Let the distribution of a random vector (X,Y) be given by density. Are X,Y independent? Let the distribution of a random vector $(X,Y)$ be given by density.
$$f(x,y)=\left\{\begin{matrix}
e^{-x-y} & x,y\geq 0\\ 
0 & x<0 \hspace{2mm} \vee y<0
\end{matrix}\right.$$
Are  $X,Y$ independent?
My take:
So I have to calculate $fx(x) \cdot fy(y)$ and check if they are equal to $f(x,y)$
$\int_{0}^{\infty}e^{-x-y}dy=e^{-x}$
$\int_{0}^{\infty}e^{-x-y}dx=e^{-y}$
$fx(x) \cdot fy(y)=f(x,y)$?
$e^{-x} \cdot e^{-y}=e^{-x-y}$
The equation is right so (X,Y) are independent
(If for example $x,y \in[1,5]$, then I would have to calculate intergral with upper $5$ and lower $1$, correct?)

You are mostly correct, but as Henry commented, you should include the support with the expressions (and using indicators to do so saves type space).   This verifies that the support for the product of the marginal functions is identical to the support for the joint function as well as the product and the joint being equal everywhere in that support. $$\begin{align}\because\qquad f_{\small X,Y}(x,y)&=\mathrm e^{-x-y}\,\mathbf 1_{0\leqslant x}\mathbf 1_{0\leqslant y}\\[2ex]f_{\small X}(x)&=\mathrm e^{-x}\mathbf 1_{0\leqslant x}\int_0^\infty \mathrm e^{-y}\,\mathrm d y\\[1ex]&=\mathrm e^{-x}\mathbf 1_{0\leqslant x}\\[2ex]f_{\small Y}(y)&=\mathrm e^{-y}\mathbf 1_{0\leqslant y}\int_0^\infty \mathrm e^{-x}\,\mathrm d x\\[1ex]&=\mathrm e^{-y}\mathbf 1_{0\leqslant y}\\[2ex]\therefore\qquad f_{\small X}(x)\,f_{\small Y}(y)&=\mathrm e^{-x}\mathbf 1_{0\leqslant x}\cdot\mathrm e^{-y}\mathbf 1_{0\leqslant y}\\[1ex]&=f_{\small X,Y}(x,y)\end{align}$$

$\qquad\mathbf 1_{E}=\begin{cases}1&:& E\text{ is true}\\0&:& \text{otherwise}\end{cases}$
