MvCalc3

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

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

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

={\frac  {8}{3}}-{\frac  {1}{3}}+y^{2}(2-1)\,

={\frac  {7}{3}}+y^{2}\,

Hence I=\int _{0}^{1}\left({\frac  {7}{3}}+y^{2}\right)dy\,

=\left[{\frac  {7}{3}}y+{\frac  {y^{3}}{3}}\right]_{0}^{1}\,

Therefore,I={\frac  {7}{3}}(1-0)+\left({\frac  {1}{3}}-0\right)={\frac  {7}{3}}+{\frac  {1}{3}}={\frac  {8}{3}}\,

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