How to prove that the density function integrates to 1 over the whole support?

Question

Show that the function $f(x,y)$ when the function is given by: $$ f(x,y) = \begin{cases} \frac{15}{2}x^2y &\text{for } -1<x<1,|x|<y<1, \\ 0 &\text{else}. \end{cases} $$ I could easily verify all the other conditions but i was having problem that it integrates to 1 on the support. edit: $$\int_{\mathbb{R}^2}f(x,y)d(x,y)=\int_{-1}^1\int_{|x|}^1\frac{15}{2}x^2y\,dydx=\frac{15}{2}\int_{-1}^1x^2\int_{|x|}^1y\,dydx=\frac{15}{2}\int_{-1}^1x^2\left[\frac{1}{2}y^2\right]_{|x|}^1\,dx=\frac{15}{2}\int_{-1}^1x^2\left(\frac{1}{2}-\frac{1}{2}x^2\right)\,dx=\frac{15}{4}\int_{-1}^1x^2-x^4\,dx=\frac{15}{4}\left(\left[\frac{1}{3}x^3\right]_{-1}^1-\left[\frac{1}{5}x^5\right]_{-1}^1\right)=\frac{15}{4}\left(\frac{10-6}{15}\right)=1$$

Thanks in advance.

Answer 1

It is always advisable to graph the support of $X,Y$, in this case it is

and thus it is easier to pose the double integral, which corresponds to $$ \int_{-1}^{0} \int_{-x}^{1} f(x,y) dydx + \int_{0}^{1} \int_{x}^{1} f(x,y) dydx $$

Answer 2

Once we remark that $\forall (x,y) \in \mathbb R^2, f(-x, y) = f(x, y)$, we can make use of the parity in $x$:

$$\begin{align}\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) dx\space dy &= 2\int_0^\infty \int_{-\infty}^\infty f(x,y) dy\space dx\\ &= 2\int_0^1 \int_x^1 \frac{15}2 x^2 y\space dy\space dx\\ &= 15\int_0^1 x^2\left(\int_x^1 y\space dy\right) dx\\ &= 15\int_0^1 x^2\left(\frac{1 - x^2}2\right)dx\\ &= \frac{15}2\int_0^1 (x^2 - x^4)dx \\ &=\frac{15}2\left(\frac 1 3 - \frac 1 5\right)\\ &= \frac {15}2\times \frac 2{15}\\& = 1\end{align}$$