A textbook example asks me to solve for $c$ in the joint distribution function;
$ f(x,y) = c(y^2-x^2)e^{-y} \ \ \ -y \le x \le y, \ \ 0 < y <\infty $
The answer given involves integrating the function from $-y$ to $y$ with respect to $x$, and getting $\frac{4}{3}cy^3e^{-y}$. Then, integrate this new function with respect to $y$ from $0$ to $\infty$.
The textbook gives the following steps: $ \frac{4}{3}c\int_0^\infty y^3e^{-y}dy = 4c\int_0^\infty y^2e^{-y}dy = 8c\int_0^\infty ye^{-y}dy = 8c = 1 $
My main issue is - how did they do these steps? How is one equal to the next?