# How do you find the average value of the function for f(x)=x^2-sqrtx, 0<=x<=3?

Oct 5, 2017

$3 - \frac{2 \sqrt{3}}{3}$

#### Explanation:

Recall the formula for determining the average value of a function $f \left(x\right)$ on an interval from $\left[a , b\right]$:

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

For this problem:

$A v g = \frac{1}{3 - 0} {\int}_{0}^{3} \left({x}^{2} - \sqrt{x}\right) \mathrm{dx}$

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

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

$= \frac{1}{3} \left({3}^{3} / 3 - \frac{2}{3} {3}^{3} / 2\right) = \frac{1}{3} \left(\frac{27}{3} - \frac{2}{3} \cdot 3 \sqrt{3}\right)$

$= \frac{1}{3} \left(9 - 2 \sqrt{3}\right) = 3 - \frac{2 \sqrt{3}}{3}$