Differentials under differentials in integrals

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  • #1
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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!
 

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  • #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:
  • #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]
 
  • #4
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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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