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Question Number 96128 by bemath last updated on 30/May/20

$$find\int {\int }_R{(x+2y)}^{2}dxdyinR=[-1,2]\times [0,2]$$

Answered by john santu last updated on 30/May/20

$$\underset{0}{\stackrel{2}{\int }}\underset{-1}{\stackrel{2}{\int }}\left[({\mathrm{x}}^2+4\mathrm{xy}+{4\mathrm{y}}^2)\mathrm{dx}\right]\mathrm{dy}=$$$$\underset{0}{\stackrel{2}{\int }}{[\frac{1}{3}{\mathrm{x}}^3+{2\mathrm{x}}^2\mathrm{y}+{4\mathrm{xy}}^2]}_{-1}^2\mathrm{dy}=$$$$\underset{0}{\stackrel{2}{\int }}(3+6\mathrm{y}+{12\mathrm{y}}^2)\mathrm{dy}=$$$${[3\mathrm{y}+{3\mathrm{y}}^2+{4\mathrm{y}}^3]}_0^2=6+12+32$$$$=50$$

Commented by bemath last updated on 30/May/20

$$thanksmuch$$

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