Question Number 27690 by abdo imad last updated on 12/Jan/18
$${find}\:\:\:{I}=\:\:\int\int_{{D}} {ln}\left(\mathrm{1}+{x}+{y}\right){dxdy}\:\:{with} \\ $$$${D}=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:\:\:/\:\:{x}+{y}\leqslant\mathrm{1}\:{and}\:{x}\geqslant\mathrm{0}\:{and}\:{y}\geqslant\mathrm{0}\:\right\}. \\ $$
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) .](https://www.tinkutara.com/question/Q27756.png)
$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}−{y}\:\:\:{and}\:\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:{so}\:\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\:\:\int_{\mathrm{0}} ^{\mathrm{1}−{y}} \:{ln}\left(\mathrm{1}+{x}+{y}\right){dx}\right){dy}\:{but}\:{the}\:{ch}.\:\mathrm{1}+{x}+{y}\:={t}\:{give} \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}−{y}} \:{ln}\left(\mathrm{1}+{x}\:+{y}\right){dx}=\:\int_{\mathrm{1}+{y}} ^{\mathrm{2}} {lnt}\:{dt}\:=\:\left[{tlnt}\:−{t}\right]_{\mathrm{1}+{y}} ^{\mathrm{2}} \\ $$$$=\:\mathrm{2}{ln}\mathrm{2}\:−\mathrm{2}\:−\left(\mathrm{1}+{y}\right){ln}\left(\mathrm{1}+{y}\right)\:+\mathrm{1}+{y} \\ $$$$=\mathrm{2}{ln}\mathrm{2}\:−\mathrm{1}\:+{y}\:−\left(\mathrm{1}+{y}\right){ln}\left(\mathrm{1}+{y}\right) \\ $$$${I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{ln}\mathrm{2}−\mathrm{1}\right){dy}\:+\int_{\mathrm{0}} ^{\mathrm{1}} {ydy}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{y}\right){ln}\left(\mathrm{1}+{y}\right){dy} \\ $$$${I}=\:\mathrm{2}{ln}\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{y}\right){ln}\left(\mathrm{1}+{y}\right){dy}\:{the}\:{ch}.\mathrm{1}+{y}={t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+{y}\right){ln}\left(\mathrm{1}+{y}\right){dy}=\:\int_{\mathrm{1}} ^{\mathrm{2}} {tln}\left({t}\right){dt} \\ $$$$=\left[\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:{lnt}\right]_{\mathrm{1}} ^{\mathrm{2}} \:\:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}{dt}\:\:=\:\mathrm{2}{ln}\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} \:=\mathrm{2}{ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{2}{ln}\mathrm{2}\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${I}=\:\mathrm{2}{ln}\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:−\mathrm{2}{ln}\mathrm{2}\:+\frac{\mathrm{3}}{\mathrm{4}}=\:\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$