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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 $x$

A.Γ.
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zickens
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1 Answers1

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Two initial remarks:

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

  2. 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 $$ \int_{c-x}^{c+x}f(t)dt=\int_{c-x}^{c}f(t)dt + \int_{c}^{c+x}f(t)dt=\int_{c}^{c+x}f(s)ds+ \int_{c}^{c+x}f(t)dt=2\int_{c}^{c+x}f(t)dt $$

Ramiro
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