`int_0^((3pi)/2) |sin(x)| dx` Evaluate the integral

Textbook Question

Chapter 5, 5.4 - Problem 46 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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tiburtius | High School Teacher | (Level 2) Educator

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We have to be careful here because of absolute value sign. We need to know where the function under absolute value changes sign (in the interval of integration), which is at `x=pi` in this case. Therefore, we have

`int_0^((3pi)/2)|sin(x)|dx=int_0^pi sin(x)dx+int_pi^((3pi)/2)(-sin(x))dx=`

`-cos(x)|_0^pi+cos(x)|_pi^((3pi)/2)=-cos(pi)+cos(0)+cos((3pi)/2)-cos(pi)=`

`-(-1)+1+0-(-1)=3` <-- The solution

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