If you take $f(t)=2t$ then the integral $\int_a^xf(t)dt=x^2-a^2$ seems to be the general antiderative of $f$ and the notation of $\int f(t)dt$ makes sense as it is saying the antideritave of $f$ is really some kind of general, varying definite integral. But $x^2-a^2$ isn't the general antiderivative of $f$ because the arbitrary constant is never positive. So for particular antiderivatives like $x^2+3$ it isn't any definite integral of $2t$, so the notation $\int f(t)dt$ doesn't seem to have any meaning. Am I simply wrong in saying that $x^2+3$ isn't any kind of definite integral or is notation sometimes just notation?

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    $\begingroup$ The indefinite integral (the antiderivative) of $f$ is the set of all functions $F$ such that $F' = f$ whereas the integral of $f$ on a segment $[a, x]$ is a number. $a^2$ is not the arbitrary constant. Compare if $F(x) = x^2 +3$, then $F(x) - F(a) = x^2 + 3 - a^2 -3 = x^2 - a^2$. If $F(x) = x^2 -6$, $F(x) - F(a) = x^2 -6 -a^2 + 6 = x^2 -a^2$. $\endgroup$ – voligno Feb 10 '14 at 20:12
  • $\begingroup$ Isn't the integral of $f$ on a segment $[a,x]$ a function of x and not a number? $\endgroup$ – maydayway Feb 10 '14 at 20:56
  • $\begingroup$ For a fixed $x$ it will be a number. $\endgroup$ – voligno Feb 10 '14 at 21:10

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