# How do you find the average value of f(x)=(x-3)^2 as x varies between [2,5]?

Feb 14, 2018

The average value is $= 1$

#### Explanation:

The average value of a function $f \left(x\right)$ over an interval $\left[a , b\right]$ is

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

Here,

$f \left(x\right) = {\left(x - 3\right)}^{2} = {x}^{2} - 6 x + 9$

and $\left[a , b\right] = \left[2 , 5\right]$

Therefore,

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

$= \frac{1}{3} {\int}_{2}^{5} \left({x}^{2} - 6 x + 9\right) \mathrm{dx}$

$= \frac{1}{3} {\left[{x}^{3} / 3 - 6 {x}^{2} / 2 + 9 x\right]}_{2}^{5}$

$= \frac{1}{3} \left(\left(\frac{125}{3} - 75 + 45\right) - \left(\frac{8}{3} - 12 + 18\right)\right)$

$= \frac{1}{3} \left(\frac{117}{3} - 36\right)$

$= \frac{1}{3} \cdot 3$

$= 1$