If $f$ continuous in $[-1,1]$ Then $g(x)= \int_{-x}^x \! \, f(t)dt$ derivative in $[-1,1]$ and $g'(x)=f(x)+f(-x)$ I have this problem its a proof/disproof problem.
For some reason I get wrong answer.
If $f$ continuous in $[-1,1]$ Then $g(x)= \int_{-x}^x \! \, f(t)dt$ derivative in $[-1,1]$ and $g'(x)=f(x)+f(-x)$
What I did: 
Using the fundamental theorem since $f$ continuous $[-1,1]$ we can conclude that $f$ derivative in $[-1,1]$ and for all $x \in [-1,1]$ :
$$f(x)=F'(x)$$
Since $f$ continuous in $[-1,1]$ is also integrability, so using the fundamental forumula :
$$\int_{-x}^x \! \, f(t)dt=F(x)-F(-x) \implies g(x)= \int_{-x}^x \! \, f(t)dt=F(x)-F(-x)$$
So :
$$g(x)=F(x)-F(-x) \implies g'(x)=F'(x)-F'(-x) = f(x)-f(-x) \neq f(x)+f(-x)$$
So the statement is false, but for some reason in the answers I see that the answer is true, and I don't find any mistake I done.
Any ideas?
Thanks!

It's as MPW said but I will post a separate answer since I cannot comment.
$g(x)=F(x)-F(-x)$
Differentiate that with respect to $x$.
Use the chain rule for $F(-x)$ and you'd get:
$(F(-x))'=F'(-x)*(-x)'=F'(-x)*(-1)=-F'(-x)$
Finally, we'd get: $g'(x)=F'(x)-(-F'(-x))=F'(x)+F'(-x)=f'(x)+f'(-x)$
