I have concerns about this problem
Let $f_e(x)$ and $f_o(x)$ represent general continuous even and odd functions on $[-L,L]$.
Prove that $\int_{-L}^{L} f_e(x) dx$ = 2 $\int_{0}^{L} f_e(x) dx$
My attempt:
$\int_{-L}^{L} f_e(x) dx$ = $\int_{-L}^{L} \frac{1}{2}[ f(x)+f(-x)] dx$ = $\frac{1}{2} \int_{-L}^{L} f(x) dx$ + $ \frac{1}{2} \int_{-L}^{L} f(-x) dx$. By the Fundamental Theorem Of Calculus, I would have
$\frac{1}{2} [F(L) -F(-L)] + \frac{1}{2} [F(-L) - F(L)]$. Is this correct?
On
This has nothing to do with the Fourier series.
It holds for every even function $f$, that
$$\int_{-L}^{0} f(x) dx = \int_{-(-L)}^{0} f(-x) d(-x) = \int_{0}^{L} f(x) dx.$$
The first step uses integration by substitution with $u(x)=-x$ and the second step the defining property of even functions $f(x)=f(-x)$.
Consider $J = \int_{x=-L}^0 f_e(x) dx$. Let $v = -x$; then $dx = -dv$ and $$ J = - \int_{v=+L}^0 f_e(-v) dv = \int_{v=0}{L} f_e(-v)dv$$ Now use the fact that $f_e$ is even to write this as $$ J = \int_{v=0}{L} f_e(v)dv = \int_{x=0}{L} f_e(x)dx$$$$
Finally, $$\int_{x=-L}^L f_e(x) dx = J + \int_{x=0}^L f_e(x) dx = 2\int_{x=0}^L f_e(x) dx $$