Question Number 36190 by prof Abdo imad last updated on 30/May/18
$${calculate}\:\:\int\int_{{D}} \left({x}+{y}\right){e}^{{x}+{y}} {dxdy}\:\:{with} \\ $$$${D}\:=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/\:\mathrm{0}<{x}<\mathrm{2}\:{and}\:\:\mathrm{1}<{y}<\mathrm{2}\:\right\} \\ $$
Commented by maxmathsup by imad last updated on 20/Aug/18
![I =∫_1 ^2 (∫_0 ^2 (x+y)e^(x+y) dx)dy =∫_1 ^2 A(y)dy with A(y)=∫_0 ^2 (x+y)e^(x+y) dx A(y) = e^(y ) ∫_0 ^2 (x+y)e^x dx =e^y ∫_0 ^2 (xe^x +y e^x )ex =e^y { ∫_0 ^2 xe^x dx +y ∫_0 ^2 e^x dx} but ∫_0 ^2 e^x dx =[e^x ]_0 ^2 =e^2 −1 and by parts ∫_0 ^2 x e^x dx =[x e^x ]_0 ^2 −∫_0 ^2 e^x dx =2e^2 −e^2 +1 =e^2 +1 ⇒A(y)=e^y (e^2 +1+y(e^2 −1)) =(e^2 +1)e^y +(e^2 −1)y e^y ⇒ I = ∫_1 ^2 { (e^2 +1)e^y +(e^2 −1)ye^y )dy =(e^2 +1)∫_1 ^2 e^y dy +(e^2 −1) ∫_1 ^2 y e^y dy but ∫_1 ^2 e^y dy =e^2 −e and ∫_1 ^2 y e^y dy =[y e^y ]_1 ^2 −∫_1 ^2 e^y dy =2e^2 −e −e^2 +e =e^2 ⇒ I =(e^2 +1)(e^2 −e) +(e^2 −1)e^2 I =e^4 −e^3 +e^2 −e +e^4 −e^2 ⇒ I =2e^(4 ) −e^3 .](https://www.tinkutara.com/question/Q42233.png)
$${I}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\left(\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\left({x}+{y}\right){e}^{{x}+{y}} {dx}\right){dy}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:{A}\left({y}\right){dy}\:{with}\:{A}\left({y}\right)=\int_{\mathrm{0}} ^{\mathrm{2}} \left({x}+{y}\right){e}^{{x}+{y}} \:{dx} \\ $$$${A}\left({y}\right)\:=\:{e}^{{y}\:} \:\int_{\mathrm{0}} ^{\mathrm{2}} \left({x}+{y}\right){e}^{{x}} {dx}\:\:={e}^{{y}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\left({xe}^{{x}} \:+{y}\:{e}^{{x}} \right){ex} \\ $$$$={e}^{{y}} \left\{\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{xe}^{{x}} \:{dx}\:\:+{y}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{e}^{{x}} {dx}\right\}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:{e}^{{x}} {dx}\:=\left[{e}^{{x}} \right]_{\mathrm{0}} ^{\mathrm{2}} \:={e}^{\mathrm{2}} −\mathrm{1}\:\:{and}\:{by}\:{parts}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}\:{e}^{{x}} {dx}\:=\left[{x}\:{e}^{{x}} \right]_{\mathrm{0}} ^{\mathrm{2}} \:−\int_{\mathrm{0}} ^{\mathrm{2}} \:{e}^{{x}} {dx}\:=\mathrm{2}{e}^{\mathrm{2}} \:−{e}^{\mathrm{2}} \:+\mathrm{1}\:={e}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow{A}\left({y}\right)={e}^{{y}} \left({e}^{\mathrm{2}} \:+\mathrm{1}+{y}\left({e}^{\mathrm{2}} −\mathrm{1}\right)\right) \\ $$$$=\left({e}^{\mathrm{2}} \:+\mathrm{1}\right){e}^{{y}} \:+\left({e}^{\mathrm{2}} −\mathrm{1}\right){y}\:{e}^{{y}} \:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left\{\:\:\left({e}^{\mathrm{2}} \:+\mathrm{1}\right){e}^{{y}} \:+\left({e}^{\mathrm{2}} \:−\mathrm{1}\right){ye}^{{y}} \right){dy} \\ $$$$=\left({e}^{\mathrm{2}} +\mathrm{1}\right)\int_{\mathrm{1}} ^{\mathrm{2}} \:{e}^{{y}} {dy}\:+\left({e}^{\mathrm{2}} −\mathrm{1}\right)\:\int_{\mathrm{1}} ^{\mathrm{2}} {y}\:{e}^{{y}} \:{dy}\:\:{but} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:{e}^{{y}} {dy}\:={e}^{\mathrm{2}} \:−{e}\:\:\:{and}\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:{y}\:{e}^{{y}} {dy}\:=\left[{y}\:{e}^{{y}} \right]_{\mathrm{1}} ^{\mathrm{2}} \:−\int_{\mathrm{1}} ^{\mathrm{2}} \:{e}^{{y}} {dy} \\ $$$$=\mathrm{2}{e}^{\mathrm{2}} \:−{e}\:−{e}^{\mathrm{2}} \:+{e}\:={e}^{\mathrm{2}} \:\Rightarrow\:{I}\:=\left({e}^{\mathrm{2}} \:+\mathrm{1}\right)\left({e}^{\mathrm{2}} \:−{e}\right)\:+\left({e}^{\mathrm{2}} \:−\mathrm{1}\right){e}^{\mathrm{2}} \\ $$$${I}\:={e}^{\mathrm{4}} \:−{e}^{\mathrm{3}} \:+{e}^{\mathrm{2}} \:−{e}\:+{e}^{\mathrm{4}} \:−{e}^{\mathrm{2}} \:\Rightarrow\:{I}\:=\mathrm{2}{e}^{\mathrm{4}\:} \:−{e}^{\mathrm{3}} \:. \\ $$