Hopefully a simple question to clear up some understanding.
Goal: calculate $f_Y(y)$
My thoughts: to calculate $f_Y(y)$ we need to integrate with respect to $dx$ so in my mind from calculus this is like considering (infinitesimally) thin vertical slices of width $dx$ where each slice has area of $dA = (x+1) - x\: dx$. In other words constant height 1 and integrating over all such $dx$.
If we imagine starting at $x=0$ and sliding the thin vertical slice from left to right we can see the left and right bounds of $x$ now have two equations for them indicating we need to split $f_Y(y)$ up.
My question is whether this is a valid way to think of setting up the question by relating to calculus with areas and slicing or whether I should be setting things up differently. The reason I am asking is that when I first did this question I set it up with $x$ limits from 0 to 1, which is incorrect.
Solution here in case needed (Q5).
So we can model $f(x,y)$ as \begin{equation} f(x,y) = K \qquad \text{if } (x,y) \in A \end{equation} and $0$ elsewhere. So, let's find $K$, \begin{equation} \int_{\mathbb{R}^2} f(x,y) \ dxdy = \int_{A} f(x,y) \ dxdy = 1 \end{equation} i.e. \begin{equation} K \int\limits_0^1 \int\limits_{x}^{x+1} \ dydx = 1 \end{equation} i.e. \begin{equation} K = 1 \end{equation} An alternative is that the area of the trapezoid is $A = \int_{A} \ dy dx = 1$. Now, let's find $f(y)$ by marginalizing out $x$,i.e. \begin{equation} f(y) = \int f(x,y) \ dx \end{equation} Notice that $y$ is defined between $0$ and $2$ so we must cover all this range to define $y$. If $0 < y < 1$, $x$ will vary from $x = 0$ till the lower side of the trapezoid defined by $x = y$, so we should get \begin{equation} f(y) = \int_0^y \ dx = y, \qquad \text{ if } 0 < y <1 \end{equation} On the other hand, if $1 < y < 2$, then $x$ will vary from the upper side of the trapezoid defined by $y = x+ 1$ (or $x = y-1$) till $x = 1$, i.e. \begin{equation} f(y) = \int_{y-1}^1 \ dx = 2-y, \qquad \text{ if } 1 < y <2 \end{equation}