\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
Consider the function \(f(x) = \frac{x^2}{9}\), defined on the interval \([0, 3]\).
Verify that \(f(x)\) is a probability density function on the interval \([0, 3]\).\FieldText{1}{From the graph, \(f(x) \geqslant 0\) for all \(0 \leqslant x \leqslant 3\). $$\begin{aligned}[t] \int_{0}^{3} f(x)\; \mathrm d x&=\int_{0}^{3} \frac{x^2}{9} \; \mathrm d x\\&=\frac{1}{9} \left[\frac{x^3}{3}\right]_{0}^{3}\\&= \frac{1}{9} \left[\frac{3^3}{3} - \frac{0^3}{3}\right]\\&= \frac{1}{9} \cdot 9\\&= 1\\\end{aligned}$$ So \(f\) is a probability density function on the interval \([0, 3]\).}