MvCalc9

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I=\int _{1}^{2}\left[\int _{1}^{x}xy^{2}dy\right]dx\,

\int _{1}^{x}xy^{2}dy=\left[{\frac  {xy^{3}}{3}}\right]_{{y=1}}^{{x}}={\frac  {x^{4}}{3}}-{\frac  {x}{3}}\,

Therefore,I=\int _{1}^{2}\left[{\frac  {x^{4}}{3}}-{\frac  {x}{3}}\right]dx\,

=\int _{1}^{2}{\frac  {x^{4}}{3}}dx-\int _{1}^{2}{\frac  {x}{3}}dx\,

=\left[{\frac  {x^{5}}{15}}\right]_{1}^{2}-\left[{\frac  {x^{2}}{6}}\right]_{1}^{2}\,

=\left[{\frac  {32}{15}}-{\frac  {1}{15}}\right]-\left[{\frac  {4}{6}}-{\frac  {1}{6}}\right]\,

={\frac  {31}{15}}-{\frac  {3}{6}}\,

={\frac  {31}{15}}-{\frac  {1}{2}}\,

={\frac  {62-15}{30}}={\frac  {47}{30}}\,

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