Question

Consider the area under one arch of the curve y(t) = ksin(wt) for t ≥ 0 where k and w are positive constants.

Answer 1

The period of the function is $\frac{2\pi}{w}$ so the first arch extends over the interval $\left[ 0,\frac{\pi}{w} \right]$. The area is

$\int_0^{\pi/w} \sin{(wt)} dt$

The integral can be found by substitution.
$u=wt \\ du=w \; dt \\ \frac{du}{u}=dt$

The limits of integration change, too.

$t=0 \Rightarrow u=0 \\ t=\frac{\pi}{w} \Rightarrow u=\pi$

$\frac{k}{w} \int_0^\pi \sin{u} \; du = \frac{k}{w} \left[-\cos{u} \right]_0^\pi = \frac{k}{w} \left[ 1-(-1) \right] = \frac{2k}{w}$