Answer
$\frac{3}{2}$
Work Step by Step
$\int^{1}_{0} \int^{2x}_{ x/2} 1dy dx+\int^{2}_{0} \int^{3-x}_{x/2} $
=$\int^{1}_{0} [y]^{2x}_{x/2}dx+\int^{2}_{0} [y]^{3-x}_{x/2}$
=$\int^{1}_{0} (\frac{3}{2}x)dx+\int^{2}_{0} (3-\frac{3}{2}x)dx$
=$[\frac{3}{4}x^2]_0^1+[3x-\frac{3}{4}x^2]^2_1$
=$\frac{3}{2}$
