The joint density of $X$ and $Y$ is given by $f_{X,Y}(x,y)=$... The joint density of $X$ and $Y$ is given by $$f_{X,Y}(x,y)=
\left\{\begin{matrix}(xy), \mbox{ } 0\leq x \leq1 , 0\leq y \leq 2
\\ 0,  \mbox{   otherwise}\end{matrix}\right.$$
a) Evaluate $f_{X}(X)$ and $f_{Y}(y)$.
b) Are $X$ and $Y$ independent? 
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. I'm not sure if the limits for the each function are correct. Thanks for reading!!
My solution:
a) $$f_{X}(x) = \int_{0}^{2}(xy)dy = \left [ \frac{xy^{2}}{2} \right ]\Big|_0^2=2x.$$
$$f_{Y}(y) = \int_{0}^{1}(xy)dx = \left [ \frac{x^{2}y}{2} \right ]\Big|_0^1=\frac{y}{2}.$$
b) $X$ and $Y$ are independent if $f_{X,Y} = f_{X}(x)f_{Y}(y)$.
From a, $$f_{X}(x)f_{Y}(y) = 2x(\frac{y}{2}) = xy.$$
From given, $$f_{X,Y}(x,y)= xy.$$
So, since $xy=xy$, $X$ and $Y$ ARE independent.  
 A: Posting as an answer since I don't have enough rep to comment.
Your solution looks fine to me. What is it you're unsure of exactly?
The only thing I might include is that, when you calculate your marginal densities, for example $f_X(x)$, you haven't mentioned it only takes the value of $2x$ for specific values of $x$, and is $0$ otherwise. Same for $f_Y(y)$.
This will help you justify your answer in the second part. Can you see why?
Additions
Generally, to find the marginal distribution $f_X(x)$ from $f_{X,Y}(x,y)$, we integrate $f_{X,Y}(x,y)$ with respect to $y$. That is, we "integrate out" $y$ from the function. We do this over the entire probability space, which (for lack of information) I will assume is the reals. Thus $y$ can be any real number, so $$f_X(x) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)\mathrm{d}y.$$
But we know $f_{X,Y}(x,y)$ can only take non-zero values for $y$ between $0$ and $2$. Thus we have $$\int_{-\infty}^{\infty}f_{X,Y}(x,y)\mathrm{d}y = \int_{0}^{2}f_{X,Y}(x,y)\mathrm{d}y,$$
and so the solution follows.
