Calculate integral $ \int_0^{100} (1+2x)^2 d\mu_F(x) $ where $$ F = \left\{ \begin{array}{l} {0}, \ x < 0\\ {1 - cos(x) }, \ x \in [0, \frac{\pi}{2}),\\ {2} , \ x \ge \frac{\pi}{2} \end{array} \right.$$
My solution is to of cource divide the integral into two parts (of which only the part related to $[0, \frac{\pi}{2})$ will remain). The integral then can be represented as $\int_0^{\frac{\pi}{2}} (1 + 2x)^2 F^{'}(x) dx $ that equals $4\pi - 3$. The question is should I also consider the gap point $\frac{\pi}{2}$ and add $(1 + 2\frac{\pi}{2})^2 \times F(\frac{\pi}{2})$ to my answer? I would also be grateful for a formal algorithm for such tasks.