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
$find I= ∫∫_D ln(1+x+y)dxdy with D= {(x,y)∈R^2 / x+y≤1 and x≥0 and y≥0 }.$
$f i n d I = \int \int_{D} l n (1 + x + y) d x d y w i t h$ $D = {(x, y) \in R^{2} / x + y ⩽ 1 a n d x ⩾ 0 a n d y ⩾ 0} .$
Commented by abdo imad last updated on 14/Jan/18

$0 ⩽ x ⩽ 1 - y a n d 0 ⩽ y ⩽ 1 s o$ $I = \int_{0}^{1} (\int_{0}^{1 - y} l n (1 + x + y) d x) d y b u t t h e c h . 1 + x + y = t g i v e$ $\int_{0}^{1 - y} l n (1 + x + y) d x = \int_{1 + y}^{2} l n t d t = {[t l n t - t]}_{1 + y}^{2}$ $= 2 l n 2 - 2 - (1 + y) l n (1 + y) + 1 + y$ $= 2 l n 2 - 1 + y - (1 + y) l n (1 + y)$ $I = \int_{0}^{1} (2 l n 2 - 1) d y + \int_{0}^{1} y d y - \int_{0}^{1} (1 + y) l n (1 + y) d y$ $I = 2 l n 2 - \frac{1}{2} - \int_{0}^{1} (1 + y) l n (1 + y) d y t h e c h . 1 + y = t g i v e$ $\int_{0}^{1} (1 + y) l n (1 + y) d y = \int_{1}^{2} t l n (t) d t$ $= {[\frac{t^{2}}{2} l n t]}_{1}^{2} - \int_{1}^{2} \frac{t}{2} d t = 2 l n 2 - \frac{1}{2} {[\frac{t^{2}}{2}]}_{1}^{2} = 2 l n 2 - \frac{1}{2} (\frac{3}{2})$ $= 2 l n 2 - \frac{3}{4}$ $I = 2 l n 2 - \frac{1}{2} - 2 l n 2 + \frac{3}{4} = \frac{1}{4} .$
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