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Second fundamental theorem of calculus.

  1. Nov 13, 2012 #1
    Let [itex]f(x)[/itex] be a non-stochastic mapping [itex]f: \mathbb{R} \to \mathbb{R}[/itex]. The second fundamental theorem of calculus states that:

    [itex]\frac{d}{dx} \int_a^x f(s)ds = f(x)[/itex].



    *QUESTION 1* Is the following true?

    [itex]\frac{d}{dx} \int_x^a f(s)ds = f(x)[/itex].

    *QUESTION 2* Related to this, how can I evaluate/simplify/express:

    [itex]d\int_x^a f(s)ds[/itex].
     
  2. jcsd
  3. Nov 13, 2012 #2
    For the first question, what happens to the integral when you reverse the bounds on integration? This should tell you that your answer is not quite correct, modulo only a sign error.

    The next answer depends on your context and there are a few ways to do this. The first is to simply use the definition of the exterior derivative. Assuming you are just working on [itex] \mathbb R [/itex] you have that for any function (0-form) F, then in a coordinate chart you have
    [tex] dF = \frac{\partial F}{\partial x} dx. [/tex]
    Alternatively, if you let [itex] \gamma: [0,1] \to \mathbb R [/itex] be a smooth path such that [itex] \gamma(0) = a, \gamma'(0) = v [/itex] then
    [tex] d_a F(v) = \left. \frac{d}{dt} \right|_{t=0} (F \circ \gamma). [/tex]
    Both will give you the same solution, the the latter is coordinate independent.
     
    Last edited by a moderator: Nov 14, 2012
  4. Nov 13, 2012 #3
    Hi,

    I'm not sure what "modulo" means?

    It would make sense if the solution to QUESTION1 is actually [itex]-f(x)[/itex], is that what you were saying?
     
  5. Nov 14, 2012 #4

    HallsofIvy

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    "modulo a sign error" meant "correct except possibly a sign error".

    Yes, [itex]\int_x^a f(t)dt= -\int_a^x f(t)dt[/itex] and then apply the fundamental theorem.
     
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