Question

Consider the function $f\left(x\right)=\frac{\ln x}{x}$shown here:

(a) Find the average value of$f\left(x\right)$ on role="math" localid="1649276701979" $\left[\frac{1}{2},2\right]$.

(b) Find a value $c\in \left[\frac{1}{2},2\right]$ at which $f\left(x\right)$ achieves its average value.

Short answer

Part (a) The value is $0.$

Part (b) The value is$c=1.$

Step-by-step solution

Part (a) Step 1. Given information.

The given function is$f\left(x\right)=\frac{\ln x}{x}$

Part (a) Step 2. Explanation.

The average value of the function will be,

$$\begin{gathered} \text{Avg.}=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx \\ =\frac{1}{2-\frac{1}{2}}{\int }_{\frac{1}{2}}^2\frac{\ln x}{x}dx \\ =\frac{2}{3}{\left[\frac{1}{2}\left(\ln \right(x){)}^2\right]}_{\frac{1}{2}}^2+c \\ =\frac{2}{3}\left[\frac{1}{2}\left[\left(\ln \right(2){)}^2\right]-{\left(\ln \left(\frac{1}{2}\right)\right)}^2\right]+c \\ =0 \end{gathered}$$

Part(b) Step 1. Explanation.

Now,

At

$$\begin{gathered} c=1, \\ f\left(x\right)=\frac{\ln 1}{1} \\ =0 \end{gathered}$$

Therefore,

The value of $c=1.$