Let $X$ and $Y$ be continuous random variables. Given the joint pdf $f(x,y)=2e^{-x}e^{-y}$ defined over the shaded region pictured below, I want to determine the marginal pdf of $Y$. Would it be $$\int_0 ^y f(x,y)dx$$ or $$\int_0 ^\infty f(x,y)dx?$$
On the one hand, $0\le x \le y$ in the shaded region. On the other hand, since the shaded region is infinite, $0\le y\le \infty$ implies $0\le x \le \infty $
On
$$\int_{0}^{\infty}f_{(X,Y)}(x,y)dx=\int_{0}^{\infty}f_{(X\mid Y=y)}(x)f_Y(y)dx\\=f_Y(y)\int_{0}^{\infty}f_{(X\mid Y=y)}(x)dx$$
$f_{(X\mid Y=y)}(x)>0$ when $0\le x \le y$ and $f_{(X\mid Y=y)}(x)=0$ when $x>y$
Then $$\int_{0}^{\infty}f_{(X\mid Y=y)}(x)dx=\int_{0}^{y}f_{(X\mid Y=y)}(x)dx+\int_{y}^{\infty}f_{(X\mid Y=y)}(x)dx=\int_{0}^{y}f_{(X\mid Y=y)}(x)dx$$
Draw a horizontal line at $Y=y$ and you will see that this cuts the shaded region where $X\in[0;y]$. This is the interval over which you must "integrate out" the other random value. As can also be seen:
The support is $\{(x,y)\in\Bbb R^2: 0\leq x\leq y\}$, or $\{(x,y)\in\Bbb R^2: y\in[0;\infty), x\in[0;y]\}$ , so:
$$f_Y(y) ~{= \int_\Bbb R f_{X,Y}(x,y)\operatorname d x\\ = \int_\Bbb R 2 \mathsf e^{-x}\mathsf e^{-y}\mathbf 1_{y\in[0;\infty)}\mathbf 1_{x\in[0;y]}\operatorname d x \\ = 2\mathsf e^{-y}\mathbf 1_{y\in[0;\infty)}\cdot \int_{[0;y]}\mathbf e^{-x}\operatorname d x \\ ~\vdots}$$