# How do you find the average value of f(x)=-x^5+4x^3-5x-3 as x varies between [-2,0]?

Dec 17, 2016

The answer is $= - \frac{2}{3}$

#### Explanation:

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

We need $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

Therefore,

The average value is $\frac{1}{0 - - 2} {\int}_{-} {2}^{0} \left(- {x}^{5} + 4 {x}^{3} - 5 x - 3\right) \mathrm{dx}$

$= \frac{1}{2} {\left[- {x}^{6} / 6 + 4 {x}^{4} / 4 - 5 {x}^{2} / 2 - 3 x\right]}_{-} {2}^{0}$

$= \frac{1}{2} \left(0 - \left(- {2}^{6} / 6 + 16 - 10 + 6\right)\right)$

$= \frac{1}{2} \left(\frac{32}{3} - 12\right)$

$= - \frac{4}{6} = - \frac{2}{3}$