Chapter 6.3, Problem 32E

To determine

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

  $y=\frac{1}{x},[e,2e]$

Answer to Problem 32E

  $\frac{1}{e}\ln 2$

Explanation of Solution

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

Calculation:

  $\begin{array}{l}av(f)=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}\\ =\frac{1}{2e-e}{\displaystyle {\int }_e^{2e}\frac{1}{x}dx}\\ =\frac{1}{e}{\left[\ln x\right]}_e^{2e}\\ =\frac{1}{e}\left[\ln (2e)-\ln (e)\right]\\ =\frac{1}{e}\left[\ln \frac{2e}{e}\right]\\ =\frac{1}{e}\left[\ln 2\right]\\ =\frac{1}{e}\ln 2\end{array}$

The average value of $y=\frac{1}{x}$ over the interval $[e,2e]$ is $\frac{1}{e}\ln 2$ .