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

Nov 5, 2017

$\frac{8}{3}$

#### Explanation:

$\text{since f(x) is continuous on the closed interval}$
$\left[- 2 , 2\right] \text{ then the average value of f(x) from x = -2}$
$\text{to x = 2 is the integral}$

•color(white)(x)1/(b-a)int_a^bf(x)dx

$\text{here } \left[a , b\right] = \left[- 2 , 2\right]$

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

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

$= \frac{1}{4} \left[\left(8 - \frac{8}{3}\right) - \left(- 8 + \frac{8}{3}\right)\right]$

$\frac{1}{4} \left(\frac{32}{3}\right) = \frac{8}{3}$