Question

Find the exact value of the area under one arch of the curve y(t) = V_0sin(wt). Assume V_0 and w are positive constants.

Answer 1

One arch is half a period.
The period of this function is 2pi/w, so you can integrate between 0 and pi/w:\[Area=\int\limits_{0}^{\frac{ \pi }{ w }}V_0\sin (wt)dt=V_0\int\limits_{0}^{\frac{ \pi }{ w }}\sin (wt)dt\]V0 can be put in front of the integral, because it is a constant.
Now you just have to find a primitive of the function sin(wt). Can you do that?

Answer 2

If not:\[-\frac{ V_0 }{ w }\left[ \cos(w \frac{ \pi }{ w })-\cos 0 \right]=-\frac{ V_0 }{ w }(-1-1)=\frac{ 2V_0 }{ w }\]