Question Number 96128 by bemath last updated on 30/May/20
![find ∫∫_R (x+2y)^2 dxdy in R=[−1,2] ×[0,2]](https://www.tinkutara.com/question/Q96128.png)
$${find}\:\int\int_{{R}} \:\left({x}+\mathrm{2}{y}\right)^{\mathrm{2}} \:{dxdy}\:{in}\:{R}=\left[−\mathrm{1},\mathrm{2}\right]\:×\left[\mathrm{0},\mathrm{2}\right]\: \\ $$
Answered by john santu last updated on 30/May/20
![∫_0 ^2 ∫_(−1) ^2 [(x^2 +4xy+4y^2 )dx]dy= ∫_0 ^2 [(1/3)x^3 +2x^2 y+4xy^2 ]_(−1) ^2 dy = ∫_0 ^2 (3+6y+12y^2 ) dy = [ 3y + 3y^2 +4y^3 ]_0 ^2 = 6+12+32 = 50](https://www.tinkutara.com/question/Q96129.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\underset{−\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\left[\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4xy}+\mathrm{4y}^{\mathrm{2}} \right)\mathrm{dx}\right]\mathrm{dy}= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} \mathrm{y}+\mathrm{4xy}^{\mathrm{2}} \:\right]_{−\mathrm{1}} ^{\mathrm{2}} \mathrm{dy}\:=\: \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\left(\mathrm{3}+\mathrm{6y}+\mathrm{12y}^{\mathrm{2}} \right)\:\mathrm{dy}\:= \\ $$$$\left[\:\mathrm{3y}\:+\:\mathrm{3y}^{\mathrm{2}} +\mathrm{4y}^{\mathrm{3}} \:\right]_{\mathrm{0}} ^{\mathrm{2}} \:=\:\mathrm{6}+\mathrm{12}+\mathrm{32} \\ $$$$=\:\mathrm{50} \\ $$
Commented by bemath last updated on 30/May/20
$${thanks}\:{much} \\ $$