Find the exact value of the area under one arch of the curve y(t) = V0sin(wt). Assume V0 and w are positive constants. You can use V_0 for V0. Note - V is upper case here.

Question

anonymous · Answer

This is an integration problem because you're trying to find area under a curve over a certain interval. First, you need to find the period of the function so you can know your bounds of integration. V0sin(wt) has a period of 
                                                     T = w * pi 

However, you're looking for area under one arch of the curve so you need to find half the period, which is
                                               T/2  = (w * pi) / 2.

Now you have your bounds of integration:  0 to (w * pi) / 2 

So the integral will look like this:
$$\int\limits_{0}^{ (\omega*pi)/2} V _{0} \sin(wt) dt$$You can compute the integral knowing that the antiderivate of $$\sin (a*t) = [- \cos (a*t)]/a $$If this isn't clear, just tell me.