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Consider the function \(f(x) = \frac{3}{4}(1 - x^2)\), defined on the interval \([-1, 1]\).
Verify that \(f(x)\) is a probability density function on the interval \([-1, 1]\).\FieldText{1}{From the graph, \(f(x) \geqslant 0\) for all \(-1 \leqslant x \leqslant 1\).
$$\begin{aligned}[t] \int_{-1}^{1} f(x)\; \mathrm d x&=\int_{-1}^{1} \frac{3}{4}\left(1-x^2\right) \; \mathrm d x\\&=\frac{3}{4} \left[x -\frac{x^3}{3}\right]_{-1}^{1}\\&= \frac{3}{4} \left[ \frac{4}{3}\right]\\&= 1\\\end{aligned}$$ So \(f\) is probability density function on the interval \([-1,1]\).}