# How do you find the average value of f(x)=x^5-4x^3+2x-1 as x varies between [-2,2]?

Aug 4, 2017

${f}_{a v e} = - 1$

#### Explanation:

The average value of a function is found using the following equation:

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

on some interval $\left[a , b\right]$

Therefore:

${f}_{a v e} = \frac{1}{4} {\int}_{- 2}^{2} \left({x}^{5} - 4 {x}^{3} + 2 x - 1\right) \mathrm{dx}$

This is a basic integral.

=>f_(a v e)=1/4(1/6x^6-x^4+x^2-x)]_(-2)^2

$= \frac{1}{4} \cdot - 4$

$= - 1$

Aug 4, 2017

Morgan has given a fine answer. I want to mention a fact that can simplify the integration needed for this question.

#### Explanation:

A function $f$ is odd if and only if $f \left(- x\right) = - f \left(x\right)$ for all $x$ in the domain of $f$.

If $f$ is an odd function, and integrable on $\left[- a , a\right]$ then ${\int}_{-} {a}^{a} f \left(x\right) \mathrm{dx} = 0$

In this question the first four terms of the polynomial form an odd function that is integrable on any closed interval, so

${\int}_{-} {2}^{2} \left({x}^{5} - 4 {x}^{3} + 2 x - 1\right) \mathrm{dx} = {\int}_{-} {2}^{2} \left(- 1\right) \mathrm{dx}$

$= {\left.- x\right]}_{-} {2}^{2} = - \left(2\right) - \left(- \left(- 2\right)\right) = - 4$