# How do you find the average value of the function for f(x)=root3(x), 0<=x<=8?

Nov 8, 2017

$\frac{1}{8 - 0} {\int}_{0}^{8} \sqrt[3]{x} \setminus \mathrm{dx} = \frac{3}{2}$

#### Explanation:

The average value of a function $f \left(x\right)$ from $a$ to $b$ is given by $\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \setminus \mathrm{dx}$.

Just input $f \left(x\right) = \sqrt[3]{x} , a = 0 , b = 8$ into the formula: $\frac{1}{8 - 0} {\int}_{0}^{8} \sqrt[3]{x} \setminus \mathrm{dx}$.

Evaluate this integral using the power rule to get $\frac{1}{8} {\left[\frac{3}{4} {x}^{\frac{4}{3}}\right]}_{0}^{8}$.

This gives $\frac{1}{8} \cdot \frac{3}{4} \cdot {8}^{\frac{4}{3}} - 0 = \frac{3}{2}$