Answer
$\int^{1}_{0}\int^{x}_{x^{2}}(1+2y) dy dx = \frac{3}{10}$
Work Step by Step
$\int^{1}_{0}\int^{x}_{x^{2}}(1+2y) dy dx = \int^{1}_{0}[y+y^{2}]^{y=x}_{y=x^{2}}dx = \int^{1}_{0}[x+x^2-x^2-(x^2)^2]dx =$
$\int^{1}_{0}(x-x^4)dx = [\frac{1}{2}x^2-\frac{1}{5}x^5]^1_0 = \frac{1}{2}-\frac{1}{5}-0+0$ $= \frac{3}{10}$
