Note that [tex]0\le k\sin x\le k[/tex] in the given closed interval, so the area is exactly given by the definite integral
[tex]\displaystyle\int_0^\pi k\sin x\,\mathrm dx[/tex]
(no absolute values needed!)
Integrating gives
[tex]-k\cos x\bigg|_{x=0}^{x=\pi}=-k(\cos\pi-\cos0)=-k(-1-1)=2k[/tex]
