# 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)=x^(2/3)?

Mar 28, 2017

Solve the equation $f \left(x\right) = \frac{1}{2 - 0} {\int}_{0}^{2} f \left(x\right) \mathrm{dx}$. Discard any solutions outside $\left[0 , 2\right]$

#### Explanation:

$\frac{1}{2 - 0} {\int}_{0}^{2} f \left(x\right) \mathrm{dx} = \frac{1}{2} {\int}_{0}^{2} {x}^{\frac{2}{3}} \mathrm{dx} = \frac{3 \left({2}^{\frac{2}{3}}\right)}{5}$

So you need to solve

${x}^{\frac{2}{3}} = \frac{3 \left({2}^{\frac{2}{3}}\right)}{5}$.

I get $x = \frac{6 \sqrt{15}}{25}$