Assume the random vector $(X,Y)$ has joint probability density function given by $$f_{X,Y}(x,y) = \begin{cases} kx(x-y) & \text{if}\,\,0 < x < 2\,\,\text{and}\,\,-x < y <x\\\\ 0 & \text{otherwise} \end{cases} $$ Determine the following:
(a) The value of the constant $k$
(b) The marginal probability density function of $X$
(c) The marginal probability density function of $Y$
(d) Are $X$ and $Y$ independent? Justify your answer.
MY SOLUTION
(a) Since $f_{X,Y}$ is a joint probability density function, it must satisfy the relation \begin{align*} \int_{-\infty}^{+\infty}\int_{-\infty}^{\infty}f_{X,Y}(x,y)\mathrm{d}y\mathrm{d}x = k\int_{0}^{2}\int_{-x}^{x}(x^{2} - xy)\mathrm{d}y\mathrm{d}x = k\int_{0}^{2}2x^{3}\mathrm{d}x = 1 \Rightarrow k = \frac{1}{8} \end{align*}
(b) According to the definition of marginal distribution, we get \begin{align*} f_{X}(x) = \int_{-\infty}^{+\infty}f_{X,Y}(x,y)\mathrm{d}y = \int_{-x}^{x}\frac{x^{2}-xy}{8}\mathrm{d}y = \frac{x^{3}}{4} \end{align*}
where $x\in[0,2]$.
(c) Analogously, we have \begin{align*} f_{Y}(y) = \int_{-\infty}^{+\infty}f_{X,Y}(x,y)\mathrm{d}x = \int_{|y|}^{2}\frac{x^{2} - xy}{8}\mathrm{d}x = \frac{1}{48}(-2|y|^{3} + 3y|y|^{2} - 12y + 16) \end{align*}
where $y\in[-2,2]$.
(d) No, because $f_{X,Y}(x,y) \neq f_{X}(x)f_{Y}(y)$.
Once again, I am having problem with the integration limits and as well as to the support for $Y$. Could someone check the results for me? Besides helping me out, is there any tip to avoid miscalculating them, such as here?