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Leibniz's notation and how suggestive it is

  1. Oct 10, 2006 #1
    With the method of substitution with integrals, Leibniz's notation comes in handy has it shows us usefull transformation with simple manipulation. However, I'm asking myself what is the proof behind all that? For substituion, wouldn't have to use the definition of integral and use limits?
     
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  3. Oct 11, 2006 #2

    Galileo

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    All valid notational 'abuse' is justified by the chain rule. Rather sloppily: [tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/tex]
    That is to say. You can prove the validity of the substitution rule in integrals and seperation of variables in ODE's with the chain rule.
     
  4. Oct 11, 2006 #3

    HallsofIvy

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    To phrase it another way, while the derivative is not a fraction, it is a limit of a fraction. We can always go back before the limit, make use of the fraction properties and then take the limit to show that derivatives has the properties of fractions. Leibniz notation, and especially, the definition of "differentials", [itex]dy= \left(\frac{dy}{dx}\right)dx[/itex], is a way of making use of that fact.
     
  5. Oct 11, 2006 #4
    I understand that it is mearly an expression of fractions, and with derivatives the conclusions are obvious... however I fail to see that they make everything obvious for integrals, for there is no proof shown that the Riemann Sum in question will converge to the same result we find by manipulating dy and dx...
     
  6. Oct 12, 2006 #5

    matt grime

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    Yes there is since it is just the chain rule.
     
  7. Oct 12, 2006 #6

    Galileo

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    Since integration is the 'reverse' of differentiation, everything that is obvious for differentiation has a mathematically trivial counterresult for integration. The substitution rule is just the integral version of the chain rule and the 'integration by parts'-rule is just the integral version of the product rule.

    Why don't you try to prove the rule yourself? I`ll state it here:

    If g' is continuous on [a,b] and f is continuous on the range of u=g(x), then:

    [tex]\int_a^bf(g(x))g'(x)dx=\int_{g(a)}^{g(b)}f(u)du[/tex]

    Start by assuming F be an antiderivative of f. Note that F(g(x)) is an antiderivative of f(g(x))g'(x) by the chain rule.
    You can fill in the rest.
     
  8. Oct 12, 2006 #7
    You're forgetting the essential link between integration and differentiation provided by the FTC. There's a reason we call the theorem "fundamental!"
     
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