The proof of Riemann’s bilinear relations

Question

I am reading the proof from http://page.math.tu-berlin.de/~bobenko/papers/2011_Bob.pdf

I am very confused the step: (from Stokes theorem)
$$\int_Rw\wedge w'=\int_{\partial F_g}w'(p)\int_{P_0}^Pw$$

Can you explains that?

Answer 1

On the simply connected domain $F_g$, the closed form $\omega$ becomes exact: $\omega = df$, for $f(P) = \int_{P_0}^P \omega$ (and note this integral is path-independent). Then $d(f\omega') = df\wedge\omega' = \omega\wedge\omega'$. The author is writing $\int_{F_g}\omega\wedge\omega' = \int_{F_g} d(f\omega') = \int_{\partial F_g} f\omega'$.