Yes it's not continuous at and you can't just integrate it.
When dealing with type things usually you can split them up into...
However since it's gonna be even in your case you can just do...
(obviously with 1-x though)
This is probably going to be one of those questions where I can't believe I was so dumb but...What I'm trying to do is calculate the fourier series of some function. This is okay, but I'm having a problem with my integral.
In the fourier series Ao is given by
The fourier series I'm attempting to calculate is for
for
Let us note that this function is even
So Ao should simplify to
We then plug in the values and obtain
evaluated from 0-->2
This of course yields
Now, I was fiddling with this. And had I not changed my integral I get a different answer.
evaluted from -2-->2
I'm sure that I have violated some law in the second evaluation, i don't remember what is is though. Is this because the equation isn't valid on the left side of the XY domain and that the derivative at 0 doesn't exist? I think I vaguely remember something about this from first year.
My book has it as equal to 0. Which is what prompted me to think about it some more. I figured it had something to do with not being contious at 0, my simple brain just couldn't rationalize why :P
Let's work it out in full
evaluated from -2-->0 and
evaluated rom 0-->-2
Yeah that equals 0. Thanks mate for helping me think back. I should probably brush up on some definitions :P