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