I'm given the family $(X, Y)$ of random variables with join density:
$$f^{X, Y}(x, y) := 2 \cdot e^{-(x+2y)} \cdot 1_{[0,\infty)}(x) \cdot 1_{[0, \infty)}(y)$$
and I'm required to calculate $P(X > Y)$. I already calculated the marginal density $f^X, f^Y$ and established that $X$ and $Y$ are independent, but didn't use this information, but calculated:
$$\int_{\{(x, y)\,\in\,\mathbb{R}^2 : x\,>\,y\}} f^{X, Y}(x,y) \,d(x,y) \\ = \int_\mathbb{R}\int_\mathbb{R} 2\cdot e^{-(x+2y)} \cdot 1_{[0, \infty)}(x) \cdot 1_{[0, x)}(y) \,dy\,dx \\= \int_\mathbb{R} 1_{[0, \infty)}(x) \int_0^x 2e^{-(x+2y)} dy\,dx \\= \int_0^\infty e^{-x} - e^{-3x}dx = 1 - \frac{1}{3} = \frac{2}{3}$$
Is this approach correct? Do I need to change $x$ in the second equality (since it's part of the bounds) or is it fine, since I integrated $dy$? Can this problem be solves quicker using $f^X, f^Y$ and independence?
You calculation is fine. You can double check your result by changing the order of integration.
$$\int_0^{\infty} \int_y^{\infty} 2\cdot e^{-(x+2y)} \, dx \, dy=\frac23$$