The random vector $(X,Y)$ has a density of $f_{X,Y} (x,y)=cxy I_D(x,y)$ where $D\subset \mathbb{R}^2$ the triangle with points $(0,0), (1,0), (1,1)$ and $c$ constant.
Calculate $c$ and $f_X$, $f_Y$
I'd start with calculating $c$. The area of the triangle is $1/2$. Therefore, $$1/2 = \int \int_D xyc \, dx\, dy$$
$D=\{(x,y)|x\in [0,1], y\in [0,1], x\geq y\}$.
Next, since I have $y\leq x$, I can compute the integral
$$1/2=\int_0^1 \int_0^{x} xyc \, dy\, dx \Rightarrow c = 4$$
Is that OK?
For the density functions I computed
$$f_X(x)=\int^{\infty}_{-\infty} f_{X,Y} dy = \begin{cases} cx/2 \quad y\in [0,1]\\ 0 \quad \text{else} \end{cases}$$
$$f_Y(x)=\int^{\infty}_{-\infty} f_{X,Y} dx = \begin{cases} cy/2 \quad x\in [0,1]\\ 0 \quad \text{else} \end{cases}$$
Am I correct? Thanks for any helpful input.
A density function always integrate to $1$.
$$1=\int_0^1 \int_0^x xy c \, dy\, dx$$
$$1=\int_0^1 \frac{x^3}2c\, dx$$
$$c=8$$
We have $$f_X(x) = \int_0^x cxy\, dy=\frac{c}2x^3=4x^3, 0 \le x \le 1$$