MvCalc26

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The limits of x and y are given as x=1 to x=2 and y=0 to y=3.

Therefore,given I=\iint _{D}(4xy-y^{2})dxdy=\int _{1}^{2}\int _{0}^{3}(4xy-y^{2})dxdy=\int _{1}^{2}[\int _{{y=0}}^{{3}}(4xy-y^{2})dy]dx\,

Now \int _{0}^{3}(4xy-y^{2})dy=[2xy^{2}-{\frac  {y^{3}}{3}}]_{0}^{3}=2x(3^{2})-{\frac  {3^{3}}{3}}=18x-9\,

Therefore I=\int _{1}^{2}(18x-9)dx=[18{\frac  {x^{2}}{2}}-9x]_{1}^{2}=[9x^{2}-9x]_{1}^{2}=[9(2^{2})-9(2)]-[9(1^{2})-9(1)]=36-18-9+9=18\,

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