In this question I was supposed to find the $E[X]$ and $\text{Var}[X]$.
I found $E[X] = 0.6$ and $\text{Var}[X] = 0.3696$ but when I checked the answer it shows that my $E[X]$ is correct but my $\text{Var}[X]$ is wrong!!!
What is the correct answer for $\text{Var}[X]$?
ThankYou in advance.

Integrating
$$\int_x^{\infty}f(x,y)dy=f_X(x)=\frac{125}{2}x^2e^{-5x}=\frac{5^3}{\Gamma(3)}x^{3-1}e^{-5x}$$
we immediately realize that
$$X\sim \text{Gamma}[3;5]$$
Thus
$$\mathbb{E}[X]=\frac{3}{5}$$
and
$$\mathbb{E}[X]=\frac{3}{5^2}$$
wihtout any other calculations