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

Dec 6, 2016

$\frac{121}{15}$

#### Explanation:

Since f(x) is continuous on the closed interval [-2 ,1] the average value is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where [a ,b] is the closed interval.

$\Rightarrow \frac{1}{1 + 2} {\int}_{-} {2}^{1} \left(- {x}^{4} + 2 {x}^{2} + 4\right) \mathrm{dx}$

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

$\frac{1}{3} \left[\left(- \frac{1}{5} + \frac{2}{3} + 4\right) - \left(- \frac{32}{5} - \frac{16}{3} - 8\right)\right]$

$= \frac{1}{3} \left[\frac{67}{15} - \left(- \frac{296}{15}\right)\right] = \frac{1}{3} \left(\frac{363}{15}\right) = \frac{121}{15}$