For , compute the following integral:

Since we are taking the integral of an absolute value, first we want to determine on which intervals is positive and on which it is negative. We have,

Now must consider two cases. First, if , then,

In the other case, that , then,

Also consider this solution:

Using the additive property we have

and applying Thm. 1.19 to the first integral we get

but is an even function and

so Def. 1.4 yields

Again, from the previous exercise we know that the integrand has the property

and since we now only have to solve an integral over with

we need only worry about . However, because the integrand changes sign mid interval (depending on ) we must split into cases. Thus

where the first case gives

and the second

Hello, Can you please explain the function (the absolute function) that you have in the 1st line? Why is the function between -2pi/3 and 2pi/3 negative ?

I think the range of the positve and negative range of the absolute value are the wrong way around

The calculations are correct but at the top, the cases of absolute value sign are the wrong way around for each range.