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I have two functions $f_x$ = $\frac{1}{2}\delta(x-5) + 1/4$ where the 1/4 corresponds to a uniform distribution from 5 to 7. I also have $f_{y|x}$ = $\frac {1}{2}\delta(y-x-4) + 1/4$ which is 1/4 in the range of x+4 to x+6. I am trying to find $f_y$ using the equation: $$ f_y = \int f_x f_{y|x} \,\mathrm dx \>. $$ My question is: How do I find the limits of this integration? I had thought something like integrating from $y-4$ to $y-6$ but that doesn't really seem correct.

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Deficiencies in your notation may be causing some confusion. Presumably "$\delta$" represents the point measure at zero. Don't, however, conflate constants with uniform distributions! For instance, your first "$1/4$" could more explicitly be written as $1/4 I_{[5,7]}(x)$ where $I_A$ is the indicator function for a set $A$. (This is a guess: I presume you intend $f_x dx$ to be a probability measure.) When you do this, the limits of integration do not matter: all integrals can be viewed as being over the entire Real line. – whuber♦ Jan 14 at 15:24
Right, sorry I'm bad about leaving out specifics sometimes. I understand that ${\bf f_x}$ is only valid between 5 and 7 and that my ${\bf f_{y|x}}$ is valid only between x+4 and x+6. I think that this would mean that my resulting answer for ${\bf f_y}$ would only be valid between 9 and 13? I'm just not totally sure how to get there. – nomad2986 Jan 14 at 17:38
I think confusion may arise in part due to your notation. When you write $f_{y|x}$, it reads as though you are thinking of $y$ and $x$ as separate random variables drawn from a joint distribution. However, your narrative suggests that you intend for $y$ to be a function of $x$. In this latter case, once $x$ is known, $y$ is known: the distribution of $y$ conditioned on $x$ is a single point, $y(x)$. Which is it? – Arthur Small Jan 14 at 19:16
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So your model could be written: $x_1 \sim Unif(5,7)$, $x_2 \sim Unif(4,6)$, $y=x_1+x_2$; and your question is, What is the distribution of $y$. Is this correct? – Arthur Small Jan 14 at 20:12
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If so, and if $x_2$ does not depend on $x_1$ in any way, then the answer's pretty obvious: $y \sim Unif(9,13)$. – Arthur Small Jan 14 at 20:15
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@whuber is correct on both counts (and should be given credit for an accepted answer, imho).

In general, to derive a marginal distribution, you integrate the joint distribution over the entire support of the variable you are integrating out. In this case, integrating wrt $x$ over the entire real line gives the correct answer.

Working from first principles: recall how a conditional distribution is defined in terms of a joint distribution:

$f(x,y) = f(y \mid x) f(x)$

The marginal distribution of $y$ is given by integrating out $x$ over its entire support (the closure of the set on which $x \ne 0$). You get the same answer if you integrate over the entire real line:

$f(y) = \int_\mathbb{R} f(x,y) dx = \int_\mathbb{R} f(y \mid x) f(x) dx$.

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