Given probability density function
$$f(x) = \begin{cases} 3x^2 &\mbox{for } 0 < x \le1 \\ 0 & \mbox{otherwise } \end{cases}$$
Show that the cumulative probability density function $F$ corresponding to $f$ is given by $$ F(x) = \begin{cases} 0 &\mbox{for } x<0 \ \\ x^3 & \mbox{for }0 \leq x \le1 \\1 & \mbox{for }x>1 \end{cases} $$
I know that the probability density function of a random variable $X$ is given by $$f(x)=f_X(x)=\frac{d}{dx}F(x) $$ , but that's about it. Help appreciated, thanks!
The CDF at $x$ is just given by the integration of the PDF from $-\infty$ to $x$. If you want justification, you can look at this answer.
Now, with this information, it is pretty easy to divide $F(x)$ and perform piecewise integration to get your answer.