# Invalid Integration Variables

1. Jul 19, 2005

### Orion1

I am inquiring as to why integration variables cannot be interpreted using this method:

$$\frac{d}{dx} (x^n) = nx^{n - 1}$$
$$\int nx^{n - 1} dx = x^n + A$$

$$\frac{d^2}{dx^2} (x^n) = n(n - 1)x^{n - 2}$$
$$\int n(n - 1)x^{n - 2} dx^2 = x^n + A + Bx$$

Last edited: Jul 20, 2005
2. Jul 19, 2005

### matt grime

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. Jul 19, 2005

### mathman

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. Jul 30, 2005

### Orion1

relative reference...

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

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

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

It certainly seems alot easier to read, I certainly would not want to write out a relative example: $$dx^{10}$$ 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?

Last edited: Jul 30, 2005
5. Jul 30, 2005

### Crosson

Differentiation is the inverse of Integration.

Antidifferentiation is the inverse of differentiation.

6. Jul 31, 2005

### Orion1

Inverse Operation...

Differentiation:
$$\frac{d}{dx} (x^n) = nx^{n - 1}$$

Indefinite Integration:
$$\int (nx^{n - 1}) dx = x^n + A$$

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

OK, then what is the difference between the operation of indefinite Integration and Anti-Differentiation?

Last edited: Jul 31, 2005
7. Aug 1, 2005

### HallsofIvy

One of the difficulties is that while $$\frac{df}{dx}= f '(x)$$ yields the differential notation df= f'(x)dx

$$\frac{d^2f}{dx^2}= f"(x)$$ 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. Aug 1, 2005

### Hurkyl

Staff Emeritus
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.