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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?

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    @RobertIsrael$a$little too complicated for my current level, but thanks nonetheless2012-10-18

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