Answer
$\frac{3}{2}$
Work Step by Step
$\int^{0}_{-1} \int^{1-x}_{-2x}dydx+\int^{2}_{0} \int^{1-x}_{-x/2}dydx$
=$\int^{0}_{-1}(1+x)dx+\int^{2}_{0}(1-\frac{x}{2})dx$
=$[x+\frac{x^2}{2}]^0_{-1}+[x-\frac{x^2}{4}]$
=$(-1+\frac{1}{2})+(2-1)=\frac{3}{2}$
