Vtyjkyuk

$$f_X, Y (x, y) =begincases
24xy & x geq 0, ygeq0, x+y leq 1\
0 & textotherwise \
endcases
$$

Compute $Var(X)$.

This question was done with a shortcut:

$f_X(x) = int_0^1-x24xydy = 12x(1-x)^2, 0 leq x leq 1$ and zero otherwise.

Recall that this is $beta(a = 2, b = 3)$, and that $V(X) = fracab(a+b)^2(a+b+1)$. So.

$V(X) = frac2 cdot 3(2 + 3)^2(2+3+1) = 0.04$.

My question is how did they get the interval $0 leq x leq 1$? $x geq 0$ and $x leq 1 - y$. :/

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 $intlimits_x=0^1 x f_X(x) dx = 2 over 5$.

The variance is:

$intlimits_x=0^1 (x - mu)^2 f_X(x) dx = 1 over 25 = 0.04$ (as you calculated).