\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
Let \( X \) be a continuous random variable following a continuous uniform distribution on \([a, b]\).
Prove that the expected value of \( X \) is:$$E(X) = \frac{a + b}{2}.$$\FieldText{1}{$$\begin{aligned}[t]E(X) &= \int_{a}^{b} x \cdot \frac{1}{b - a} \, \mathrm{d}x \\&= \frac{1}{b - a} \left[ \frac{x^2}{2} \right]_{a}^{b} \\&= \frac{1}{b - a} \left( \frac{b^2}{2} - \frac{a^2}{2} \right) \\&= \frac{b^2 - a^2}{2 (b - a)} \\&= \frac{(b - a)(b + a)}{2 (b - a)} \\&= \frac{a + b}{2}.\end{aligned}$$}