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calculate-D-e-x-y-dxdy-with-D-x-y-R-2-x-lt-1-and-0-y-1-




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}
calculateDexydxdywithD={(x,y)R2/x∣<1and0y1}
Commented by maxmathsup 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=DexydxdyI=11exdx.01eydy=[ex]11.[ey]01=(ee1)(1e1)=e1e1+e2I=e11e+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 )
0111exydxdy=01exy11dy=01(e1ye1y)dy=e1y+e1y01=[1+e2][e+e1]=1+1e2+e1e=e2+1+e3ee2

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