Question

The area of the region in the first quadrant between the graph of y = k*sin(x) and x-axis (where k > 0 ) on the closed interval [0, \pi] is?

Answer 1

Note that $0\le k\sin x\le k$ in the given closed interval, so the area is exactly given by the definite integral

$\displaystyle\int_0^\pi k\sin x\,\mathrm dx$

(no absolute values needed!)

Integrating gives

$-k\cos x\bigg|_{x=0}^{x=\pi}=-k(\cos\pi-\cos0)=-k(-1-1)=2k$