Question Number 34292 by math khazana by abdo last updated on 03/May/18
$${calculate}\:\int\int_{{w}} \:\left({x}+{y}\right){e}^{{x}−{y}} {dxdy}\:{with} \\ $$$${w}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/\:\mid{x}\mid\:\leqslant\mathrm{1}\:\:{and}\:\mid{y}+\mathrm{1}\mid\leqslant\mathrm{3}\:\right\} \\ $$
Commented by math khazana by abdo last updated on 07/May/18
![∣y+1∣≤3 ⇔−3≤ y+1≤3 ⇔ −4 ≤y≤4 so ∫∫_w (x+y) e^(x−y) dxdy =∫_(−4) ^4 ( ∫_(−1) ^1 (x+y)e^(x−y) dx)dy but A(y) = ∫_(−1) ^1 (x+y) e^(x−y) dx =e^(−y) ∫_(−1) ^1 (x+y) e^x dx = e^(−y) { [(x+y)e^x ]_(−1) ^1 −∫_(−1) ^1 e^x dx} = e^(−y) { (1+y)e −(−1+y)e^(−1) −(e −e^(−1) )} =e(1+y) e^(−y) +e^(−1) (1−y) e^(−y) −(e−e^(−1) )e^(−y) ∫∫_w (x+y)e^(x−y) dxdy = e∫_(−4) ^4 (1+y)e^(−y) dy +e^(−1) ∫_(−4) ^4 (1−y)e^(−y) −(e−e^(−1) )∫_(−4) ^4 e^(−y) dy by parts ∫_(−4) ^4 (1+y)e^(−y) dy =[−(1+y)e^(−y) ]_(−4) ^4 +∫_(−4) ^4 e^(−y) dy =−3 e^4 −5 e^(−4) + [ −e^(−y) ]_(−4) ^4 = −3 e^4 −5 e^(−4) + e^4 −e^(−4) we follow the?same method for the other integrals ...](https://www.tinkutara.com/question/Q34482.png)
$$\mid{y}+\mathrm{1}\mid\leqslant\mathrm{3}\:\Leftrightarrow−\mathrm{3}\leqslant\:{y}+\mathrm{1}\leqslant\mathrm{3}\:\Leftrightarrow\:\:−\mathrm{4}\:\leqslant{y}\leqslant\mathrm{4}\:\:{so} \\ $$$$\int\int_{{w}} \:\left({x}+{y}\right)\:{e}^{{x}−{y}} {dxdy}\: \\ $$$$=\int_{−\mathrm{4}} ^{\mathrm{4}} \:\left(\:\int_{−\mathrm{1}} ^{\mathrm{1}} \left({x}+{y}\right){e}^{{x}−{y}} {dx}\right){dy}\:{but} \\ $$$${A}\left({y}\right)\:=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\left({x}+{y}\right)\:{e}^{{x}−{y}} {dx}\:={e}^{−{y}} \:\int_{−\mathrm{1}} ^{\mathrm{1}} \left({x}+{y}\right)\:{e}^{{x}} {dx} \\ $$$$=\:{e}^{−{y}} \left\{\:\:\left[\left({x}+{y}\right){e}^{{x}} \right]_{−\mathrm{1}} ^{\mathrm{1}} \:−\int_{−\mathrm{1}} ^{\mathrm{1}} \:{e}^{{x}} {dx}\right\} \\ $$$$=\:{e}^{−{y}} \left\{\:\:\left(\mathrm{1}+{y}\right){e}\:−\left(−\mathrm{1}+{y}\right){e}^{−\mathrm{1}} \:\:−\left({e}\:−{e}^{−\mathrm{1}} \right)\right\} \\ $$$$={e}\left(\mathrm{1}+{y}\right)\:{e}^{−{y}} \:\:+{e}^{−\mathrm{1}} \left(\mathrm{1}−{y}\right)\:{e}^{−{y}} \:−\left({e}−{e}^{−\mathrm{1}} \right){e}^{−{y}} \\ $$$$\int\int_{{w}} \left({x}+{y}\right){e}^{{x}−{y}} {dxdy}\: \\ $$$$=\:{e}\int_{−\mathrm{4}} ^{\mathrm{4}} \left(\mathrm{1}+{y}\right){e}^{−{y}} {dy}\:\:+{e}^{−\mathrm{1}} \:\int_{−\mathrm{4}} ^{\mathrm{4}} \left(\mathrm{1}−{y}\right){e}^{−{y}} \:−\left({e}−{e}^{−\mathrm{1}} \right)\int_{−\mathrm{4}} ^{\mathrm{4}} \:{e}^{−{y}} {dy} \\ $$$${by}\:{parts} \\ $$$$\int_{−\mathrm{4}} ^{\mathrm{4}} \left(\mathrm{1}+{y}\right){e}^{−{y}} {dy}\:=\left[−\left(\mathrm{1}+{y}\right){e}^{−{y}} \right]_{−\mathrm{4}} ^{\mathrm{4}} \:+\int_{−\mathrm{4}} ^{\mathrm{4}} \:{e}^{−{y}} {dy} \\ $$$$=−\mathrm{3}\:{e}^{\mathrm{4}} \:−\mathrm{5}\:{e}^{−\mathrm{4}} \:\:+\:\left[\:−{e}^{−{y}} \right]_{−\mathrm{4}} ^{\mathrm{4}} \\ $$$$=\:−\mathrm{3}\:{e}^{\mathrm{4}} \:\:−\mathrm{5}\:{e}^{−\mathrm{4}} \:\:+\:{e}^{\mathrm{4}} \:−{e}^{−\mathrm{4}} \\ $$$${we}\:{follow}\:{the}?{same}\:{method}\:{for}\:{the}\:{other}\:{integrals} \\ $$$$… \\ $$