Question

Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 seconds. The maximum rate of air flow into the lungs is about 0.5 L/s. This explains, in part, why the function f(t)=(1/2)sin((2pi*t)/5) has often been used to model the rate of air flow into the lungs. Use this model to find the volume (in liters) of inhaled air in the lungs at time t.

This was my answer, but it's wrong...
-(5/(4pi))(cos((2pi*((5(t))/2))/5))+(5/(4pi))(cos((2pi*(0)/5)))

Use wolframalpha website, it helps!

Thank you for your time!

Answer 1

it says that
(1/2)sin (2π/5)t is the rate of air flow
then the volume would be the integral of the above

V = (-5/(4π) cos(2πt/5) + c , where c is a constant

Answer 2

I've recently had a similar problem to this. For mine it listed the initial volume as 0 at t=0. If this is the case for anyone else you'll need to solve for c (the constant) then you'll have

V = -5/(4pi) cos((2pi)/5) + 5/(4pi)