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Consider the function \(f(x) = \sin(x)\), defined on the interval \([0, \frac{\pi}{2}]\).
Verify that \(f(x)\) is a probability density function on the interval \([0, \frac{\pi}{2}]\).\FieldText{1}{From the graph, \(f(x) \geqslant 0\) for all \(0 \leqslant x \leqslant \frac{\pi}{2}\). $$\begin{aligned}[t] \int_{0}^{\frac{\pi}{2}} f(x)\; \mathrm d x&=\int_{0}^{\frac{\pi}{2}} \sin(x) \; \mathrm d x\\&=\left[-\cos(x)\right]_{0}^{\frac{\pi}{2}}\\&= -\cos\left(\frac{\pi}{2}\right) - (-\cos(0))\\&= 0 + 1\\&= 1\\\end{aligned}$$ So \(f\) is a probability density function on the interval \([0, \frac{\pi}{2}]\).}