# How do you find the point c in the interval 0<=x<=2 such that f(c) is equation to the average value of f(x)=sqrt(2x)?

Apr 14, 2017

#### Explanation:

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

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

So we need to find the average value, then solve the equation $f \left(x\right) = \text{average value}$ on the interval $\left[a , b\right]$. If the equation has more than one solution then $c$ can be any one of the solutions in $\left[a , b\right]$

In this question we get

Solve: $\sqrt{2 x} = \frac{1}{2 - 0} {\int}_{0}^{2} \sqrt{2 x} \mathrm{dx}$

Evaluating the integral gets us:

Solve: $\sqrt{2 x} = \frac{4}{3}$.

The only solution is $x = \frac{8}{9}$