Let $f: ~\mathbb R\rightarrow \mathbb R$ be a continuous function such that $\int_{-1}^{x}f(t)dt=0$ for all $x \in [-1,1]$ I was thinking about the problem that says:
Let $f: ~\mathbb R\rightarrow \mathbb R$ be a continuous function such that $\int_{-1}^{x}f(t)dt=0$ for all $x \in [-1,1]$. Then which of the following option(s) is/are correct?
(A) $f$ is identically $0$,
(B) $f$ is a non-zero odd function,
(C) $f$ is a non-zero even function,
(D) $f$ is a non-zero periodic function.
Please help. Thank you in advance for your time.
 A: Let $F(x)$ be the function defined by the integral. Use the Fundamental Theorem of Calculus to calculate $F'(x)$.
A: Let $$F(x)=\int_{-1}^{x}f(t)dt$$
By The Fundumental Theorem of Calculus $F^{\prime}(x)=f(x)$ for $x\in [-1,1]$. But $F$ is identically $0$ and so $f=F^{\prime}$ is identically $0$ on $[-1,1]$ as well.
A: The fundamental theorem of calculus is probably the easiest way, but you could also use the following property of the integral
$$\int_a^b f + \int_b^c f = \int_a^c f$$
and the version below of Lebesgue's theorem
$$f(x) = \lim_{r \to 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} f.$$
Cheers!
A: By way of contradiction assume w.l.o.g. that $f(x)<0$ for some $x\in[-1,1]$, then there is a $c<0$ and an interval $(a_x,b_x)$ around $x$ such that $f(x)<c$ for all $x\in (a_x,b_x)$. But then $0>c(b_x-a_x)\geq\int ^{b_{x}}_{a_{x}}fdx=\int ^{b_{x}}_{-1}fdx-\int ^{a_{x}}_{-1}fdx=0$, which is a contradiction. 
A: None of the $B, C, D$ is true since $A$ is a possibility. To show $A$ is true take $F(x)=0.$ Then $\forall~x\in[-1,1], F'(x)=0\implies f(x)=0$
