Example 5.4.4. Continuous Expected Value.
Consider \(f(x) = x^2/3\) over \(R\) = [-1,2] with value function given by \(v(x) = e^x - 1\text{.}\) Then, the expected value for \(v(x)\) is given by
\begin{equation*}
E = \int_{-1}^2 (e^x-1) \cdot x^2/3 = -1/9 \cdot (e + 15) \cdot e^{-1} + 2/3 \cdot e^2 - 8/9 \approx 3.3129
\end{equation*}