# Can you bring out the delta x constant?

RandomMystery
Is this legal? I assumed that delta x is a constant that is approaching zero. So I brought it out. When I took the derivative, the delta x canceled out with the denominator. So I'm left with the derivative of the intergral of f(x) with respect to itself. So I'm essentially trying to find the rate of change of the sum of all the f(x) values which should equal f(x).

http://latex.codecogs.com/gif.latex...ace;f(x)=&space;d\int&space;f(x)=&space;f(x)"

This link should work ^.

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## Answers and Replies

Mentor
Is this legal?
No. For one thing, it's not $\Delta x$ in the integral: it's dx. For another thing, something that is constant doesn't change, so it can't be approaching some number.
I assumed that delta x is a constant that is approaching zero. So I brought it out. When I took the derivative, the delta x canceled out with the denominator. So I'm left with the derivative of the intergral of f(x) with respect to itself. So I'm essentially trying to find the rate of change of the sum of all the f(x) values which should equal f(x).

http://latex.codecogs.com/gif.latex...ace;f(x)=&space;d\int&space;f(x)=&space;f(x)"

This link should work ^.

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RandomMystery

No. For one thing, it's not $\Delta x$ in the integral: it's dx. For another thing, something that is constant doesn't change, so it can't be approaching some number.

Sorry, I'm assuming that the change in x is approaching zero. So I retyped it:

http://latex.codecogs.com/gif.latex...x)=&space;\int&space;f(x)&space;=&space;f(x)"

If a constant number can't be approaching some number, then why can we use e and pi? Since in calculating them we always approach some number although not zero. I know that e = the limit of (1+r/n)^n as n approaches infinity. Why can't I do the same thing for a function that approaches zero?

Actually doesn't dx equal zero in this scenario?
According to Wikipedia " The fundamental Theorem of Calculus"

(limit as h approaches zero will get the actual integral)

Which corresponds to the graph found at
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus" [Broken]

And if you look at the graph
h = the instantaneous change in x = dx = 0 (we know from derivatives that h is zero since it's instantaneous)

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RandomMystery
What I want to know is if this argument is true:
http://latex.codecogs.com/gif.latex?d\int&space;f(x)&space;=&space;f(x)"
It seems to make sense that the derivative of the anti derivative of a function is the function (with the exclusion of the C constant)

What I think that the equation up there ^ says is that the rate of change of the sum of f(x) with respect to (f(x) maybe?)
- I think this has to do with the absolute derivative or absolute integral which I'm not too sure about.

Another point I don't understand is the need for a dx at the end of the integral. What would happen if we took it off? Would it not simply change from area into the sum of all the lengths (f(x)'s)?

If this is true, then why does the instantaneous rate of change of the sum of all the f(x)'s equal f(x)?

(Edit: Oh, is it because the f(x) is the actual length being added at that instant! Hence! instantaneous rate of change of the sum! Thank you!)

What if we did the integral of f(x) except we subbed in df(x) instead of dx?
(Edit: Nevermind the statement above ^, the sentence above it answers it :b...)

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Mentor

Sorry, I'm assuming that the change in x is approaching zero. So I retyped it:

http://latex.codecogs.com/gif.latex...x)=&space;\int&space;f(x)&space;=&space;f(x)"

If a constant number can't be approaching some number, then why can we use e and pi?
e and pi are constants, much the same as 2 and 0 are.
Since in calculating them we always approach some number although not zero.
I know that e = the limit of (1+r/n)^n as n approaches infinity.
Actually that limit converges to er, not e. I think you might be confusing two different things: an expression that is changing (such as (1 + r/n)^n as n changes, and the limit of that expression.
Why can't I do the same thing for a function that approaches zero?

Actually doesn't dx equal zero in this scenario?
No.
According to Wikipedia " The fundamental Theorem of Calculus"

(limit as h approaches zero will get the actual integral)

Which corresponds to the graph found at
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus" [Broken]

And if you look at the graph
h = the instantaneous change in x = dx = 0 (we know from derivatives that h is zero since it's instantaneous)

While it's true that the derivative of the antiderivative of an integrable function is the function itself, the proof is more complicated than what you are doing.

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RandomMystery

e and pi are constants, much the same as 2 and 0 are.
Actually that limit converges to er, not e. I think you might be confusing two different things: an expression that is changing (such as (1 + r/n)^n as n changes, and the limit of that expression.

I'm using [delta x = dx] which is the expression (x2 - x1). The limit of the change in x approaches zero, so it converges into zero. (edit: the derivative is this limit)
(What do you mean that e converges to er? I thought it converged to 2.718281...)

The integral is a modification of the summation or sigma notation right? If you notice the sigma notation, every f(x) value is being multiplied by delta x or dx. Since all of these delta x and dx approach the same limit, I can treat them all as the same constant, just like e, and factor it outside of the sigma.

What's left is the summation of every f(x) value which is infinity times the delta x or dx which is zero. So,in a way, this infinity is essentially being multiplied by zero ( delta x times the sum of f(x) ).

So when we do the derivative of, the sum of the f(x) values times delta x or dx, with respect to x - the delta x or dx cancels out, and your left with the derivative of the sum of f(x) values.

We can interpret [the derivative of the sum of f(x) values] as the instantaneous rate of change in the sum of all the f(x) values, which we know is f(x) because it is the instantaneous sum (or value) being added, or in other words the instantaneous change.

So can I bring out the delta x or dx in the integral? From my interpretation the integral function does the sum of all the values between two points, we then multiply it by delta x to get the "area" (actually the integral). Since delta x can be treated as the limit of some function (just like e is the limit of (1 + r/n)^n) it can also be treated as constant. Since it is a constant, we can put it inside the integral, and put it right behind f(x).

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