# How do you find the average value of f(x)=-x^5+3x^3 as x varies between [0,1]?

Nov 25, 2016

$\frac{7}{12}$

#### Explanation:

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

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

Which here gives us the average value

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

Integrating term by term using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$ gives

$= {\left[- {x}^{6} / 6 + \frac{3 {x}^{4}}{4}\right]}_{0}^{1}$

The $0$ part of this will still be $0$, leaving us with just

$- \frac{1}{6} + \frac{3}{4} = \frac{7}{12}$