Find $c$ such that $cy^3(1-y)^8$ for $0 \leq y \leq 1$ is a PDF

Question

I have a function $$f(y) = \left\{\begin{array}{cc} cy^3(1-y)^8 & 0 \leq y \leq 1 \\ 0 & \text{ Otherwise.}\end{array}\right. .$$ I want to figure out what value of $c$ makes this function a legal probability density function.

I'd imagine that we start out by setting $$1 = \int_0^1 cy^3(1-y)^8 dy$$ $$\implies 1 = c \int_0^1 y^3(1-y)^8 dy.$$ However, the degrees seems to make this integral somewhat hard. I tried plugging this into Wolfram, but I couldn't seem to get any meaningful information as to how to calculate that. Any suggestions?

Answer 1

Hint. One may recall the Euler beta integral (or here), $$ \int_0^1 y^{a-1}(1-y)^{b-1} dy=\frac{\Gamma(a)\cdot \Gamma(b)}{\Gamma(a+b)} $$ giving $$ \int_0^1 y^3(1-y)^8 dy=\frac{\Gamma(4)\cdot \Gamma(9)}{\Gamma(13)}=\frac{1}{1980}. $$

Answer 2

You can always expand $(1-y)^8$ with the binomial theorem and integrate term by term. Somewhat more elegant is to substitute $u=1-y$.