MvCalc2

From Example Problems
Jump to: navigation, search

Here the order of integration is first with respect to x and then with respect to y.Therefore,

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

Now \int _{1}^{2}(x^{2}+y^{2})dx=[{\frac  {x^{3}}{3}}+xy^{2}]_{{x=1}}^{{2}}={\frac  {1}{3}}(8-1)+y^{2}(2-1)={\frac  {7}{3}}+y^{2}\,

Hence I=\int _{0}^{2}({\frac  {7}{3}}+y^{2})dy={\frac  {7}{3}}[y]_{0}^{2}+{\frac  {1}{3}}[y^{3}]_{0}^{2}={\frac  {7}{3}}(2)+{\frac  {1}{3}}(8)={\frac  {22}{3}}\,


Main Page