Based on multivariable density find singlevariable one. Random variable has a density function of $$g(x,y)=1_{\{0 \le y \le 1-|x|\}}$$
Now I need to find a density of those two variables (X and Y separately) , however I have doubts about integration. What should be the limits of those integrals how they precisely should look?
 A: Hint: The domain of $g(x,y)$ is 
$$-1\le x \le 1 \qquad 0\le y\le 1-|x|$$ or if you solve the absolute value $$(-1\le x\le 0,\, 0\le y\le 1+x) \qquad \text{and}\qquad (0< x\le 1, \, 0\le y\le 1-x)$$

So $$g_X(x)=\int_{Y}g(x,y)dy=\begin{cases}\int_{0}^{1-x}dy, & \text{if } 0<x\le 1\\[0.3cm]\int_{0}^{1+x}dy, & \text{if } -1<x\le 0\end{cases}$$ and similarly for $g_Y(y)$.
A: Let our random variables be $X$ and $Y$.
The key is to draw the region where the joint density function of $(X,Y)$ is non-zero.
For $x\ge 0$, we want $0\le y\le 1-x$. Draw the line $y=1-x$. We are looking at the part of the first quadrant that is below that line.
For $x\lt 0$, we want $0\le y\le 1-(-x)=1+x$. Draw the line $y=1+x$. We are looking at the part of the second quadrant which is below the line $y=1+x$.
So our region is the triangle with corners $(1,0)$, $(0,1)$, and $(-1,0)$.
Now for the density function of $Y$, we "integrate out" $x$. For any fixed $y$ between $0$ and $1$, $x$ travels from the line $y=1+x$ to the line $y=1-x$, so we integrate from $x=y-1$ to $x=1-y$.
For the density function of $X$, we integrate out $y$. For any fixed $x$ between $-1$ and $0$, $y$ goes from $0$ to the line $y=1+x$. For $x$ between $0$ and $1$, $y$ travels from $0$ to $1-x$.
Remark: We used integration because the question asked for that. However, in this case the cdf of $X$ and the cdf of $Y$ can be found by a simple area calculation.
