Question Number 82871 by abdomathmax last updated on 25/Feb/20
![find ∫∫_([0,1]^2 ) ∣x^2 −y^2 ∣dxdy](https://www.tinkutara.com/question/Q82871.png)
$${find}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\mid{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \mid{dxdy} \\ $$
Commented by mathmax by abdo last updated on 25/Feb/20
![I =∫∫_([0,1]^2 ) ∣x+y∣∣x−y∣dxdy =∫∫_([0,1]^2 ) (x+y)∣x−y∣dxdy =∫_0 ^1 ( ∫_0 ^x (x+y)∣x−y∣dy +∫_x ^1 (x+y)∣x−y∣dy)dx =∫_0 ^1 (∫_0 ^x (x+y)(x−y)dy +∫_x ^1 (x+y)(y−x)dy)dx =∫_0 ^1 (∫_0 ^x (x+y)(x−y)dy)dx +∫_0 ^1 (∫_x ^1 (y+x)(y−x)dy)dx we have ∫_0 ^x (x+y)(x−y)dy =∫_0 ^x (x^2 −y^2 )dy =x^3 −[(1/3)y^3 ]_0 ^x =x^3 −(x^3 /3) =(2/3)x^3 ⇒∫_0 ^1 (....dy)dx =(2/3)∫_0 ^1 x^3 dx =(2/3)×(1/4) =(1/6) ∫_x ^1 (y+x)(y−x)dy =∫_x ^1 (y^2 −x^2 )dy =[(y^3 /3)]_x ^1 −x^2 (1−x) =(1/3)−(x^3 /3)−x^2 +x^3 =(1/3)−x^2 +(2/3)x^3 ⇒ ∫_0 ^1 (....dy)dx =∫_0 ^1 ((1/3)−x^2 +(2/3)x^3 )dx =[(x/3)−(1/3)x^3 +(2/(12))x^4 ]_0 ^1 =(1/6) ⇒I =(1/6)+(1/6) =(1/3)](https://www.tinkutara.com/question/Q82918.png)
$${I}\:=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\mid{x}+{y}\mid\mid{x}−{y}\mid{dxdy}\:=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\left({x}+{y}\right)\mid{x}−{y}\mid{dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\:\:\int_{\mathrm{0}} ^{{x}} \left({x}+{y}\right)\mid{x}−{y}\mid{dy}\:+\int_{{x}} ^{\mathrm{1}} \:\left({x}+{y}\right)\mid{x}−{y}\mid{dy}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{{x}} \left({x}+{y}\right)\left({x}−{y}\right){dy}\:+\int_{{x}} ^{\mathrm{1}} \left({x}+{y}\right)\left({y}−{x}\right){dy}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\int_{\mathrm{0}} ^{{x}} \left({x}+{y}\right)\left({x}−{y}\right){dy}\right){dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\int_{{x}} ^{\mathrm{1}} \left({y}+{x}\right)\left({y}−{x}\right){dy}\right){dx} \\ $$$${we}\:{have}\:\int_{\mathrm{0}} ^{{x}} \left({x}+{y}\right)\left({x}−{y}\right){dy}\:=\int_{\mathrm{0}} ^{{x}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy} \\ $$$$={x}^{\mathrm{3}} −\left[\frac{\mathrm{1}}{\mathrm{3}}{y}^{\mathrm{3}} \right]_{\mathrm{0}} ^{{x}} \:={x}^{\mathrm{3}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \left(….{dy}\right){dx}\:=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{3}} \:{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\int_{{x}} ^{\mathrm{1}} \left({y}+{x}\right)\left({y}−{x}\right){dy}\:=\int_{{x}} ^{\mathrm{1}} \left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){dy}\:=\left[\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\right]_{{x}} ^{\mathrm{1}} −{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \:\:=\frac{\mathrm{1}}{\mathrm{3}}−{x}^{\mathrm{2}} \:+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(….{dy}\right){dx}\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}−{x}^{\mathrm{2}} \:+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \right){dx}\:=\left[\frac{{x}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{12}}{x}^{\mathrm{4}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow{I}\:=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}}\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$