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Question Number 57319 by turbo msup by abdo last updated on 02/Apr/19

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

calculate∫∫Dex−ydxdy withD={(x,y)∈R2/∣x∣<1and0⩽y⩽1}

Commented bymaxmathsup by imad last updated on 02/Apr/19

let I =∫∫_D e^(x−y) dxdy ⇒I =∫_(−1) ^1 e^x dx .∫_0 ^1 e^(−y) dy =[e^x ]_(−1) ^1 .[−e^(−y) ]_0 ^1 =(e−e^(−1) )(1−e^(−1) ) =e−1−e^(−1) +e^(−2) ⇒I =e−1−(1/e) +(1/e^2 ) .

letI=∫∫Dex−ydxdy⇒I=∫−11exdx.∫01e−ydy =[ex]−11.[−e−y]01=(e−e−1)(1−e−1)=e−1−e−1+e−2 ⇒I=e−1−1e+1e2.

Answered by kaivan.ahmadi last updated on 02/Apr/19

∫_0 ^1 ∫_(−1) ^1 e^(x−y) dxdy=∫_0 ^1 e^(x−y) ∣_(−1) ^1 dy= ∫_0 ^1 (e^(1−y) −e^(−1−y) )dy= −e^(1−y) +e^(−1−y) ∣_0 ^1 =[−1+e^(−2) ]−[−e+e^(−1) ]= −1+(1/e^2 )+e−(1/e)=((−e^2 +1+e^3 −e)/e^2 )

∫01∫−11ex−ydxdy=∫01ex−y∣−11dy= ∫01(e1−y−e−1−y)dy= −e1−y+e−1−y∣01=[−1+e−2]−[−e+e−1]= −1+1e2+e−1e=−e2+1+e3−ee2

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