function is:

  $\begin{array}{l}{h}_{average}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}h(u)du}\\ \text{ =}\frac{1}{\left(1-\left(1\right)\right)}{\displaystyle \underset{-1}{\overset{1}{\int }}{(3-2u)}^{-1}}du\\ Assume3-2u=x\\ \Rightarrow Apply\text{ derivative on both sides, then }\\ \Rightarrow 0-2du=dx\\ \Rightarrow \text{-2du = dx }\\ \Rightarrow du=\frac{dx}{-2}\\ Limits\text{ : If u=-1 then x=5 and if u=1 then x=1}\text{.}\\ Substitute\text{ the above in average equation,}\\ h_{average}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}h(u)du}\\ \text{ =}\frac{1}{\left(1-\left(1\right)\right)}{\displaystyle \underset{-1}{\overset{1}{\int }}{(3-2u)}^{-1}}du\\ \text{ =}\frac{1}{2}{\displaystyle \underset{5}{\overset{1}{\int }}{x}^{-1}\frac{dx}{-2}}\\ \text{ = -}\frac{1}{4}\left[{\displaystyle \underset{5}{\overset{1}{\int }}\frac{1}{x}\text{ dx}}\right]\\ $ \text{ = -}\frac{1}{4}{[\log (x)]}_5^1$\text{ = -}\frac{1}{4}\left[\log (1)-\log (5)\right]\\ \text{ = -}\frac{1}{4}\left[0-0.7\right]\\ \text{ = }\frac{0.7}{4}\\ \text{ = 0}\text{.175}\end{array}$

Hence, the average of a given function is: $h_{average}=0.175$