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The Fundamental Theorem of Calculus

  1. Nov 24, 2005 #1

    dx

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    Id like to know if the following argument is valid.
    Take an arbitrary function [tex]f(x)[/tex]. [tex]f(x)dx[/tex] can be thought of an infinitesimal area of a certain form (I emphasise this because I use it later in the argument) determined by the form of the function [tex]f(x)[/tex]. Lets denote its integral by [tex]Y[/tex].

    [tex]\int{ f(x)} dx = Y[/tex]

    Now, I argue that an infinitesimal of the whole formed by summing up infinitesimals of a certain form must be an infinitesimal of the same form. ie,

    [tex] d\int{ f(x)} dx = f(x)dx[/tex]

    Then the fundamental theorem of calculus follows.

    [tex] f(x) dx = dY [/tex]

    [tex] f(x) = \frac{dY}{dx} [/tex]

    [tex] f(x) = \frac{d}{dx}\int{ f(x)} dx[/tex]

    If this argument is valid. Can it be made rigorous?
     
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  3. Nov 24, 2005 #2

    matt grime

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    No, it is not valid. If f is arbitary there is no reason for its integral to be differentiable, and indeed trivial examples prove this.
     
  4. Nov 24, 2005 #3

    dx

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    But im not differentiating it. Im just saying that im taking an infinitesimal part of [tex]Y[/tex]. Any way, lets say both f(x) and its integral are differentiable. Then is it valid?

    EDIT : even if it is not completely valid, is it atleast an argument that suggests the fundamental theorem of calculus? I am talking about his particular statement.

    "Now, I argue that an infinitesimal of the whole formed by summing up infinitesimals of a certain form must be an infinitesimal of the same form."

    Is it ok to think this way, or is there something really really wrong with the way I understand it?
     
    Last edited: Nov 24, 2005
  5. Nov 24, 2005 #4

    matt grime

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    By it we are referring to the integral, and you are definitely differentiating that.


    As for the rest, I struggle to decipher the words that have necessary meaning mathematically.
     
  6. Nov 24, 2005 #5

    dx

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    Im trying to say that when you concatenate a 'd' on the left side of some quantity, you are making it an infinitesimal. So, we have an integral which is summing up infinitesimal quantities. So, in a way, the integral in made up of infinitesimals of a certain form ([tex]x^{2}dx[/tex],or [tex]e^{x}dx[/tex] etc.) When you concatenate a d to the integral, you are undoing the summing up and getting back the infinitesimal that you originally summed up. so [tex] d\int{f(x)}dx = f(x)dx [/tex].
     
  7. Nov 24, 2005 #6

    matt grime

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    you might be, i'm not sure the rest of mathematics would support that view in the way you wish it to.


    no, we don't, not really, you're conflating an analogy with the actual thing itself

    again, no it isn't.

    no you're not.
     
  8. Nov 24, 2005 #7
    Let's say that f(x) is Riemann integrable, ok? EDIT: In some interval (a,b)

    You take the limit of a Riemann sum to obtain its integral, with all that this implies, not by summing up 'infinitesimals of certain form'.
     
    Last edited: Nov 25, 2005
  9. Nov 24, 2005 #8

    dx

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    ok, thanks.
     
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