Can someone help me give a proof for this? I know there are theorems about integrals of odd and even functions, but i kept wondering about integrals that share symmetry around an axis $x=c$. I've been trying to give a proof for this but can't seem to get around it; could someone help me prove/disprove this?
$$
\large \int_{c-x}^{c+x}f(x)dx=2\int_{c}^{c+x}f(x)dx
$$
Hypothesis--------------------
$$
\large f(c-x)=f(c+x)
$$
$f(x)$ is symmetric around $x=c$ for all $x$

Two initial remarks: 


*

*For clarity, you should not use the same letter in the limits of integragion and as the integration variable itself. 

*I assume your hypothesis is that, for all $x$, $f(c-x)=f(c+x)$.
Let us prove that 
$$
\large \int_{c-x}^{c+x}f(t)dt=2\int_{c}^{c+x}f(t)dt
$$
Proof: 
First, let us consider:
$$ \int_{c-x}^{c}f(t)dt$$
Let $s=2c-t$, then we have: $t=2c-s$, $dt=-ds$ and
$$ \int_{c-x}^{c}f(t)dt=-\int_{c+x}^{c}f(2c-s)ds=\int_{c}^{c+x}f(c+(c-s))ds=\int_{c}^{c+x}f(s)ds$$
In the last step above, we used the hypothesis to concluded that $f(c+(c-s))=f(c-(c-s))=f(s)$.
Now we have 
$$
\int_{c-x}^{c+x}f(t)dt=\int_{c-x}^{c}f(t)dt + \int_{c}^{c+x}f(t)dt=\int_{c}^{c+x}f(s)ds+ \int_{c}^{c+x}f(t)dt=2\int_{c}^{c+x}f(t)dt
$$
