# How do you find the average value of the function for f(x)=4-x^2, -2<=x<=2?

Feb 25, 2017

The average value of the function $f \left(x\right)$ on $a \le x \le b$ is:

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

So here the average value is:

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

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

$A = \frac{1}{4} \left(4 \left(2\right) - \frac{1}{3} {\left(2\right)}^{3}\right) - \frac{1}{4} \left(4 \left(- 2\right) - \frac{1}{3} {\left(- 2\right)}^{3}\right)$

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

$A = \left(2 - \frac{2}{3}\right) + \left(2 - \frac{2}{3}\right)$

$A = 4 - \frac{4}{3}$

$A = \frac{8}{3}$