Integral total and partial of a function?

[Jhenrique]

Like we have the total differential of a function:

I was thinking, why not take the "total integral" of a function too? Thus I did some algebraic juggling and, how I haven't aptitude for be a Ph.D. in math, I bring my ideia for the experients from here evaluate... Anyway, the ideia is the follows:

Let y = f(x), so: \int y dx = \int f dx \int y \frac{dx}{dx} = \int f \frac{1}{dx}dx \int y = \int f \frac{1}{dx}dx \;\;\;\Rightarrow \;\;\; \int y du = \int f \frac{du}{dx}dx Generalizing...

Let w = f(x,y,z), so: \int w = \int f \frac{1}{dx}dx + \int f \frac{1}{dy}dy + \int f \frac{1}{dz}dz

I don't venture take the 2nd integral of y because I think that will arise one d²x in the denominator...

What you think about? All this make sense?

 

[Mark44]

Like we have the total differential of a function:

Presumably dx² means dx * dx, but what does d²x mean?

I was thinking, why not take the "total integral" of a function too? Thus I did some algebraic juggling and, how I haven't aptitude for be a Ph.D. in math, I bring my ideia for the experients from here evaluate... Anyway, the ideia is the follows:

Let y = f(x), so: \int y dx = \int f dx \int y \frac{dx}{dx} = \int f \frac{1}{dx}dx \int y = \int f \frac{1}{dx}dx \;\;\;\Rightarrow \;\;\; \int y du = \int f \frac{du}{dx}dx Generalizing...

The steps in the middle make no sense to me. Dividing by dx is not a valid step. You started with f as a function of x. Is it somehow transformed to become a function of u later on?

Let w = f(x,y,z), so: \int w = \int f \frac{1}{dx}dx + \int f \frac{1}{dy}dy + \int f \frac{1}{dz}dz

I don't venture take the 2nd integral of y because I think that will arise one d²x in the denominator...

What you think about? All this make sense?

 

[Jhenrique]

Presumably dx² means dx * dx, but what does d²x mean?The steps in the middle make no sense to me.

Yeah, dx²=dxdx

d²x is the 2nd differential of x wrt nothing. Wrt to something it's become: \frac{d^2y}{du^2}=\frac{d^2f}{dx^2}\left ( \frac{dx}{du} \right )^2+\frac{df}{dx}\frac{d^2x}{du^2}

Dividing by dx is not a valid step.

humm...

You started with f as a function of x. Is it somehow transformed to become a function of u later on?

The first three equations was a attempt for show what would an integral of f (like an differential of f) and the implication shows the utility of the integral of f as an chain rule.

If you get the last equation, ∫w, and multiply the equation by an arbitrary differential du, you'll have an chain rule of integrals in tree-dimensions:
\int w du= \int f \frac{du}{dx}dx + \int f \frac{du}{dy}dy + \int f \frac{du}{dz}dz

 

[Mark44]

So d²x would be d(dx). AFAIK, this doesn't mean anything. At least it's not anything I've ever seen. Also, as I mentioned earlier, you can't divide by dx, and you can't divide by d²x. The "fractions" dy/dx and d²y/dx² are more notation than fractions that you can manipulate.

Instead of merely manipulating symbols, as you seem to like to do, make up a function w = f(x, y, z), and see if your formula for $\int wdu$ has any relation to reality.

 

[Jhenrique]

hummm... so, tell me you, why no exist total and partial integral like in differentiation, that has partial and total differential. Why my analogy no make sense?

 

[Mark44]

hummm... so, tell me you, why no exist total and partial integral like in differentiation, that has partial and total differential. Why my analogy no make sense?

Just off the top of my head, possibly it's because differentiation and integration aren't exactly inverse operations.