Invalid Integration Variables

  • Thread starter Orion1
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  • #1
Orion1
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I am inquiring as to why integration variables cannot be interpreted using this method:

[tex]\frac{d}{dx} (x^n) = nx^{n - 1}[/tex]
[tex]\int nx^{n - 1} dx = x^n + A[/tex]

[tex]\frac{d^2}{dx^2} (x^n) = n(n - 1)x^{n - 2}[/tex]
[tex]\int n(n - 1)x^{n - 2} dx^2 = x^n + A + Bx[/tex]

 
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Answers and Replies

  • #2
matt grime
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because that isn't how they behave. why should they? integration and differentiation are NOT inverse operators since integrals are only defined up to additive constants.
 
  • #3
mathman
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To Orion1

There are two problems in your approach. First (as mattgrime states) you need a constant of integration, so your first integral is xn+A. Second, the standard notation for iterated integrals is different from what you have for the second integral, which would be xn+A+Bx.
 
  • #4
Orion1
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relative reference...


What about defining the integration variable as 'relative', such that:

[tex]\int f(x) \; dx^2 = \int \left( \int [ f(x) dx] \right) dx[/tex]

The second integration with respect to [tex]dx[/tex]:
[tex]\int n(n - 1)x^{n - 2} dx^2 = \int \left( \int n(n - 1)x^{n - 2} \; dx \right) \; dx = x^n + A + Bx[/tex]

It certainly seems alot easier to read, I certainly would not want to write out a relative example: [tex]dx^{10}[/tex] using the politically correct immediate RHS method...

If integration and differentiation are NOT inverse operators, then what are the inverse operators of integration and differentiation?
 
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  • #5
Crosson
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If integration and differentiation are NOT inverse operators, then what are the inverse operators of integration and differentiation?

Differentiation is the inverse of Integration.

Antidifferentiation is the inverse of differentiation.

Hey, you asked...
 
  • #6
Orion1
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Inverse Operation...

Crosson said:
Differentiation is the inverse of Integration.
Antidifferentiation is the inverse of differentiation.

Differentiation:
[tex]\frac{d}{dx} (x^n) = nx^{n - 1}[/tex]

Indefinite Integration:
[tex]\int (nx^{n - 1}) dx = x^n + A[/tex]

Anti-Differentiation:
[tex]\frac{d^{-1}}{dx^{-1}} (nx^{n - 1}) = x^n + A[/tex]

OK, then what is the difference between the operation of indefinite Integration and Anti-Differentiation?
 
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  • #7
HallsofIvy
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One of the difficulties is that while [tex]\frac{df}{dx}= f '(x)[/tex] yields the differential notation df= f'(x)dx

[tex]\frac{d^2f}{dx^2}= f"(x)[/tex] does not yield d2f= f"(x)dx2. One of the reasons for having the "2" in different places in "numerator" and "denominator" is to make it clear that "2nd order" differentials don't work that way!


(There is no difference between "indefinite integral" and "anti-derivative". They are different names for the same thing. {Well, sometimes people refer to "an" anti-derivative to mean a specific choice of constant. I have never heard that done with "indefinite integral".})
 
  • #8
Hurkyl
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Differentiation is not an invertible operation on differentiable functions. :tongue2: Remember, for instance, that an invertible function must be one-to-one, and differentiation clearly is not.
 

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