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To avoid the confusing (to me) notation, let's call $g(x)$ the antiderivative of $f(x)$, such that $g'(x) = f(x)$. I've seen $F(c)$ defined as the definite integral of $f(x)$ from $0$ to $c$ with respect to $x$ in some situations, and as $g(x)$ in other situations. While these definitions are equivalent if $g(0) = 0$, they are different if $g(0)$ is not $0$ since the former definition is equivalent to $g(c) - g(0)$, while the latter is equivalent to $g(c)$. Assume that $f(b)$ for some value $b$ is known, so there is only one correct $g(c)$ (as opposed to infinite antiderivatives) and thus I am able to compute $g(c)$.

Without an explicit definition, does $F(c)$ take on the value of $g(c) - g(0)$ (the former definition) or $g(c)$ (the latter definition). In other words, is $F(c)$ the definite integral from $0$ to $c$ of $f(x)$ with respect to $x$, or the one indefinite integral of $f(x)$ evaluated at $c$?

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    $\begingroup$ Well, afaik $F$ it's just any antiderivative, just pick one specific. $\endgroup$ – enedil Mar 12 '19 at 1:55
  • $\begingroup$ When you say that $g$ is the antiderivative, you already run into problems with circular reasoning. Which antiderivative are you talking about? (You can always add $C$, you know.) $\endgroup$ – Hans Lundmark Mar 12 '19 at 7:40
  • $\begingroup$ @HansLundmark From the question, "Assume that f(b) for some value b is known, so there is only one correct g(c) (as opposed to infinite antiderivatives) and thus I am able to compute g(c)." $\endgroup$ – Mario Ishac Mar 12 '19 at 16:53
  • $\begingroup$ You mean that $g(b)$ is known? From $f(b)$ you can't determine the constant of integration. Anyway, as the first comment said, $F$ usually just means some antiderivative (doesn't matter which one). $\endgroup$ – Hans Lundmark Mar 12 '19 at 21:25

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