When one writes: [itex]\int^{t}_{t_{0}} f(s) ds[/itex] Do they generally mean "the antiderivative of f(t), and ignore the arbitrary constant/pick t naught so that it is 0?"
Antiderivative - yes, call it F(t). The integral is F(t) - F(t_{0}), t_{0} can be anything - not necessarily 0.
Of course it can be anything, but I was asking if nothing else is said, then I could assume they mean an antiderivative with no arbitrary constant.
I'm not sure you're writing what you meant to. The above is the definite integral of f over the interval [t_{0}, t]. If F is an antiderivative of f, then the value of the integral is F(t) - F(t_{0}. t_{0} might or might not be zero, and F(t_{0}) might or might not be zero. If you're talking about this, however, $$ \frac{d}{dt}\int^{t}_{t_{0}} f(s) ds$$ then that evaluates to f(t).
Actually, I'm talking about what I wrote. I am aware that f(t0) may be 11, 42, grahams number, or batman riding a trex. I am asking if the designation of "tee naught" is commonly taken as an obvious intent to notate an antiderivative with no arbitrary constant. I don't know a better way to ask my question.
http://en.wikipedia.org/wiki/Integrating_factor Here is a very straight forward and common use of the notation. I'm asking if this is a routine and acceptable way to say "take the antiderivative and don't give me an arbitrary constant" since as far as I know, there is no other way to say that. I'm asking if I wrote that in a proof, people would know what I am talking about, but judging by the responses, the answer is no.
The problem is a "floating pronoun". You ask if, in [tex]\int_{x_0}^x f(t) dt[/tex], "ignore the arbitrary constant/pick t naught so that it is 0?" What does "it" refer to? If F(t) is an anti-derivative of f(t), then the integral is F(x)- F(x_0) so, at [itex]x= x_0[/itex], the value of the function is 0. But you certainly cannot "pick t naught so that it is 0?" You cannot pick [itex]t_0[/itex], it is given in the integral.
Near the top in the wiki article, they have this. (I made one change, from P(s) to p(s). You'll see why in a minute.) $$ M(x) = e^{\int_{s_0}^x p(s)ds}$$ Let's assume that P(s) is an antiderivative of p(s). Then the exponent on e is $$ \left. P(s)\right|_{s_0}^x = P(x) - P(s_0)$$ So M(x) = e^{P(x) - P(s0)} = e^{P(x)}/e^{P(s0)} Since P(s_{0}) is just a constant, we can write M(x) = Ke^{P(x)}, where K = 1/e^{P(s0)}. If you have an integrating factor, then a constant multiple of it will also work, so we can ignore the K.