Question

What is the net area between `f(x) = cos2x-sinx` and the x-axis over `x in [0, 3pi ]`?

Answer 1

`f(x)=cos2x-sinx`
area between `f(x)` and X-axis is
`y=int_0^(3pi)(cos2x-sinx )dx`
`y=[(sin2x)/2-(-cosx)]_0^(3pi)`
`y=[(sin2x)/2+cosx]_0^(3pi)`
`y=[(sin2(3pi))/2+cos(3pi)]-[(sin2(0))/2+cos(0)]`
`y=[(sin6pi)/2+cos(3pi)]-[1]`
`y=[(0)/2+cos(2pi+pi)]-[1]`
`y=[(cos(pi)]-[1]`
`y=[-1]-[1]`
`y=-2`

Area is Positive
Area`=2 unit^2`