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calculate-D-xe-x-siny-dy-with-D-is-the-triangle-OAB-O-0-0-A-1-0-B-0-1-

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Question Number 89312 by abdomathmax last updated on 16/Apr/20
calculate ∫∫_D xe^(−x) siny dy with D is the triangle OAB O(0,0) A(1,0) B(0,1)
calculate∫∫Dxe−xsinydywithDisthetriangleOABO(0,0)A(1,0)B(0,1)
Commented by mathmax by abdo last updated on 17/Apr/20
the equation of line (AB) is x+y =1 ⇒y =1−x ⇒ ∫∫_D xe^(−x) siny dxdy =∫_0 ^1 (∫_0 ^(1−x) sinydy)xe^(−x) dx we have ∫_0 ^(1−x) siny dy =[−cosy]_0 ^(1−x) =1−cos(1−x) ⇒ I =∫_0 ^1 (1−cos(1−x))xe^(−x) dx =∫_0 ^1 xe^(−x) dx−∫_0 ^1 xe^(−x) cos(1−x)dx but ∫_0 ^1 xe^(−x) dx =[−xe^(−x) ]_0 ^1 +∫_0 ^1 e^(−x) dx =−e^(−1) +[−e^(−x) ]_0 ^1 =−e^(−1) +1−e^(−1) =1−2e^(−1) ∫_0 ^1 xe^(−x) cos(1−x)dx =Re(∫_0 ^1 xe^(−x) e^(i(1−x)) dx) ∫_0 ^1 xe^(−x+i−ix) ex =e^i ∫_0 ^1 x e^(−(1+i)x) dx =e^i { [ (x/(−(1+i))) e^(−(1+i)x) ]_0 ^1 −∫_0 ^1 (1/(−(1+i)))e^(−(1+i)x) dx} =e^i {−(1/((1+i)))(e^(−(1+i)) −1)+(1/(1+i))[−(1/(1+i))e^(−(1+i)x) ]_0 ^1 } =−(e^i /(1+i))(e^(−(1+i)) −1)−(e^i /((1+i)^2 ))(e^(−(1+i)) −1) =−(((1−i))/2)(e^(−1) −e^i )−((e^(−1) −e^i )/(2i)) =((e^i −e^(−1) )/2)(1−i +(1/i)) =((e^i −e^(−1) )/2)(1−2i) =(1/2)(1−2i)(cos(1)+isin(1)−e^(−1) ) =(1/2){cos(1)+isin(1)−e^(−1) −2icos(1)+2sin(1)+2ie^(−1) } ⇒ ∫_0 ^1 x e^(−x) cos(1−x)dx =((cos(1)+2sin(1)−e^(−1) )/2) ⇒ I =1−2e^(−1) −(1/2){cos(1)+2sin(1)−e^(−1) }
theequationofline(AB)isx+y=1⇒y=1−x⇒∫∫Dxe−xsinydxdy=∫01(∫01−xsinydy)xe−xdxwehave∫01−xsinydy=[−cosy]01−x=1−cos(1−x)⇒I=∫01(1−cos(1−x))xe−xdx=∫01xe−xdx−∫01xe−xcos(1−x)dxbut∫01xe−xdx=[−xe−x]01+∫01e−xdx=−e−1+[−e−x]01=−e−1+1−e−1=1−2e−1∫01xe−xcos(1−x)dx=Re(∫01xe−xei(1−x)dx)∫01xe−x+i−ixex=ei∫01xe−(1+i)xdx=ei{[x−(1+i)e−(1+i)x]01−∫011−(1+i)e−(1+i)xdx}=ei{−1(1+i)(e−(1+i)−1)+11+i[−11+ie−(1+i)x]01}=−ei1+i(e−(1+i)−1)−ei(1+i)2(e−(1+i)−1)=−(1−i)2(e−1−ei)−e−1−ei2i=ei−e−12(1−i+1i)=ei−e−12(1−2i)=12(1−2i)(cos(1)+isin(1)−e−1)=12{cos(1)+isin(1)−e−1−2icos(1)+2sin(1)+2ie−1}⇒∫01xe−xcos(1−x)dx=cos(1)+2sin(1)−e−12⇒I=1−2e−1−12{cos(1)+2sin(1)−e−1}

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