If the function $f$ has the property $f(a+x) = -f(a-x)$ $\forall x$, where a is a constant, then
$$\int_{(a-2)}^{(a+2)} f(x)dx = 0$$.
I have to deside whether this statement is true or false, and to give a proof. I understand that here $f$ is an odd function hence the statement is true, however while prooving it got the following results:
[
Which shows the opposite (that f is an even function). Where did I made a mistake? Thanks in advance.
P.S. I'm sorry I don't have the fancy sign for the defined integral, I hope you understand what my equation looks like from the pinned photo.
Going from your second to your third line, there's a missing minus sign.
When you change $\int_0^2 f(x+a) dx$ to $\int_0^{-2} f(-x+a) dx$, you're really doing a substitution $u=-x$ (and then renaming $u$ as $x$, which is fine). This changes your limits of integration from $0..2$ to $0..-2$, but it also changes the $dx$ to $-du$, so an extra minus sign should appear to account for that. This will fix your problem.