# What is the average value of a function y=6/x on the interval [1,e]?

Jul 30, 2016

$\frac{6}{e - 1} \approx 3.4918602412 \ldots$

#### Explanation:

The average value of the function $f \left(x\right)$ on the interval $\left[a , b\right]$ can be expressed as:

$\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

So, where the function is $f \left(x\right) = \frac{6}{x}$ and the interval is $\left[1 , e\right]$, the average value of the function is:

$\frac{1}{e - 1} {\int}_{1}^{e} \frac{6}{x} \mathrm{dx}$

The $6$ can be brought out of the integrand:

$= \frac{6}{e - 1} {\int}_{1}^{e} \frac{1}{x} \mathrm{dx}$

Note that since the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$, the integral of $\frac{1}{x}$ is $\ln \left(x\right)$.

$= \frac{6}{e - 1} {\left[\ln \left(x\right)\right]}_{1}^{e}$

Now evaluating:

$= \frac{6}{e - 1} \left[\ln \left(e\right) - \ln \left(1\right)\right]$

$= \frac{6}{e - 1} \left(1 - 0\right)$

$= \frac{6}{e - 1}$