A device contains two circuits. The second circuit is a backup for the first, so the second is used only when the first has failed. The device fails when and only when the second circuit fails. Let $X$ and $Y$ be the times at which the first and second circuits fail, respectively. $X$ and $Y$ have joint probability density function
$$f_{X,Y}(x,y)=\begin{cases} 6e^{-x} e^{-2y} & 0<x<y<\infty \\[1ex] 0 & \text{otherwise} \end{cases}$$
What is the expected time at which the device fails?
I understand the steps of the solution. The $f_Y(y)$ needs to be derived first and then used for the E(Y) calculation.
$$\begin{align}f_{Y}(y) & =\int_{0}^{\color{Red}{y}} (6e^{-x} e^{-2y})\;\mathrm dx \\[1ex] & =6e^{-2y}-6 e^{-3y}\\[2ex]\mathsf E(Y) & =\int_{\color{Red}{0}}^{ \infty} y(6e^{-2y}-6 e^{-3y})\;\mathrm dy\end{align}$$
I have a question to the the intervals used in the first and second integral. The interval for the first integral (for $X$) was set from $0$ to $y$, however the $x$ was skipped in the second integral (for $Y$) and it was set from zero to $0$ to $\infty$ (why not for example from $x$ to $\infty$ analogically to the first case?)
I know that technically applying $x$ in the second integral doesn't give any final number but I just don't get why the variable was skipped in one case not in the other. $X$ could be set from $0$ to $\infty$ either following this logic.
For the first integral, you must answer: what are the minimum and maximum values of $X$ for a particular value of $Y$; specifically where $Y=y$. This is $0\leq X\leq y$.
For the second integral, you must answer: what are the minimum and maximum values $Y$ for all values of $X$; (since $X$ has been marginalised). This is $0\leq Y < \infty$.