MvCalc4

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I=\int _{0}^{3}[\int _{1}^{2}xy(x+y)dx]dy\,

\int _{1}^{2}xy(x+y)dx=\int _{1}^{2}x^{2}ydx+xy^{2}dx\,

=[{\frac  {x^{3}y}{3}}+{\frac  {x^{2}y^{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\,

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