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Homework Help: Differentials under differentials in integrals

  1. Jul 17, 2006 #1
    Hello.

    Are the following integrals equivalent:

    Integral from 0 to 5 of dx / x

    and

    Integral from 0 to 5 of dx / (x + dx)

    What about

    Integral from 0 to 5 of dx / (x + 2dx + dx^2)

    ???

    If they are all equivalent, why? (I have an intuitive answer, but it has 0 mathematical foundation). Or are they not all the same? Can someone explain this strongly?

    Thank you!
     
  2. jcsd
  3. Jul 17, 2006 #2

    quasar987

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    Take note that you should not take my answer seriously, but I am tempted to propose an answer, and see how people react to it.

    The way I would interpret

    [tex]\int_0^5 \frac{dx}{x+dx+(dx)^2}[/tex]

    is by treating the dx on top as the dx that's part of the notation for "the integral of a function", and the rest of them I would treat as constant (or more precisely, as parameters), so the answer would be

    [tex]\ln (5+dx+(dx)^2)-ln(dx+(dx)^2)[/tex]


    With this interpretation, you can see that your three integrals are not equal.
     
    Last edited: Jul 17, 2006
  4. Jul 17, 2006 #3

    Hurkyl

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    I would say that it's nonsense unless you defined what it meant.


    Boldly doing manipulations without any solid justification (which is the only kind we can do when we don't have a definition for what we're manipulating)...

    [tex]
    \frac{dx}{x + dx + (dx)^2}
    = \frac{1}{x + dx + (dx)^2} dx
    \approx \frac{1}{x} \left(1 - \frac{dx + (dx)^2}{x}\right) \, dx
    \approx \frac{dx}{x}
    [/tex]
     
  5. Jul 18, 2006 #4
    I am interested primarily because of a curious step in this derivation of the Tsiolkovsky "rocket equation": http://ed-thelen.org/rocket-eq.html

    Namely, where he says, "we are looking at the area under the curve of 1/(x+dx) where in the process we make dx arbitrarily small so that we are looking at the area under the curve of 1/x in the limit. In some presentations, this simplification is done before the integration is performed."
     
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