I have the joint density function given by $$f(x,y) = 1 \hspace{2mm}\mathrm{for}\hspace{2mm} 0 \leq x \leq 2, \max\{0,x-1\} \leq y \leq \min \{1,x\}, $$ and $f(x,y) = 0$ otherwise. And I want to find the joint distribution function.
My initial idea was to split it into two cases, where $x < 1$ (where we then have $0 < y < x$) and $x > 1$ (where we then have $x-1 < y < 1$), and solve the integral $$\int_0^1 \int_0^x 1\hspace{1mm} dy dx + \int_1^2 \int_{x-1}^1 1 \hspace{1mm} dydx, $$
but apparently that is completely wrong. The solution is divided into six cases, where the first case is $F(x,y) = 1$ for $x > 2, y > 1$, and that does make sense to me. But after that I'm completely lost. I don't even understand why they are dividing the solutions into these particular cases. For example, the second case is $F(x,y) = y$ for $0 < y < 1$ and $x-1 > y$. But I have no idea why I should be considering those specific bounds or overall how to attack this problem.
Look at the following picture
We seek to find $F(x_0,y_0)=Pr(X<x_0,Y<y_0)$. As you can see if $x_0<0$ or $y_0<0$ the probability is zero. Therefore we assume $x_0,y_0>0$. For $x_0<1$we have:$$F(x_0,y_0)=\dfrac{1}{2}\{\min(x_0,y_0)\}^2$$and for $1<x_0<2$ if $y_0<x_0-1$ we have $$F(x_0,y_0)=y_0$$ and for $y_0>x_0-1$ $$F(x_0,y_0)=y_0-\dfrac{1}{2}(y_0-x_0+1)^2$$ so for $1<x_0<2$$$F(x_0,y_0)=y_0-\dfrac{1}{2}\max\lbrace 0,y_0-x_0+1\rbrace^2$$if $x_0>2$ it is clear that $F(x_0,y_0)=\min(y_0,1)$