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Question Number 57320 by turbo msup by abdo last updated on 02/Apr/19

calculate ∫∫_D xy e^(−x^2 −y^2 )  dxdy  with D={(x,y)∈R^2 / 0≤x≤2 and  1≤y≤3}

$${calculate}\:\int\int_{{D}} {xy}\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \:{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\:{and}\right. \\ $$$$\left.\mathrm{1}\leqslant{y}\leqslant\mathrm{3}\right\} \\ $$

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) } .

$$\int\int_{{D}} {xy}\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}\:{e}^{−{x}^{\mathrm{2}} } {dx}\:.\int_{\mathrm{1}} ^{\mathrm{3}} \:{y}\:{e}^{−{y}^{\mathrm{2}} } {dy}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}\:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{e}^{−\mathrm{4}} \:−\mathrm{1}\right\}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−{e}^{−\mathrm{4}} \right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \:{y}\:{e}^{−{y}^{\mathrm{2}} } {dy}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{y}^{\mathrm{2}} } \right]_{\mathrm{1}} ^{\mathrm{3}} \:=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{e}^{−\mathrm{9}} \:−{e}^{−\mathrm{1}} \right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{e}^{−\mathrm{1}} \:−{e}^{−\mathrm{9}} \right\}\:\Rightarrow \\ $$$$\int\int_{{D}} {xy}\:{e}^{−{x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} } {dxdy}\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−{e}^{−\mathrm{4}} \right)\left({e}^{−\mathrm{1}} \:−{e}^{−\mathrm{9}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{{e}^{−\mathrm{1}} \:−{e}^{−\mathrm{9}} \:−{e}^{−\mathrm{5}} \:+{e}^{−\mathrm{13}} \right\}\:. \\ $$

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) )

$$\int_{\mathrm{1}} ^{\mathrm{3}} \int_{\mathrm{0}} ^{\mathrm{2}} {xye}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}=\frac{−\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} \:\:{ye}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \mid_{\mathrm{0}} ^{\mathrm{2}} {dy}=\: \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} \left({ye}^{−\mathrm{4}−{y}^{\mathrm{2}} } −{ye}^{−{y}^{\mathrm{2}} } \right){dy}= \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{4}−{y}^{\mathrm{2}} } +\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{y}^{\mathrm{2}} } \right)_{\mathrm{1}} ^{\mathrm{3}} = \\ $$$$\frac{−\mathrm{1}}{\mathrm{4}}\left[\left({e}^{−\mathrm{4}−\mathrm{9}} +{e}^{−\mathrm{9}} \right)−\left({e}^{−\mathrm{4}−\mathrm{1}} +{e}^{−\mathrm{1}} \right)\right]= \\ $$$$\frac{−\mathrm{1}}{\mathrm{4}}\left({e}^{−\mathrm{13}} +{e}^{−\mathrm{9}} −{e}^{−\mathrm{5}} −{e}^{−\mathrm{1}} \right) \\ $$

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