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Calculus notation question

  1. Apr 18, 2013 #1
    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?"
     
  2. jcsd
  3. Apr 18, 2013 #2

    mathman

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    Antiderivative - yes, call it F(t).
    The integral is F(t) - F(t0), t0 can be anything - not necessarily 0.
     
  4. Apr 18, 2013 #3
    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.
     
  5. Apr 18, 2013 #4

    Mark44

    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.

    If you're talking about this, however,
    $$ \frac{d}{dt}\int^{t}_{t_{0}} f(s) ds$$
    then that evaluates to f(t).
     
  6. Apr 18, 2013 #5

    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
  7. Apr 18, 2013 #6
    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.
     
  8. Apr 18, 2013 #7

    HallsofIvy

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    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.
     
  9. Apr 18, 2013 #8

    Mark44

    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.
     
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