Answer
$\frac{13}{3}$
Work Step by Step
$\int^{1}_{0} \int^{\sqrt{x}}_{-x}1dydx +\int^{4}_{0} \int^{\sqrt{x}}_{x-2} 1dydx $
=$\int^{1}_{0} [y]^{\sqrt{x}}_{-x}dx+\int^{4}_{0} [y]^{\sqrt{x}}_{x-2}dx$
=$\int^{1}_{0} (\sqrt{x}+x)dx+\int^{4}_{0} (\sqrt{x}-x+2)dx$
=$[\frac{2}{3}x^{3/2}+\frac{1}{2}x^2]^1_0+[\frac{2}{3}x^{3/2}-\frac{1}{2}x^2+2x]^4_1$
=$\frac{13}{3}$
