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.