# Calculus notation question

1. Apr 18, 2013

### 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?"

2. Apr 18, 2013

### mathman

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

3. Apr 18, 2013

### 1MileCrash

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.

4. Apr 18, 2013

### Staff: Mentor

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).

5. Apr 18, 2013

### 1MileCrash

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.

Last edited: Apr 18, 2013
6. Apr 18, 2013

### 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.

7. Apr 18, 2013

### 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.

8. Apr 18, 2013

### Staff: Mentor

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.