Question Number 49645 by maxmathsup by imad last updated on 08/Dec/18
![calculate ∫∫_C ∣x+y∣dxdy with C=[−1,1]×[−1,1]](Q49645.png)
$${calculate}\:\int\int_{{C}} \:\mid{x}+{y}\mid{dxdy}\:\:{with}\:{C}=\left[−\mathrm{1},\mathrm{1}\right]×\left[−\mathrm{1},\mathrm{1}\right] \\ $$
Answered by ajfour last updated on 08/Dec/18
![I = 2∫_(−1) ^( 1) [∫_(−x) ^( 1) (x+y)dy]dx = 2∫_(−1) ^( 1) (xy+(y^2 /2))∣_(−x) ^1 dx = 2∫_(−1) ^( 1) (x+(1/2)+x^2 −(x^2 /2))dx = 2∫_0 ^( 1) (x^2 +1)dx = (8/3) .](Q49649.png)
$${I}\:=\:\mathrm{2}\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left[\int_{−{x}} ^{\:\:\mathrm{1}} \left({x}+{y}\right){dy}\right]{dx} \\ $$$$\:\:\:=\:\mathrm{2}\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left({xy}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right)\mid_{−{x}} ^{\mathrm{1}} {dx} \\ $$$$\:\:\:=\:\mathrm{2}\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left({x}+\frac{\mathrm{1}}{\mathrm{2}}+{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right){dx} \\ $$$$\:\:\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{1}\right){dx}\:=\:\frac{\mathrm{8}}{\mathrm{3}}\:. \\ $$
Commented by maxmathsup by imad last updated on 09/Dec/18
$${correct}\:{answer}\:{thanks}\:{sir}. \\ $$