I am stuck at this problem of probability. $f(x,y)=ce^{-(x+y)}$ when $0< x< y+1$ if $x$ didn't belong to $0< x< y+1$ then $f(x,y)=0$
The questions are : 
Find $c$ 
Determine the distribution function $F(y)$
 A: You have the right idea, it's just that the integral you set up (as I read from your comment) appears to be wrong. So you recognize that integrating $f(x,y)$ over $\Bbb R^2$ should equal $1$ since it is a probability distribution, and since $f(x,y)$ is only nonzero whenever $(x,y)\in A$, you only need integrate over $A$. From the question we know that $A=\{(x,y)\in\Bbb R^2:0<x<y+1\}$, and $f(x,y)=ce^{-x-y}$ when $(x,y)\in A$ and $f(x,y)=0$ otherwise. So we only integrate over $A$, which does not have bounds $x-1<y<\infty$ and $0<x<y+1$, but rather has bounds $0<x<y+1$ and $-1<y<\infty$ where we integrate with respect to $x$ first. If these bounds are unclear, I suggest graphing the lines $x=0$ and $x=y+1$ (which are the bounds for $x$), and seeing which values of $y$ are in that domain (i.e. the bounds on $y$). So the problem then becomes solving for $c$ in:
$$\int_{-1}^{\infty}\int_0^{y+1}ce^{-x-y}\,dxdy=1$$
Now that you see the nonzero domain of $f(x,y)$ and its bounds, I will leave finding $F(y)$ to you. Just to get you started, I am assuming $F(y)$ is the marginal distribution of $y$ (since $F$ is only a function of $y$ I assume this, although in my classes we use different notation), so you would use the formula:
$$F(y)=\int_xf(x,y)dx$$
which means you integrate $f(x,y)$ over $x$ (what are the bounds on $x$?).
