Question

What is the net area between `f(x) = sinx - cos^2x` and the x-axis over `x in [0, 3pi ]`?

Answer 1

The net area is `2-(3pi)/2`

Explanation:

A small area is `dA=ydx` `=>` `dA=(sinx-cos^2x)dx`
so the area is `A=int_0^(3pi)(sinx-cos^2x)dx`
To integrate `cos^2x`, we use `cos2x=2cos^2x-1`
`=>` `cos^2x=(1+cos2x)/2`
So, `A=int_0^(3pi)(sinx-(1+cos2x)/2)dx`
`=[-cosx-x/2-(sin2x)/4 ] _0^(3pi)`
`=1-(3pi)/2+1=2-(3pi)/2`