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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
in-context