Confusion - Density that is symmetric about $ξ$

Question

So the problem asks: If $X$ is a continuous random variable with a density that is symmetric about some point, $ξ$ , show that $E(X) = ξ$ , provided that $E(X)$ exists.

The solution given is:

But I am confused that why is $X-ξ$ is symmetric about $0$? Isn't $X-ξ$ the distance between $X-ξ$?

Answer 1

But I am confused that why is $X-ξ$ is symmetric about $0$?

That $X$ is symmetrically distributed about $\xi$ means exactly that $f_X(\xi+s)=f_X(\xi-s)$ for all $s\in\Bbb R^+$.

$Y=X-\xi$ is a linear transformation; a shift of axis.   As such $f_Y(y)=f_X(y+\xi$).

Then clearly $f_Y(y)=f_Y(-y)$, so therefore $Y$ is symmetric about $0$.