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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.

5 Answers 5

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

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

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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.

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    @SugataAdhya Well yes. I think the correct question would have $f:[-1,1]\to \mathbb{R}$2012-12-21
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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!

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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) 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.