Question on Integrals

[Ergot]

I understand that [tex]\int{f^{\prime}(x)dx} = f(x)[/tex] but I don't get what happens to [tex]dx[/tex] or what it is :(

[rabbit]

dx means "derivative with respect to x".  The integral (crazy S = [tex]\int[/tex]) cancels it out.  What you have written is technically wrong, since [tex]\int{f^{\prime}(x)dx} = f(x)[/tex] is the same as [tex]f^{\prime}(x)dx = f^{{\prime}{\prime}}(x) = f(x)\frac{d^2}{dx^2}[/tex], so what you have is saying that the integral of the 2nd derivative of f(x) is f(x), when in actuality it's the first derivative.  dx doesn't mean all that much by itself, but slapped next to a function it means a whole lot.

Ask your teacher to do this one:
[tex]\int^{\pi}_{\frac{{\pi}}{2}}\int^{x^2}_0\frac{x}{2}cos(\frac{y}{x})dydx[/tex]

[Ergot]

uhh then are you saying that all the problems in the book are wrong? Because they are all written in [tex]\int{f^{\prime}(x)dx}[/tex] form.

[Ender]

dx means "derivative with respect to x".  The integral (crazy S = [tex]\int[/tex]) cancels it out.  What you have written is technically wrong, since [tex]\int{f^{\prime}(x)dx} = f(x)[/tex] is the same as [tex]f^{\prime}(x)dx = f^{{\prime}{\prime}}(x) = f(x)\frac{d^2}{dx^2}[/tex], so what you have is saying that the integral of the 2nd derivative of f(x) is f(x), when in actuality it's the first derivative.  dx doesn't mean all that much by itself, but slapped next to a function it means a whole lot.

Ask your teacher to do this one:
[tex]\int^{\pi}_{\frac{{\pi}}{2}}\int^{x^2}_0\frac{x}{2}cos(\frac{y}{x})dydx[/tex]

Hm?

What he has written is right (unless he's edited it since). The integral of the derivative of f with respect to x is f. Another way of looking at this is that f'(x) = df/dx, so INT f'(x) dx = INT df/dx  * dx = INT df = f(x). One might say that he didn't put the +C, but the way I think of it is that f represents a family of functions since it is arbitrary, unlike sin(x) for example, in which case you'd put the +C.

Integrating over the plane is just taking the sum of infinitely many, infinitely narrow rectangles under the curve. The quantity dx is an infinitesimal, that is it is infinitely small. It represents the width of one of the infinitely many rectangles you are summing. Area of rectangle = height * width = f(x) * dx. If you choose finitely many rectangles arbitrarily, then you would only approximate the area under the curve, not get it exactly. [You can however choose finitely many rectangles with care to get the exact area, but we won't get into that now (basically it has to do with averages).] So what makes the calculation of a definite integral exact is that there are infinitely many, infinitely narrow rectangles.

I have glossed over the distinction between definite and indefinite integrals. With definite integrals you are summing the rectangles from point x=a to point x=b, so you end up calculating a value. With indefinite integrals, there are no bounds from which you are summing your integrals, so you don't end up with a value since you don't know where to start and where to end. So with indefinite integrals we add a constant + C. What you wrote is an indefinite integral, but as explained in the first paragraph I think it's okay to omit the C in this case (but be wary!)

I made the assumption that you are integrating over the plane, but I think that's a safe assumption.

[Ergot]

Yea I forgot the +C, whoops sorry, just starting out :P. Thanks for the reply. Thanks to Sidoh too. It was driving crazy :O, seriously.

[Ender]

An example of another type of integral is a path integral. Path integrals integrate over a path, whereas before we integrated over the plane. Suppose you hike up a mountain. You end up doing something like a spiral around the mountain. So you take that path. Suppose we want to calculate the work you do against friction. We want to get the component of friction in the direction of your path, that is the projection of friction onto your path, so we dot friction and our infinitesimal path length dl. So,

[tex]\displaystyle W = \int \vec{f} \cdot \vec{dl}[/tex]

Just shows you that the differential is more than just the width of a rectangle on the plane.

[Ergot]

* Ergot explodes

[rabbit]

The way I learned it, [tex]f^{\prime}(x) = f(x)dx[/tex]

[Camel]

That is not correct rabbit; you can't just slap a 'dx' at the end of something to make it a derivative. The 'dx' is a variable that approaches zero in the limit:

[tex]f^{\prime}(x) := \frac{d}{dx}[f(x)][/tex]

[Sidoh]

[tex]\int^{\pi}_{\frac{{\pi}}{2}}\int^{x^2}_0\frac{x}{2}cos(\frac{y}{x})dydx[/tex]

[tex]\displaystyle \frac{\pi^2-\pi-2}{2}[/tex]

Damn you for making me do integration by parts. :P

[Ender]

Yes, rabbit's definition of the derivative is incorrect.

Camel, you forgot an integral in your third line, e.g. [tex]\int_{a}^{b} f(x) = ...[/tex], not f(x) = ...

I am going to be AFK for over a week, very busy.

[rabbit]

That is not correct rabbit; you can't just slap a 'dx' at the end of something to make it a derivative. The 'dx' is a variable that approaches zero in the limit:

I was being lazy.dx

[tex]\int^{\pi}_{\frac{{\pi}}{2}}\int^{x^2}_0\frac{x}{2}cos(\frac{y}{x})dydx[/tex]

[tex]\displaystyle \frac{\pi^2-\pi-2}{2}[/tex]

Damn you for making me do integration by parts. :P

Parts is annoying :D

[Sidoh]

There's also an assumption that Camel is making (which becomes understood, but it's important to know).

Where he says

[tex]\displaystyle \lim_{x \to \infty}\sum_{i=1}^{n}{f\left(a+\frac{b-a}{n}i\right)\frac{b-a}{n}[/tex].

It is assumed that the widths of all of the rectangles whose widths are approaching zero have exactly the same width.  This sort of implied by the way it is written here, though.  However, it isn't as general as it can be.

The integral is more precisely defined as (I'm not correcting anyone; just clarifying)

[tex]\displaystyle \lim_{||P|| \to 0}\sum_{k=1}^{n}{f\left(c_k\right)\Delta x_k[/tex]

Where [tex]c_k[/tex] is the kth subinterval and [tex]\Delta x_k[/tex] is the width of that subinterval.  P is the partition and is defined as the set of all subintervals between [a,b].

||P|| is called the norm of the partition and is defined as the largest subinterval in P.  If we let [tex]||P|| \to 0[/tex], we obviously avoid the aforementioned problem.  Making all subintervals the same "width" works too, but this is more general, I think. :)

And as Ender mentioned, the assumption here is that we're integrating on a plane. :)

I was being lazy.dx

I think there's still something wrong with your notation.  Multiplying a derivative by a differential doesn't turn it into its "parent function."  I think what you mean to say is

[tex]\displaystyle F^{\prime}(x) = \frac{d}{dx}\int_{a}^{x}{f(t)\,dt}=f(x)[/tex]

Which is the Fundamental Theorem of Calculus, Part 1.

[Camel]

Sidoh: your definition, while more verbose, is not more complete. The definition of a convention does not require proof, or its derivation (no pun intended). If you substitute all of the things you had to define for their mathematical definition, your definition reduces to my definition.

It would have been prudent of me to qualify [tex]a[/tex] and [tex]b[/tex] as the bounds of the integration.

Yes, rabbit's definition of the derivative is incorrect.

Maybe I'm blind, but I'm not seeing a definition of a derivative in any of rabbits posts? I see a definition of the \prime notation, which by the way is incorrect.

Camel, you forgot an integral in your third line, e.g. [tex]\int_{a}^{b} f(x) = ...[/tex], not f(x) = ...

If you look closely, you'll see a \dot above the function; \dot is to integration as \prime is to derivation.

[Rule]

dx shouldn't be restricted to solely mean the limit in the definition of a derivative, but rather, it should be thought of as an "infinitesimal" quantity.  Intuitively that might be easy to accept, but it turns out to be very difficult to precisely define what this means.

"dx" is a "differential" and stands for "the differential of x".  For one interpretation, trying googling "differential forms".

I can think of two possible reasons for the confusion in Rabbit's post:
1) The differential is an operator.  The differential of f(x), df, = df/dx dx.  So in this sense,
"d" can be heuristically thought of as d/dx *dx. [Very limited and often incorrect interpretation]
2) Capital "d", D, stands for the derivative of a function.  Df(x) = df/dx.  This notation is often used when
referring to either the derivative operator (matrix) in many variables, or the directional derivative.


Also, note, contrary to Rabbit's post, [tex]f(x)\frac{d^2}{dx^2} \ne \frac{d^2f(x)}{dx^2}[/tex]

And, Camel, the "dot" is often understood to mean the time derivative (although it can generally be used as an alternative to the ' notation).  Newton used the "dot" notation in the same way most people now use the "prime" notation.

e.g. x (dot) = dx/dt