How do you find the average value of f(x)=x-2sqrtx as x varies between [0,2]?

Apr 9, 2018

Average value is given by

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

Therefore

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

$A = \frac{1}{2} {\left[\frac{1}{2} {x}^{2} - \frac{4}{3} {x}^{\frac{3}{2}}\right]}_{0}^{2}$

$A = \frac{1}{2} \left[2 - \frac{4}{3} \sqrt{8}\right]$

$A = 1 - \frac{2}{3} \left(2\right) \sqrt{2}$

$A = 1 - \frac{4}{3} \sqrt{2} \approx - 0.89$

Hopefully this helps!