what is the CDF of $f(x)=\frac{3x^2}{2}$?

Question

This is probably a dumb question but I just want to make sure. The pdf is $f(x)=\frac{3x^2}{2}$ if $-1 \leq 0 \leq 1$. The CDF is $F(x)=\frac{x^3}{2}$ but with what bounds? sorry if this is an easy and obvious answer.

Answer 1

Actually, the CDF isn't quite equal to $\frac{x^3}{2}$. Instead, $$ F(x)=\int_{-\infty}^xf(t)\;dt $$ This integral is equal to zero if $x<-1$, is equal to $$ \int_{-1}^x\frac{3t^2}{2}\;dt=\frac{x^3}{2}+\frac{1}{2} $$ if $-1\leq x\leq 1$, and is equal to $1$ if $x>1$.