MvCalc27

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The region is bounded by y=x,y=2x,x=1\, from these equations,the limits of y are from x to 2x and x from 0 to 1 (since first quadrant)

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

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

=x^{3}+{\frac  {8}{3}}x^{3}-{\frac  {1}{3}}x^{3}=x^{3}+{\frac  {7}{3}}x^{3}={\frac  {10x^{3}}{3}}\,

Hence I=\int _{0}^{1}({\frac  {10x^{3}}{3}})dx=[{\frac  {10x^{4}}{12}}]_{0}^{1}={\frac  {10}{12}}={\frac  {5}{6}}\,


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