I am asked to compute the integral $\int_{(0,3a]}x\,dF(x)$ with $a > 0$ where $$F(x) = \begin{cases} \pi & 0\leq x < a\\ 4+a-x & a\leq x < 2a \\ (x-2a)^2 & 2a\leq x \end{cases} $$
The answer should be $\frac{8}{3}a^3 + \frac{1}{2}a^2 - (4+\pi)a$.
Here is my attempt. I don't know if this is the correct way to do it. (In particular, the way I partition the interval). I would appreciate it if someone could comment on it.
$$\int_{(0,3a]}xdF(x) = \int_{(0,a)}xdF(x) + a(F(a)-F(a^-)) + \int_{(a,2a)}xdF(x) + 2a(F(2a)-F(2a^-)) + \int_{(2a,3a]}xdF(x)$$ $$\int_{(0,3a]}xdF(x) = 0 + a(4-\pi) - \frac{3a^2}{2} - 8a + 2a^2 + \frac{8a^3}{3} = -4(a+\pi) + \frac{a^2}{2} +\frac{8a^3}{3}$$