# What is the default definition of $F(x)$ given $f(x)$?

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 $$c$$?

• Well, afaik $F$ it's just any antiderivative, just pick one specific. – enedil Mar 12 '19 at 1:55
• 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.) – Hans Lundmark Mar 12 '19 at 7:40
• @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)." – Mario Ishac Mar 12 '19 at 16:53
• 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). – Hans Lundmark Mar 12 '19 at 21:25