I have included a pdf attachment I made with mathematica to help try and explain my problem.

Merlyn.

2. Originally Posted by mercou
I have included a pdf attachment I made with mathematica to help try and explain my problem.

Merlyn.
I assume $\displaystyle f(x) = \cos x$ ....?

Then $\displaystyle \int_0^{2 \pi} \cos^2 x \, dx = \int_0^{2 \pi} \frac{1}{2} [ \cos (2x) + 1] \, dx$

substituting from the usual double angle formula

$\displaystyle = \frac{1}{2} \int_0^{2 \pi} \cos (2x) + 1 \, dx = \pi$.

If you're studying Fourier Series it's expected you're familiar with basic techniques of integration.

3. Originally Posted by mercou
I have included a pdf attachment I made with mathematica to help try and explain my problem.

Because of the symmetries of sinusoids the integral of $\displaystyle \sin^2$ or $\displaystyle \cos^2$ over a complete period is half of the area of the rectangle of height $\displaystyle 1$ and length equal to the period.
Look at your diagram and you will see that this is obvious (draw a rectangle around the plot of $\displaystyle \cos^2(x)$ from $\displaystyle 0$ to $\displaystyle 2 \pi$).