Answer
$= \frac{32}{3}$
Work Step by Step
$\int^{0}_{-2} (4-x^{2}) dx + \int^{2}_{0} (4-x^{2}) dx$
$= [4x - \frac{x^{3}}{3} |^{0}_{-2}] + [4x - \frac{x^{3}}{3} |^{2}_{0}]$
$= [(4(0) - \frac{0^{3}}{3}) - (4(-2) - \frac{(-2)^{3}}{3})] + [(4(2) - \frac{2^{3}}{3}) - (4(0) - \frac{(0)^{3}}{3})]$
$= [(0) - (-8 - \frac{-8}{3})] + [(8 - \frac{8}{3}) - (0)]$
$= [- (-\frac{16}{3})] + [\frac{16}{3}]$
$= [\frac{16}{3}] + [\frac{16}{3}]$
$= [\frac{16}{3}] + [\frac{16}{3}]$
$= \frac{32}{3}$
