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Evaluate$$ \iint_R \left ( e^{-x-y} \right )dxdy, $$ where $R$ is the region in the first quadrant in which $x+y\leq 1$.

I think the first step is $$\int_{0}^{1}\int_{0}^{1-y}\left ( e^{-x-y} \right )dxdy $$but I am not sure about whether this step is correct or not.

Please tell whether my answer is correct or not and please solve it. Thank you.

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Hint:$\int\int e^{-x-y}dxdy=\int e^{-x}\int e^{-y}dydx$ – BigM Mar 16 '14 at 15:05
Does the interval that I evaluated correct? – Biboo Chung Mar 16 '14 at 15:06
Yeah.its correct. – BigM Mar 16 '14 at 15:08
Thank you very much. – Biboo Chung Mar 16 '14 at 15:10
Thank you very much – Biboo Chung Mar 16 '14 at 15:19

2 Answers 2

up vote 1 down vote accepted

Being rigorous in the notation will help you. The domain is $x\ge 0$, $y\ge 0$ and $x+y\leq 1$ (or $0\le y\le 1-x$), i.e. a triangle. You see that when $x$ ranges in $[0,1]$, and $y$ ranges in $[0,1-x]$.

$\int^1_{x=0} e^{-x}(\int^{1-x}_{y=0}{e^{-y}dy})dx=\int^1_{x=0} e^{-x}(-e^{-y}|^{1-x}_{y=0})dx=\int^1_{x=0}e^{-x}(1-e^{x-1})dx=\int^1_{x=0}(e^{-x}-\frac{1}{e})dx=(-e^{-x}-\frac{x}{e})|^1_{x=0}=1-\frac{2}{e}$

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Hint. $$ \int_{0}^{1}\int_{0}^{1-y}\left ( \mathrm{e}^{-x-y} \right )dx\,dy =\int_0^1\left( \int_0^{1-y}\mathrm{e}^{-x}\,dx\right)\, \mathrm{e}^{-y}\,dy= \int_0^1\left(1-\mathrm{e}^{y-1}\right)\, \mathrm{e}^{-y}\,dy. $$

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Thank you very much – Biboo Chung Mar 16 '14 at 15:17

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