How to find CDF of PDF i'm doing exercise (i'm a student) for probability and there is something that I don't understand, How do we manage to find bound of $CDF$ given a $PDF$.
the PDF : $f(x,y) = k$ , is obviously uniformly distributed have the constraint of $0<x<y<1$.
well we know that $\int_{}^{}\int_{}^{}f(x,y)dydx = 1$
$\int_{0}^{1}\int_{x}^{1}kdydx = 1$
$k/2 = 1$ then $k = 2$ we can verify that $k = 2$ by integrating the functions in all of it's domains and it is giving $1$
When I want to find the CDF I do this :
$\int_{0}^{x}\int_{0}^{y} 2dvdu = \int_{0}^{x} 2y du = 2xy$
I know that i'm wrong because $x$ and $y$ are dependant each other and here i'm counting also the part where where $x>y$ but I dont know to bound the integrals to respect the dependance. I followed the formula of a CDF given PDF : $F(x,y) = \int_{-\infty}^{x}\int_{-\infty}^{y} f(u,v)dvdu$
Thanks you in advance.

The joint cdf is given by
$P(X\leq u,Y\leq v)$ . As $f(x,y)>0$ for $x<y<1$. We have to look at whether $u<v$ or $u>v$ and separate the cases .
if $u\geq v$ . Then
$P(X\leq u, Y\leq v) = P(X\leq v, Y\leq v) = \int_{0}^{v}\int_{0}^{y}2\,dx\,dy=v^{2}$.
If $u<v$ . Then we have $P(X\leq u,Y\leq v)=2(uv-\frac{1}{2}u^{2})$
So $$F(x,y)= \begin{cases} y^{2}\cdot\mathbf{1}_{\{x\geq y,\,0\leq y\leq 1\}}\\2(xy-\frac{x^{2}}{2})\cdot\mathbf{1}_{\{x<y,\,0\leq y\leq1\}}\\2(x-\frac{x^{2}}{2})\cdot\mathbf{1}_{\{y>1\,,\,0\leq x\leq 1\}}\\1\cdot \mathbf{1}_{\{x>1,y>1\}} \\0,\,\text{elsewhere} \end{cases}$$
