## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{2}{3}$
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, $\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{2}{3}$