Question Number 37279 by abdo.msup.com last updated on 11/Jun/18
![cslculate ∫∫_([0,1]^2 ) (x−y)e^(−x−y) dxdy .](Q37279.png)
$${cslculate}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\left({x}−{y}\right){e}^{−{x}−{y}} {dxdy}\:. \\ $$
Commented by prof Abdo imad last updated on 18/Jun/18
![let use the changement (u,v)→(x,y)=(x−y,x+y)=(ϕ_1 (u,v),ϕ_2 (u,v)) we have 0≤x≤1 and 0≤y≤1 ⇒ −1≤−y≤0 ⇒ −1≤x−y≤1 and 0≤x+y≤2 ⇒ −1≤u≤1 and 0≤v≤2 and M_j = ((((∂ϕ_1 /∂u) (∂ϕ_1 /∂v))),(((∂ϕ_2 /∂u) (∂ϕ_2 /∂v))) ) = ((),() ) we have x−y=u and x+y=v ⇒2x=u+v and 2y = −u +v ⇒ x= (1/2) u +(1/2)v and y =−(1/2)u +(1/2)v ⇒ M_j = ((((1/2) (1/2))),((−(1/2) (1/2))) ) and det(M_j )= (1/2) ∫∫_([0,1]^2 ) (x−y)e^(−(x+y)) dxdy =∫∫_(−1≤u≤1 and 0≤v≤2) u e^(−v) (1/2)du dv =(1/2) ∫_(−1) ^1 u du.∫_0 ^2 e^(−v) dv =(1/2)[ (u^2 /2)]_(−1) ^1 .[ −e^(−v) ]_0 ^2 =0](Q37829.png)
$${let}\:{use}\:{the}\:{changement} \\ $$$$\left({u},{v}\right)\rightarrow\left({x},{y}\right)=\left({x}−{y},{x}+{y}\right)=\left(\varphi_{\mathrm{1}} \left({u},{v}\right),\varphi_{\mathrm{2}} \left({u},{v}\right)\right) \\ $$$${we}\:{have}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{1}\leqslant−{y}\leqslant\mathrm{0}\:\Rightarrow\:−\mathrm{1}\leqslant{x}−{y}\leqslant\mathrm{1}\:{and} \\ $$$$\mathrm{0}\leqslant{x}+{y}\leqslant\mathrm{2}\:\Rightarrow\:−\mathrm{1}\leqslant{u}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{v}\leqslant\mathrm{2}\:{and} \\ $$$${M}_{{j}} \:=\:\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial{v}}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial{v}}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{}\\{}\end{pmatrix} \\ $$$${we}\:{have}\:{x}−{y}={u}\:{and}\:{x}+{y}={v}\:\Rightarrow\mathrm{2}{x}={u}+{v} \\ $$$${and}\:\mathrm{2}{y}\:=\:−{u}\:+{v}\:\Rightarrow\:{x}=\:\frac{\mathrm{1}}{\mathrm{2}}\:{u}\:+\frac{\mathrm{1}}{\mathrm{2}}{v}\:{and} \\ $$$${y}\:=−\frac{\mathrm{1}}{\mathrm{2}}{u}\:+\frac{\mathrm{1}}{\mathrm{2}}{v}\:\:\Rightarrow \\ $$$${M}_{{j}} \:=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix}\:\:\:\:\:{and}\:{det}\left({M}_{{j}} \right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\left({x}−{y}\right){e}^{−\left({x}+{y}\right)} {dxdy} \\ $$$$=\int\int_{−\mathrm{1}\leqslant{u}\leqslant\mathrm{1}\:{and}\:\:\mathrm{0}\leqslant{v}\leqslant\mathrm{2}} \:{u}\:{e}^{−{v}} \:\frac{\mathrm{1}}{\mathrm{2}}{du}\:{dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:{u}\:{du}.\int_{\mathrm{0}} ^{\mathrm{2}} \:{e}^{−{v}} \:{dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{1}} ^{\mathrm{1}} .\left[\:−{e}^{−{v}} \right]_{\mathrm{0}} ^{\mathrm{2}} \:\:=\mathrm{0} \\ $$
Answered by ajfour last updated on 11/Jun/18
![A=∫_0 ^( 1) (1/e^y )[∫_0 ^( 1) xe^(−x) dx−y∫_0 ^( 1) e^(−x) dx]dy =∫_9 ^( 1) (1/e^y )[−(1+x)e^(−x) +ye^(−x) ]∣_0 ^1 dy =∫_0 ^( 1) (1/e^y )[((y/e)−(2/e))−(y−1)]dy =∫_0 ^( 1) [((1/e)−1)y+(1−(2/e))]e^(−y) dy ={−[((1/e)−1)y+(1−(2/e))]e^(−y) −((1/e)−1)e^(−y) }∣_0 ^1 =−(1/e)[(1/e)−1+1−(2/e)]+1−(2/e) −(1/e)((1/e)−1)+(1/e)−1 = 0 .](Q37313.png)
$${A}=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{\mathrm{1}}{{e}^{{y}} }\left[\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {xe}^{−{x}} {dx}−{y}\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {e}^{−{x}} {dx}\right]{dy} \\ $$$$=\int_{\mathrm{9}} ^{\:\:\mathrm{1}} \frac{\mathrm{1}}{{e}^{{y}} }\left[−\left(\mathrm{1}+{x}\right){e}^{−{x}} +{ye}^{−{x}} \right]\mid_{\mathrm{0}} ^{\mathrm{1}} {dy} \\ $$$$=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{\mathrm{1}}{{e}^{{y}} }\left[\left(\frac{{y}}{{e}}−\frac{\mathrm{2}}{{e}}\right)−\left({y}−\mathrm{1}\right)\right]{dy} \\ $$$$=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left[\left(\frac{\mathrm{1}}{{e}}−\mathrm{1}\right){y}+\left(\mathrm{1}−\frac{\mathrm{2}}{{e}}\right)\right]{e}^{−{y}} {dy} \\ $$$$=\left\{−\left[\left(\frac{\mathrm{1}}{{e}}−\mathrm{1}\right){y}+\left(\mathrm{1}−\frac{\mathrm{2}}{{e}}\right)\right]{e}^{−{y}} \right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:−\left(\frac{\mathrm{1}}{{e}}−\mathrm{1}\right){e}^{−{y}} \right\}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{{e}}\left[\frac{\mathrm{1}}{{e}}−\mathrm{1}+\mathrm{1}−\frac{\mathrm{2}}{{e}}\right]+\mathrm{1}−\frac{\mathrm{2}}{{e}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{{e}}\left(\frac{\mathrm{1}}{{e}}−\mathrm{1}\right)+\frac{\mathrm{1}}{{e}}−\mathrm{1} \\ $$$$=\:\mathrm{0}\:. \\ $$
Commented by prof Abdo imad last updated on 18/Jun/18
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{Ajfour}\:{thanks}. \\ $$
Commented by ajfour last updated on 18/Jun/18
$${thanks}\:{for}\:{confirming}\:{my}\:{answer} \\ $$$${Sir};\:{makes}\:{me}\:{happy}. \\ $$