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Question Number 36190 by prof Abdo imad last updated on 30/May/18
calculate ∫∫_D (x+y)e^(x+y) dxdy with D = {(x,y)∈R^2 / 0<x<2 and 1<y<2 }
calculate∫∫D(x+y)ex+ydxdywithD={(x,y)∈R2/0<x<2and1<y<2}
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 .
I=∫12(∫02(x+y)ex+ydx)dy=∫12A(y)dywithA(y)=∫02(x+y)ex+ydxA(y)=ey∫02(x+y)exdx=ey∫02(xex+yex)ex=ey{∫02xexdx+y∫02exdx}but∫02exdx=[ex]02=e2−1andbyparts∫02xexdx=[xex]02−∫02exdx=2e2−e2+1=e2+1⇒A(y)=ey(e2+1+y(e2−1))=(e2+1)ey+(e2−1)yey⇒I=∫12{(e2+1)ey+(e2−1)yey)dy=(e2+1)∫12eydy+(e2−1)∫12yeydybut∫12eydy=e2−eand∫12yeydy=[yey]12−∫12eydy=2e2−e−e2+e=e2⇒I=(e2+1)(e2−e)+(e2−1)e2I=e4−e3+e2−e+e4−e2⇒I=2e4−e3.

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