Question

Verify for alpha = 3, and beta = 3 that the integral of the beta density, from 0 to 1 is
equal to 1.

Answer 1

Using the beta distribution function I end up with the integral:
\[\int\limits_{0}^{1}30x^2(1-x)^2dx\]

Answer 2

From here the answer should be 1 as stated here: http://www.wolframalpha.com/input/?i=integrate+30x%5E2%281-x%29%5E2dx+from+x%3D0+to+1
but I'm having trouble getting that. I think I've integrated somewhere wrong.By using the tabular method I get the following:
\[(1-x)^2(30x^3/3)(+1)(2(1-x)^1(30x^4/12)(-1))\]
which should end up being:
\[(1-x)^2\frac{30x^3}{3}-2(1-x)\frac{30x^4}{12}]_{0}^{1}\]right?

Answer 3

oh wait I think that sign should be a plus not minus.

Answer 4

beta density is for probability