Compute the $Cov(X,Y)$ of the two random variables Let $X,Y$ be absolutely continuous random variables. $X$ is $Uniform[0,12]$ and $f_{Y|X}(y|x)=\frac{1}{x}$, for $y\in [0,x]$ and $0$ otherwise.
We are asked to find $Cov(X,Y)$.

$Cov(X,Y)=E(XY)-E(X)E(Y)$.
Usually in these types of problems, I'm given $Y=X^2, X+3$, or a function of $X$. Here I am given the conditional function.
So, $f_{Y|X}=\dfrac{f_{XY}(x,y)}{f_{X}(x)}$, by definition.
We know the $f_X(x)=\dfrac{1}{12}; 0\leq x \leq 12$, as it is uniform.
Then, we get that $\dfrac{1}{x}=\dfrac{f_{XY}(x,y)}{(1/12)}\to f_{XY}(x,y)=\dfrac{1}{12x}$
This is as far as I got.
To find $f_Y$, I need to integrate $f_{XY}$ over a range ($x$ values). What is this range?
We have firstly that $0 \leq y \leq x$, and then we have $0 \leq x \leq 12$. How can I combine these two?
I don't know what to integrate $f_Y$ over.
 A: Hint:
You know $0 \le x \le 12$ and $0 \le y \le x$, which you can combine into $0 \le y \le x \le 12$
Write the joint density with an indicator function taking the value $1$ when the condition is met and $0$ otherwise
$$f_{XY}(x,y)=\dfrac{1}{12x} I{[0 \le y \le x \le 12]}$$ 
Then you want to integrate this over $x$ to get the marginal density for $Y$ and then adjust the limits of integration to respect the indicator function
$$f_{Y}(y)=\int_{x=-\infty}^{\infty} \dfrac{1}{12x} I{[0 \le y \le x \le 12]} \, dx= \int_{x=y}^{12} \dfrac{1}{12x} I{[0 \le y \le 12]} \, dx$$
and that integral is not particularly difficult: you will be left with a function of $y$ multiplied by an indicator function depending on $y$ showing where the marginal density is non-zero

Alternative hint:
You know that $E[Y \mid X=x]=\frac{x}{2}$ so $E[Y \mid X]=\frac{X}{2}$
This will give you $E[Y]=\frac12E[X]$ and $E[XY]=\frac12E[X^2]$ 
So $\text{Cov}(X,Y)=\frac12E[X^2]-\frac12(E[X])^2 = \frac12\text{Var}(X)$ which you can calculate
