Talk:Fundamental Theorem of Calculus

Does the lower bound of the first part have to be $a$ of the interval, or can it be any constant$\in [a..b]$? Or does such a distinction not matter, because we can always pick a subset of the interval? --GFauxPas 07:53, 17 January 2012 (EST)
 * The constant you can always add to an indefinite integral originates at least partly from this free choice of $a$. However, in this particular case, $a$ is convenient, as it means that swapping the bounds with a minus sign is not necessary. --Lord_Farin 07:59, 17 January 2012 (EST)