#### Answer

$\dfrac{2}{3}$

#### Work Step by Step

Re-arrange the given integral as follows:
$\int_{0}^{1}[x-\dfrac{x^3}{6}-\dfrac{y^2}{2}]_{0}^{1}]dy=\int_{0}^{1}[(\dfrac{5}{6} -\dfrac{y^2}{2} ) dy$
This implies that
$\int_{0}^{1}[(\dfrac{5}{6} -\dfrac{y^2}{2} ) dy= (\dfrac{5y}{6}-\dfrac{y^3}{6})_{0}^{1}$
Thus, we have, $\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{2}{3}$