Question Number 57320 by turbo msup by abdo last updated on 02/Apr/19
$${calculate}\:\int\int_{{D}} {xy}\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \:{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\:{and}\right. \\ $$$$\left.\mathrm{1}\leqslant{y}\leqslant\mathrm{3}\right\} \\ $$
Commented by maxmathsup by imad last updated on 02/Apr/19
![∫∫_D xy e^(−x^2 −y^2 ) dxdy =∫_0 ^2 x e^(−x^2 ) dx .∫_1 ^3 y e^(−y^2 ) dy but ∫_0 ^2 x e^(−x^2 ) dx =[−(1/2)e^(−x^2 ) ]_0 ^2 =−(1/2){e^(−4) −1}=(1/2){1−e^(−4) ) ∫_1 ^3 y e^(−y^2 ) dy =[−(1/2)e^(−y^2 ) ]_1 ^3 =−(1/2){e^(−9) −e^(−1) } =(1/2){ e^(−1) −e^(−9) } ⇒ ∫∫_D xy e^(−x^2 −y^2 ) dxdy =(1/4)(1−e^(−4) )(e^(−1) −e^(−9) ) =(1/4){e^(−1) −e^(−9) −e^(−5) +e^(−13) } .](https://www.tinkutara.com/question/Q57341.png)
$$\int\int_{{D}} {xy}\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}\:{e}^{−{x}^{\mathrm{2}} } {dx}\:.\int_{\mathrm{1}} ^{\mathrm{3}} \:{y}\:{e}^{−{y}^{\mathrm{2}} } {dy}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}\:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{e}^{−\mathrm{4}} \:−\mathrm{1}\right\}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−{e}^{−\mathrm{4}} \right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \:{y}\:{e}^{−{y}^{\mathrm{2}} } {dy}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{y}^{\mathrm{2}} } \right]_{\mathrm{1}} ^{\mathrm{3}} \:=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{e}^{−\mathrm{9}} \:−{e}^{−\mathrm{1}} \right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{e}^{−\mathrm{1}} \:−{e}^{−\mathrm{9}} \right\}\:\Rightarrow \\ $$$$\int\int_{{D}} {xy}\:{e}^{−{x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} } {dxdy}\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−{e}^{−\mathrm{4}} \right)\left({e}^{−\mathrm{1}} \:−{e}^{−\mathrm{9}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{{e}^{−\mathrm{1}} \:−{e}^{−\mathrm{9}} \:−{e}^{−\mathrm{5}} \:+{e}^{−\mathrm{13}} \right\}\:. \\ $$
Answered by kaivan.ahmadi last updated on 02/Apr/19
![∫_1 ^3 ∫_0 ^2 xye^(−x^2 −y^2 ) dxdy=((−1)/2)∫_1 ^3 ye^(−x^2 −y^2 ) ∣_0 ^2 dy= ((−1)/2)∫_1 ^3 (ye^(−4−y^2 ) −ye^(−y^2 ) )dy= ((−1)/2)(((−1)/2)e^(−4−y^2 ) +(1/2)e^(−y^2 ) )_1 ^3 = ((−1)/4)[(e^(−4−9) +e^(−9) )−(e^(−4−1) +e^(−1) )]= ((−1)/4)(e^(−13) +e^(−9) −e^(−5) −e^(−1) )](https://www.tinkutara.com/question/Q57333.png)
$$\int_{\mathrm{1}} ^{\mathrm{3}} \int_{\mathrm{0}} ^{\mathrm{2}} {xye}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}=\frac{−\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} \:\:{ye}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \mid_{\mathrm{0}} ^{\mathrm{2}} {dy}=\: \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} \left({ye}^{−\mathrm{4}−{y}^{\mathrm{2}} } −{ye}^{−{y}^{\mathrm{2}} } \right){dy}= \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{4}−{y}^{\mathrm{2}} } +\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{y}^{\mathrm{2}} } \right)_{\mathrm{1}} ^{\mathrm{3}} = \\ $$$$\frac{−\mathrm{1}}{\mathrm{4}}\left[\left({e}^{−\mathrm{4}−\mathrm{9}} +{e}^{−\mathrm{9}} \right)−\left({e}^{−\mathrm{4}−\mathrm{1}} +{e}^{−\mathrm{1}} \right)\right]= \\ $$$$\frac{−\mathrm{1}}{\mathrm{4}}\left({e}^{−\mathrm{13}} +{e}^{−\mathrm{9}} −{e}^{−\mathrm{5}} −{e}^{−\mathrm{1}} \right) \\ $$