Is It Acceptable to Ignore Arbitrary Constants in Calculus Notation?

[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?"

 

[mathman]

Antiderivative - yes, call it F(t).
The integral is F(t) - F(t₀), t₀ can be anything - not necessarily 0.

 

[1MileCrash]

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.

 

[Mark44]

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.

If you're talking about this, however,
$$ \frac{d}{dt}\int^{t}_{t_{0}} f(s) ds$$
then that evaluates to f(t).

 

[1MileCrash]

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.

 

[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.

 

[HallsofIvy]

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.

 

[Mark44]

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) = e^{P(x) - P(s₀)} = e^P(x)/e^P(s₀)

Since P(s₀) 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.