Question

How do you evaluate the integral of `int abssinx dx` from 0 to 3pi/2?

Answer 1

3

Explanation:

Remark that
`|sinx|=sinx` for `0 <= x <=pi`
`|sinx|=-sinx` for `pi<=x<=(3pi)/2`

Therefore we can rewrite the expression as
`=int_0^pi sinx*dx-int_pi^((3pi)/2) sinx*dx`
`=-cos x|_0^pi` `+cosx|_pi^((3pi)/2)`
`=-[-1-(1)]+[0-(-1)]=2+1=3`