$$f_{X, Y} (x, y) =\begin{cases} 24xy & x \geq 0, y\geq0, x+y \leq 1\\ 0 & \text{otherwise }\\ \end{cases} $$
Compute $Var(X)$.
This question was done with a shortcut:
$f_X(x) = \int_{0}^{1-x}24xydy = 12x(1-x)^2, 0 \leq x \leq 1$ and zero otherwise.
Recall that this is $beta(a = 2, b = 3)$, and that $V(X) = \frac{ab}{(a+b)^2(a+b+1)}$. So.
$V(X) = \frac{2 \cdot 3}{(2 + 3)^2(2+3+1)} = 0.04$.
My question is how did they get the interval $0 \leq x \leq 1$? $x \geq 0$ and $x \leq 1 - y$. :/
On
Well, let's take a look how the conditions look on (x,y) plane
The blue line is the equation $x+y=1$. Now since we are interested in the positive quadrant $x>0$ and $y>0$, I only drew that part. So the area surface where your function $f(x,y)$ is non-zero is within the triangle where $x$ moves along the red arrow and $y$ moves along the black one.
Notice that $0 < x < 1$ and since $y$ is below the line then $x+y<1$ or $y<1-x$.
Here is the distribution:
Here is $f_X(x) = 12 x (1 - x)^2$:
The mean is shown in red and $\pm \sigma$ in purple.
The mean is $\int\limits_{x=0}^1 x f_X(x)\ dx = {2 \over 5}$.
The variance is:
$\int\limits_{x=0}^1 (x - \mu)^2 f_X(x)\ dx = {1 \over 25} = 0.04$ (as you calculated).