Fourier Series Problem: Finding Coefficients for Piecewise Function f(t)

[Drake25]

Homework Statement

one period of a function f(t) is given by the piecewise function f(t)= 2t if -pi/2 <t <0 and -1 if 0< t < pi/2
I need to find a sub 0, a sub n, a sub 1,2,3,& 4, and b sub 1,2,3 &4
If I could just get a sub 0 the rest are easy, however, my answer does not look correct.

Homework Equations

a sub 0= 1/p * integral from p to -p f(t)dt
a sub n= 1/p * integral from p to -p f(t)cos (n*pi*t)/p dt
b sub n is the same except sin replaces cos

The Attempt at a Solution

I worked through the equation for a sub o and got -pi/2 -1
Now, it's the -1 that I don't think belongs? But i don't know for sure if it is wrong but if it is that will mess up all of the other answers. I just wish someone could explain to me how I have done it wrong. Thank you!

 

[Aureolux]

$$\int_{0}^{\frac{\pi}{2}}{-1dt}$$ is $$-\pi/2$$, not $$-1$$.

Also keep in mind that the equations as you've written them are wrong; either you want a $$1/2p$$ in the beginning or you want to integrate from $$-p/2$$ to $$p/2$$. The point is to divide by the "length of integration", but if you integrate from -p to p that length is actually 2p. $$a_0$$ should represent the average value of the function on that interval.

 

[Drake25]

ok, that makes sense! thank you!
i wish my computer would allow me to type symbols like you did, where can i find those so that my question is easier to read?