I am trying to do the following question:
A continuous random variable $X$ has the pdf $f(x)=c(1-x)x^2; 0<x<1$.
(a) Find the constant $c$.
Do I find the CDF $F(x)$ and use the property that $\lim_{x\to\infty}F(x)=1$? Because I am stuck. I found that $F(x) = \frac{cx^{3}}{3}-\frac{cx^{4}}{4}$. How do I find what $c$ is?
You want $\int_0^1 f(x)dx=1$ (i.e. the probability your RV takes ANY value in its domain is $1$). So you want your $F(1)=1$. That is $c/3-c/4=1$.