$\displaystyle f(x,y)=\left\{

\begin{array}{lr}

\frac{2x+y}{4}&0\le x\le 1;0\le y\le 2\\

0&otherwise

\end{array}

\right.$

Question: Findcdf(remember to consider all cases).

Answer:

$\displaystyle x \le 0 \, , F(x,y)=0$

$\displaystyle y \le 0 \, , F(x,y)=0$

$\displaystyle x > 1\, , y > 2, \, F(x,y)=1$

$\displaystyle 0 \le x \le 1, 0 \le y \le 2$ :

$\displaystyle F(x,y)=\int_{0}^{x}\int_{0}^{y}\frac{2u+v}{4}\,dvd u = \frac{xy}{8}(2x+y)$

Shouldn't that be it? What's with these next two cases if x and y are never greater than 1 and 2 respectively?

$\displaystyle 0 \le x \le 1 \, , y \ge 2 \, : F(x,y)=\frac{x}{2}(x+1)$

$\displaystyle 0 \le y \le 2 \, , x \ge 1 \, , F(x,y)=\frac{y}{8}(2+y)$