# 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.

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.

• @JavaMan I added some extra detail and completed the proof. Should I remove this answer? – Nameless Dec 8 '12 at 9:23
• @JavaMan I think that this complements André's answer and provides more detail. – Daryl Dec 8 '12 at 9:25
• @Nameless: The question didn't say $f$ is identically $0$ on $[-1,1]$ e.g. consider $f:\mathbb R\to \mathbb R$ such that $f(x)=0$ on $(-\infty,1]$ & $f(x)=x-1$ for $x>1$ – Sugata Adhya Dec 21 '12 at 7:11
• @SugataAdhya The question said $F$ is identically $0$ – Nameless Dec 21 '12 at 8:18
• @Nameless: I wanted to tell, $F$ is identically $0$ on $[-1,1]\implies f$ is identically $0$ on $[-1,1]$, but doesn't implies $f=0$ on the whole of $\mathbb R.$ – Sugata Adhya Dec 21 '12 at 11:41

Let $F(x)$ be the function defined by the integral. Use the Fundamental Theorem of Calculus to calculate $F'(x)$.

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!

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.

None of the $B, C, D$ is true since $A$ is a possibility. To show $A$ is true take $F(x)=0.$ Then $\forall~x\in[-1,1], F'(x)=0\implies f(x)=0$