# What is the average value of the function f(x)=(x-1)^2 on the interval from x=1 to x=5?

Mar 24, 2016

The average value is $\frac{16}{3}$

#### Explanation:

The average value of a function $f$ on an interval $\left[a , b\right]$ is

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

So the value we seek is

$\frac{1}{5 - 1} {\int}_{1}^{5} {\left(x - 1\right)}^{2} \mathrm{dx} = \frac{1}{4} {\left[{\left(x - 1\right)}^{3} / 3\right]}_{1}^{5}$

$= \frac{1}{12} \left[{\left(4\right)}^{3} - {\left(0\right)}^{3}\right]$

$= \frac{16}{3}$