Area under half cycle of a sine wave

1. Area under half cycle of a sine wave

I forget math. Given a sine wave with offset 0, amplitude a, and frequency f (Hz), the area under a half cycle would be...

area = a * 0.637 * 0.5 / f

or simplified:

area = a / (Π * f)

... right? Because the area under a half cycle of a 1/2 hz wave would just be 1 * 0.637 (width * height of equivalent rectangle). Just want to double check. Thanks! •

2. Re: Area under half cycle of a sine wave

Why don't you check the calculation through a simple integral? You know the sine wave expression, you know the endpoints, so the definite integral would give you the right answer, irrespective of the simple formulae you're using :-D

Cheers •

3. Re: Area under half cycle of a sine wave

Hey Jasonc2,

If you integrate a*sin(2pi*f*t)*dt within the limits 0 and 1/2f you will get the area to be

a/(pi*f), thats how you get it

hope it helps

1 members found this post helpful. •

4. Re: Area under half cycle of a sine wave

Thanks! I briefly attempted to check my reasoning myself with the integral but ended up just regretting all the times I slept through calculus. :lol: --[[ ]]--