MvCalc15

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

Now \int _{{0}}^{{1-x}}(x^{2}+y^{2})dy\,

=[x^{2}y+{\frac  {y^{3}}{3}}]_{{y=0}}^{{1-x}}\,

=x^{2}(1-x)+{\frac  {(1-x)^{3}}{3}}\,

=x^{2}-x^{3}+{\frac  {1}{3}}[1-3x+3x^{2}-x^{3}]\,

Therefore,I=\int _{0}^{1}[x^{2}-x^{3}+{\frac  {1}{3}}(1-3x+3x^{2}-x^{3})]dx\,

=[{\frac  {x^{3}}{3}}-{\frac  {x^{4}}{4}}+{\frac  {1}{3}}(x-{\frac  {3x^{2}}{2}}+x^{3}-{\frac  {x^{4}}{4}}]_{0}^{1}\,

=[{\frac  {1}{3}}-{\frac  {1}{4}}+{\frac  {1}{3}}-{\frac  {1}{2}}+{\frac  {1}{3}}-{\frac  {1}{12}}]\,

=1-{\frac  {1}{4}}-{\frac  {1}{2}}-{\frac  {1}{12}}\,

=1-[{\frac  {3+6+1}{12}}]=1-{\frac  {10}{12}}=1-{\frac  {5}{6}}={\frac  {1}{6}}\,

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