Probability density function (PDF) problem Problem
$X$ and $Y$ random variables, the common probability density function of $X$ and $Y$ is given as follows:
$$f(x,y)=
\begin{cases}
ke^{-x-y},&\textrm{when  } x\geqslant y\geqslant 0\\
0\;,&\textrm{otherwise  }
\end{cases}
$$
a) Find the constant $k$

Proposed Solution
$$\int_{y=0}^{\infty}\int_{x=0}^{\infty} ke^{-x-y} dxdy = 1$$


Is any of my work correct? Any feedback is much appreciated, and if you think I should add more details to my calculations, please point it out and I will edit my work accordingly.
Thank you for your time.
 A: a) $\int_0^{\infty} \int_y^{\infty} k\cdot e^{-x-y} \, dx \, dy$
inner integral: $\int_y^{\infty} k\cdot e^{-x-y} \, dx=k\cdot e^{-y}\int_y^{\infty} e^{-x} \, dx$
$$=k\cdot e^{-y}\cdot \left(-e^{-x}\bigg|_y^{\infty}\right)=k\cdot e^{-y}\cdot(-0-(-e^{-y}))=k\cdot e^{-2y}$$
Now the outer integral: $\int_0^{\infty} k\cdot e^{-2y} \, dy=k\cdot \int_0^{\infty} e^{-2y} \, dy$
$$=k\cdot  \left(-\frac12 \cdot e^{-2y}\bigg|_0^{\infty}\right)=k\cdot(-0-(-\frac12))=\frac{k}2=1$$
$\Rightarrow k=...$
A: Reprising this post
After understanding that $k=2$, your joint density is the following
$$f_{XY}(x,y)=2e^{-x}e^{-y}\mathbb{1}_{[0;+\infty)}(x)\mathbb{1}_{[0;x]}(y)$$
or equivalently
$$f_{XY}(x,y)=2e^{-x}e^{-y}\mathbb{1}_{[0;+\infty)}(y)\mathbb{1}_{[y;+\infty)}(x)$$
now to get the marginals  you have only to integrate the opposite variable "copying" the integral extremes in the $f(x,y)$ formula
Thus
$$f_X(x)=\int_0^x [2e^{-x}e^{-y} ]dy=2e^{-x}(1-e^{-x})\mathbb{1}_{[0;+\infty)}(x)$$
$$f_y(y)=\int_y^{+\infty} [2e^{-x}e^{-y} ]dx=2e^{-2y}\mathbb{1}_{[0;+\infty)}(y)$$
$Y\sim Exp(2)$
