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Suppose we are given $f(x)$ such that $f(-x)=-f(x)$, then if we consider the integral $$F(x)=\int_{0}^{x} f(t)dt$$ Will $F(x)$ be an even function?

There are a couple of examples that strengthen this claim, such as $x^3$ and $\sin(x)$, but I'm not sure as to how one would go about proving this claim (if at all, it is true). Maybe, we could do something like this:

We have to show that $$F(-x)=F(x)\Rightarrow \int_{0}^{x} f(t)dt=\int_{0}^{-x} f(t)dt$$ If we substitute, $t=-t'\Rightarrow dt=-dt'$, in the second integral we get $$\int_{0}^{-x} f(t)dt=-\int_{0}^{x} f(-t')dt'=\int_{0}^{x} f(t')dt'\Rightarrow \int_{0}^{x} f(t)dt=\int_{0}^{-x} f(t)dt\Rightarrow F(x)=F(-x)$$ Is this argument correct?

Student
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    Your argument is correct. – copper.hat Sep 30 '16 at 14:30
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    @Ranc: I'm not sure I'm following your reasoning. Why can't there (a priori) be a non-even function whose derivative happens to be odd? To the OP: I don't agree with your "We have to show that..." line - I think you should change $F(x) = F(-x)\implies$ to $f(x) = -f(-x)$. But, like copper hat, I agree with your argument. – Jason DeVito - on hiatus Sep 30 '16 at 14:32

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