MvCalc16

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I=\int _{0}^{a}[\int _{{{\frac  {x^{2}}{a}}}}^{{2a-x}}xydy]dx\,

Now \int _{{{\frac  {x^{2}}{a}}}}^{{2a-x}}xydy=[{\frac  {xy^{2}}{2}}]_{{{\frac  {x^{2}}{a}}}}^{{2a-x}}\,

=[{\frac  {x(2a-x)^{2}}{2}}-{\frac  {x^{5}}{2a^{2}}}]\,

=[{\frac  {1}{2}}(x(4a^{2}-4ax+x^{2})-{\frac  {x^{5}}{2a^{2}}}]\,

=[2a^{2}x-2ax^{2}+{\frac  {x^{3}}{2}}-{\frac  {x^{5}}{2a^{2}}}]\,

Therefore I=\int _{0}^{a}[2a^{2}x-2ax^{2}+{\frac  {x^{3}}{2}}-{\frac  {x^{5}}{2a^{2}}}]dx\,

=[a^{2}x^{2}-{\frac  {2ax^{3}}{3}}+{\frac  {x^{4}}{8}}-{\frac  {x^{6}}{12a^{2}}}]_{0}^{a}\,

=a^{4}-{\frac  {2a^{4}}{3}}+{\frac  {a^{4}}{8}}-{\frac  {a^{4}}{12}}\,

={\frac  {48a^{4}-32a^{4}+6a^{4}-4a^{4}}{48}}={\frac  {18a^{4}}{48}}={\frac  {3a^{4}}{8}}\,

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