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find-I-D-ln-1-x-y-dxdy-with-D-x-y-R-2-x-y-1-and-x-0-and-y-0-

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Question Number 27690 by abdo imad last updated on 12/Jan/18
find I= ∫∫_D ln(1+x+y)dxdy with D= {(x,y)∈R^2 / x+y≤1 and x≥0 and y≥0 }.
findI=∫∫Dln(1+x+y)dxdywithD={(x,y)∈R2/x+y⩽1andx⩾0andy⩾0}.
Commented by abdo imad last updated on 14/Jan/18
0≤x≤1−y and 0≤y≤1 so I = ∫_0 ^1 ( ∫_0 ^(1−y) ln(1+x+y)dx)dy but the ch. 1+x+y =t give ∫_0 ^(1−y) ln(1+x +y)dx= ∫_(1+y) ^2 lnt dt = [tlnt −t]_(1+y) ^2 = 2ln2 −2 −(1+y)ln(1+y) +1+y =2ln2 −1 +y −(1+y)ln(1+y) I= ∫_0 ^1 (2ln2−1)dy +∫_0 ^1 ydy −∫_0 ^1 (1+y)ln(1+y)dy I= 2ln2 −(1/2) − ∫_0 ^1 (1+y)ln(1+y)dy the ch.1+y=t give ∫_0 ^1 (1+y)ln(1+y)dy= ∫_1 ^2 tln(t)dt =[ (t^2 /2) lnt]_1 ^2 −∫_1 ^2 (t/2)dt = 2ln2 −(1/2)[(t^2 /2)]_1 ^2 =2ln2−(1/2)((3/2)) = 2ln2 −(3/4) I= 2ln2 −(1/2) −2ln2 +(3/4)= (1/4) .
0⩽x⩽1−yand0⩽y⩽1soI=∫01(∫01−yln(1+x+y)dx)dybutthech.1+x+y=tgive∫01−yln(1+x+y)dx=∫1+y2lntdt=[tlnt−t]1+y2=2ln2−2−(1+y)ln(1+y)+1+y=2ln2−1+y−(1+y)ln(1+y)I=∫01(2ln2−1)dy+∫01ydy−∫01(1+y)ln(1+y)dyI=2ln2−12−∫01(1+y)ln(1+y)dythech.1+y=tgive∫01(1+y)ln(1+y)dy=∫12tln(t)dt=[t22lnt]12−∫12t2dt=2ln2−12[t22]12=2ln2−12(32)=2ln2−34I=2ln2−12−2ln2+34=14.

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