Example 5.3.4. Continuous Probability Function.
Consider \(f(x) = x^2/c\) for some positive real number c and presume \(R\) = [-1,2]. Then f(x) is nonnegative (and only equals zero at one point). To make \(f(x)\) a probability density function 5.3.2, we must have
\begin{equation*}
\int_{x \in R} f(x) dx = 1.
\end{equation*}
In this instance you get
\begin{equation*}
1 = \int_{-1}^2 x^2/c = \frac{x^3}{3c} |_{-1}^2 dx = \frac{8}{3c} - \frac{-1}{3c} = \frac{3}{c}
\end{equation*}
Therefore, \(f(x)\) is a probability density function over \(R\) provided \(c = 3\text{.}\)