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I have the statement: A continuous function $f: \left(0, 1\right) \to \mathbb{R}$ has an antiderivative $F: \left(0, 1\right) \to \mathbb{R}$. Is this true or false?

I'm confused because we stated the fundamental theorem of calculus only for functions on a closed interval.

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Fix $a\in (0,1)$ and try $$ F(x):=\int_a^xf(t)\,\mathrm dt$$ (Just to clarify: If $a>x$, this is commonly understood to equal $-\int_x^af(t)\,\mathrm dt$)

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  • $\begingroup$ Since a, x are both in (0, 1), f is also continuous on [a, x], and by the FTC, F is its primitive function? Is my reasoning correct? $\endgroup$ – Darks May 14 '18 at 13:31
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It is true. Actually, here's one :

$$F_{\frac{1}{2}} : x \mapsto\int^x_{\frac{1}{2}}f(t)dt$$

As you can see, for every closed interval in the form of $[\frac{1}{2}, x]$ or $[x, \frac{1}{2}]$ for $x \in (0, 1)$, $f$ is continuous on that closed interval, thus it is integrable and $F_{\frac{1}{2}}$ is well-defined on $(0, 1)$.

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