Probability density function

**n3mo** · October 2nd 2010, 01:48 PM

I am struggling to prove the following problem. Could anyone help demonstrate how this can be done:
Find k so that $f(x) = kx(1-x)$ when $0<x<1$ , or $f(x) = 0$ otherwise.

I understand that the integral of a probability distribution function needs to equal 1, but I cannot demonstrate that it does myself...

Thanks for any help on this

**Plato** · October 2nd 2010, 01:54 PM

Just solve $\int_0^1 {kx\left( {1 - x} \right)dx} = 1$ for $k$ .

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