MvCalc1

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\int _{0}^{2}\int _{0}^{1}(2x+y)^{8}dxdy\,

Perform the inner integral:

\int _{0}^{1}(2x+y)^{8}dx={\frac  {(2x+y)^{9}}{9}}{\Bigg |}_{{x=0}}^{{x=1}}\,

={\frac  {(2(1)+y)^{9}}{18}}-{\frac  {(2(0)+y)^{9}}{18}}\,

={\frac  {(2+y)^{9}}{18}}-{\frac  {y^{9}}{18}}\,

Perform the outer integral:

\int _{0}^{2}\left[{\frac  {(2+y)^{9}}{18}}-{\frac  {y^{9}}{18}}\right]dy={\frac  {(2+y)^{{10}}}{180}}-{\frac  {y^{{10}}}{180}}{\Bigg |}_{{y=0}}^{{y=2}}\,

\left[{\frac  {(2+2)^{{10}}}{180}}-{\frac  {2^{{10}}}{180}}\right]-\left[{\frac  {2^{{10}}}{180}}-0\right]\,

={\frac  {4^{{10}}-2^{{11}}}{180}}={\frac  {261,632}{45}}\,


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