Consider \(f(x) = x^2/3\) over \(R\) = [-1,2]. Then, for \(-1 \le x \le 2\text{,}\)
\begin{equation*} F(x) = \int_{-1}^x u^2/3 du = x^3/9 + 1/9. \end{equation*}
Notice, \(F(-1) = 0\) since nothing has yet been accumulated over values smaller than -1 and \(F(2) = 1\) since by that time everything has been accumulated. In summary:
Table 5.3.9. Continuous Distribution Function Example
X F(x)
\(x \lt -1\) 0
\(-1 \le x \lt 2\) \(x^3/9 + 1/9\)
\(2 \le x\) 1