Question

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

Answer 1

The area is `=(3pi)/2(3pi-1)`

Explanation:

The area is `int_0^(3pi)(x-sin^2x)dx`
`cos2x=1-2sin^2x`
`:. sin^2x=(1-cos2x)/2`
The area is `int_0^(3pi)(x-sin^2x)dx=int_0^(3pi)(x-1/2+cos(2x)/2)dx`

`=[x^2/2-x/2+sin(2x)/4]_0^(3pi)`
`=((9pi^2)/2-(3pi)/2+sin (6pi)/4)-(0)`

`=(9pi^2)/2-(3pi)/2=(3pi)/2(3pi-1)`