MvCalc12

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\int _{0}^{1}[\int _{{{\sqrt  {y}}}}^{{2-y}}x^{2}dx]dy\,

Now \int _{{{\sqrt  {y}}}}^{{2-y}}x^{2}dx\,

=[{\frac  {x^{3}}{3}}]_{{{\sqrt  {y}}}}^{{2-y}}\,

={\frac  {1}{3}}[(2-y)^{3}-({\sqrt  {y}})^{3}]\,

={\frac  {1}{3}}[8-12y+6y^{2}-y^{3}-y{\sqrt  {y}}]\,

Therefore I=\int _{0}^{1}{\frac  {1}{3}}[8-12y+6y^{2}-y^{3}-y{\sqrt  {y}}]dy\,

={\frac  {1}{3}}[8y-{\frac  {12y^{2}}{2}}+{\frac  {6y^{3}}{3}}-{\frac  {y^{4}}{4}}-{\frac  {2y^{{{\frac  {5}{2}}}}}{5}}]_{0}^{1}\,

={\frac  {1}{3}}[8-6+2-{\frac  {1}{4}}-{\frac  {2}{5}}]\,

={\frac  {1}{3}}[{\frac  {80-5-8}{20}}]\,

={\frac  {1}{3}}[{\frac  {67}{20}}]={\frac  {67}{60}}\,

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