In the proof of the four vertex Theorem (A Pressley, Elementary Differential Geometry) we have the following:
Given $k_{s}$, the signed curvature of the unit speed simple closed convex curve $\gamma$ . We obtain the following $$ \int_{0}^{l} k_{s}'(t) \gamma(s) ds=0$$ First using the continuity of $k_{s}$ we get there exist a maximum and minimum at $P$ and $Q$ on curve. To claim there are more, we join the points P and Q and divide the curve into two arcs. Then takes two vectors $a$, unit vector along $PQ$ and $b$ obtained by rotating $a$ along $\frac{\pi}{2}$ anti clock wise. Then $$ \int_{0}^{l} k_{s}'(t) \gamma(s)b ds=0$$
We get $k_{s}' >0$ on one segment and $k_{s}' < 0$ on the other.
The he claims that it gives $$ \int_{0}^{l} k_{s}'(t) \gamma(s)b ds> 0$$ in one segment and $$ \int_{0}^{l} k_{s}'(t) \gamma(s)b ds< 0$$ on the other which is a contradiction.
I am not clear how he obtains these inequalities. Even though $k_{s}'>0$ we arrive at inequality if $\gamma(s)b > 0$ or $\gamma(s)b < 0$ throughout the segment. How to claim that ? Or is there any point I missed in the proof ? I understood the rest of the proof.