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The random variable \(X\) has the density \(f(x) = \frac{1}{2}\), on the interval \([0, 2]\). Calculate \(E(X)\).\FieldText{1}{$$\begin{aligned}[t] E(X)&=\int_{0}^{2} x \frac{1}{2} \, dx\\ &=\frac{1}{2} \left[ \frac{x^2}{2} \right]_{0}^{2}\\&= \frac{1}{2}\left[\frac{2^2}{2} - \frac{0^2}{2}\right]\\&= 1\\\end{aligned}$$}