MvCalc4

$I=\int_0^3[\int_1^2 xy(x+y)dx]dy\,$

$\int_1^2 xy(x+y)dx=\int_1^2 x^2y dx+xy^2 dx\,$

= $[\frac{x^3y}{3}+\frac{x^2y^2}{2}]_{x=1}^{2}\,$

= $[\frac{8y}{3}-\frac{y}{3}]+[\frac{4y^2}{2}-\frac{y^2}{2}]\,$

= $\frac{7y}{3}+\frac{3y^2}{2}\,$

Therefore, $I=\int_0^3 [\frac{7y}{3}+\frac{3y^2}{2}]dy\,$

= $\int_0^3\frac{7y}{3}dy+\int_0^3\frac{3y^2}{2}dy\,$

= $[\frac{7y^2}{6}]_0^3+[\frac{y^3}{2}]_0^3\,$

= $[\frac{7\cdot 9}{6}]+[\frac{3^3}{2}]\,$

= $\frac{21}{2}+\frac{27}{2}=\frac{21+27}{2}=\frac{48}{2}=24\,$