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

1 Answer
Mar 24, 2016

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