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I want to solve the integral

$$\int_0^1 \int_0^{1-x} \exp \left(\frac{y}{x+y} \right) \; \mathrm dy \; \mathrm dx$$

and I am given the hint to use the substitution $x+y=u$ and $y = uv$. Now, I've never done such multiple substitutions before and I figured I'd just try and if I'm not mistaken, after substitution the integral looks as follows:

$$\iint u \cdot e^v \; \mathrm dv \; \mathrm du.$$

Now - given I am correct - how do I have to adjust the integral bounds? For $v$ I set $uv = 1-x$ and after some substitutions, I ended up with $v=\frac{1-u}{2u}$. However, when I try to find an integral bound for $u$, I fail, because $u = \frac{1}{1-v}$ and I don't know how to substitute $v$ such that the bound becomes a number (it should be one, or shouldn't it?).

So my question is: What are the correct integral bounds and how do I find them? Also, is my integral without bounds correct? Thanks for any answers in advance.

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1  
And don't forget the Jacobian. –  mixedmath Jul 13 '11 at 20:57
    
If you have never done multiple substitutions before, maybe now is the time to look it up in your textbook! –  GEdgar Jul 13 '11 at 21:13

1 Answer 1

up vote 4 down vote accepted

The integral domain is the triangle determined by $(0,0)$, $(0,1)$ and $(1,0)$. From $u=x+y$ one can see $u$ varies from $0$ to $1$. For each fixed $u_0$, think of a line parallel to $x+y=1$, with the intercept being $u_0$. Since $v=y/u_0$ and $y$ varies from $0$ to $u_0$ (only consider the segment within the triangle), one can see $v$ also varies from $0$ to $1$.

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