$$-100x = \sum_{n=1}^{\infty}b_n\sin n\pi x$$

$$b_n=2 \int_0^1{(-100x)\sin (n \pi x)}dx = -200 \int_0^1 x \sin (n \pi x) dx =(-1)^n \frac{200}{n \pi}$$

I know it is simple and stupid, but I do not understand how the last integration of this equation yields to $(-1)^n \frac{200}{n \pi}$ (only the last part).

Can someone explain how this integration is made, and how will be the solution if removing $x$ from the initial equation as

$$-100 = \sum_{n=1}^{\infty}b_n\sin n\pi x$$