Answer
$\dfrac{1}{6}$
Work Step by Step
$$I= \int_{0}^{1} [x^2y+\dfrac{y^3}{3}]_0^{1-x} dx \\= \int_{0}^{1 } x^2(1-x) y+\dfrac{1}{3} (1-x)^3dx \\= \int_{0}^{1 } \dfrac{1-4x^3-3x+6x^2}{3} dx \\= [ \dfrac{x-x^4-\dfrac{3x^2}{2}+2x^3}{3}]_0^1 \\=\dfrac{1}{6}$$
