- To find c, we use the fact that the area under the PDF curve integrates to 1.
$$1=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}f(x,y) \ dxdy=\int_{0}^{1} \int_{0}^{1-x}(cx+1) \ dydx=\int_{0}^{1} (cx+1)(1-x) ,dx
=\frac{1}{2}+\frac{1}{6} c$$
So c=3.
- To answer we need to find the marginal for both X and Y.
$$f_{X}(x)= \int_{0}^{1-x} (3x+1) \ dy=(3x+1)(1-x)
$$
for x $\in$ [0,1]
$$f_{X}(x)=\begin{cases}(3x+1)(1-x) & \text{for } 0 \leq x \leq 1\0 & \text{otherwise}\end{cases}$$
For Y marginal distribution:
$$f_Y(y) = \int_{0}^{1-y} (3x+1) dx = 3x^2/2 + x |_{0}^{1-y} = 3(y^2 - 2y +1)/2 + 1 - y = (3/2)y^2 - 4y + 5/2=\frac{1}{2}(1-y)(5-3y)
$$
for y $\in$ [0,1]
$$f_{Y}(y)=\begin{cases}\frac{1}{2}(1-y)(5-3y) & \text{for } 0 \leq y \leq 1\0 & \text{otherwise}\end{cases}$$
Now clearly $f_{XY}(X,Y) \neq f_{X}(X) f_{Y}(Y)$. For example, consider the point $(x,y) = (2/3,2/3)$. Then $F_X(2/3) > 0$ and $F_Y(2/3) > 0$, but since $2/3+ 2/3 > 1$, $f_{XY}(2/3,2/3) = 0$.