# Calculus notation question

by 1MileCrash
Tags: calculus, notation
 P: 1,292 When one writes: $\int^{t}_{t_{0}} f(s) ds$ Do they generally mean "the antiderivative of f(t), and ignore the arbitrary constant/pick t naught so that it is 0?"
 Sci Advisor P: 6,056 Antiderivative - yes, call it F(t). The integral is F(t) - F(t0), t0 can be anything - not necessarily 0.
P: 1,292
 Quote by mathman Antiderivative - yes, call it F(t). The integral is F(t) - F(t0), t0 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.

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P: 21,249
Calculus notation question

 Quote by 1MileCrash When one writes: $\int^{t}_{t_{0}} f(s) ds$ Do they generally mean "the antiderivative of f(t), and ignore the arbitrary constant/pick t naught so that it is 0?"
I'm not sure you're writing what you meant to. The above is the definite integral of f over the interval [t0, t].

If F is an antiderivative of f, then the value of the integral is F(t) - F(t0. t0 might or might not be zero, and F(t0) might or might not be zero.

$$\frac{d}{dt}\int^{t}_{t_{0}} f(s) ds$$
then that evaluates to f(t).
P: 1,292
 Quote by Mark44 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.
 P: 1,292 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,488 The problem is a "floating pronoun". You ask if, in $$\int_{x_0}^x f(t) dt$$, "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 $x= x_0$, the value of the function is 0. But you certainly cannot "pick t naught so that it is 0?" You cannot pick $t_0$, it is given in the integral.
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P: 21,249
 Quote by 1MileCrash 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.
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) = eP(x) - P(s0) = eP(x)/eP(s0)

Since P(s0) is just a constant, we can write M(x) = KeP(x), where K = 1/eP(s0).

If you have an integrating factor, then a constant multiple of it will also work, so we can ignore the K.

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