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 . 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.

[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.

Parts is annoying

[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

[rabbit]

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

Also, note, I didn't say that.

[Rule]

[tex]f^{{\prime}{\prime}}(x) = f(x)\frac{d^2}{dx^2}[/tex]

Where I assume, f''(x) means the second derivative of f with respect to x?
f''(x)=[tex]\frac{d^2f(x)}{dx^2}[/tex]

[Camel]

a) The dot goes above the function's name, not next to it.
b) I've never heard of \dot notation used for anything other than integration, or for letters of a couple of foreign languages.

[tex]f^{{\prime}{\prime}}(x) = f(x)\frac{d^2}{dx^2}[/tex]

Where I assume, f''(x) means the second derivative of f with respect to x?
f''(x)=[tex]\frac{d^2f(x)}{dx^2}[/tex]

It's easier to understand why this is true, but why rabbit's statement is not, if you think of [tex]\displaystyle\frac{d^2}{dx^2}[/tex] as a function of its own, as in:

[tex]\displaystyle f^{\prime\prime}(x) = \frac{d^2}{dx^2}[f(x)][/tex]

However, since the statement [tex]\displaystyle\frac{d^2}{dx^2}[/tex] is indeed a quotient - be it shorthand for a more complex division than it appears to be - it's legal to put the target function on the numerator.

[Rule]

a) The dot goes above the function's name, not next to it.
b) I've never heard of \dot notation used for anything other than integration, or for letters of a couple of foreign languages.

I've never seen it used for integration, but it is almost always used, especially in physics and in differential equations, to mean a derivative -- usually the time derivative.  It's not like you did anything wrong, you just deviated from well-established conventions; if you use the dot notation to mean integration, there will likely be some confusion.  See here.

I was being lazy in my notation, but I was referring to the way you used dot, e.g. above the function.

I don't think of d/dx (by itself) so much as quotient as it is a linear operator.
e.g.
[g(x) * d/dx ]*f(x) = g(x)*df/dx != [d/dx*g(x)]*f(x) = dg/dx*f(x)

[Camel]

The [tex]\frac{d}{dx}[/tex] notation doesn't really fit snugly in to any mathematical concept established before calculus, but if you look at the definition of it:

[tex]\displaystyle \frac{d}{dx}[f(x)] := \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}[/tex] you can plainly see that is is a quotient. Example:

[tex]\displaystyle f^\prime(x) = \frac{df}{dx}[/tex] where [tex]df[/tex] is the "rate of change in f" at value x, and [tex]dx[/tex] is the infinitesimal rate of change in x.

[Rule]

[tex]\frac{d}{dx}[/tex] is just an operator in the same way a matrix is an operator.

It is a non-commutative linear operator:
d/dx (af + bg) = a*df/dx + b*dg/dx
d/dx(f) != f*(d/dx),
where a and b are constants, f and g are functions of x.

It operates on a function in such a way that its output is the limit of a quotient.

[Ender]

Camel, you are saying that d/dx is a quotient because df/dx is a quotient. But d/dx =/= df/dx. In fact, it doesn't make sense, since d/dx is an operator, and df/dx is not.

It's like saying the division operator '/' is a quotient because a/b is a quotient.

[Camel]

Camel, you are saying that d/dx is a quotient because df/dx is a quotient. But d/dx =/= df/dx. In fact, it doesn't make sense, since d/dx is an operator, and df/dx is not.

It's like saying the division operator '/' is a quotient because a/b is a quotient.

I never said [tex]\displaystyle\frac{d}{dx} = \frac{df}{dx}[/tex], I said [tex]\displaystyle\frac{df}{dx} = \frac{d}{dx}[f(x)][/tex]

"[tex]\display\frac{d}{dx}[/tex]" is not an operator per se; it is incomplete, but is still a quotient. Just because it operates on a function does not make it a function its self, although it's reasonable to think of it as one - I defined it above the way one would define a function. Saying "[tex]\display\frac{d}{dx}[/tex]" is the same as saying "the change in [what is to follow] as x changes."


[edit] tex

[nslay]

Camel, you are saying that d/dx is a quotient because df/dx is a quotient. But d/dx =/= df/dx. In fact, it doesn't make sense, since d/dx is an operator, and df/dx is not.

It's like saying the division operator '/' is a quotient because a/b is a quotient.

I never said [tex]\displaystyle\frac{d}{dx} = \frac{df}{dx}[/tex], I said [tex]\displaystyle\frac{df}{dx} = \frac{d}{dx}[f(x)][/tex]

"[tex]\display\frac{d}{dx}[/tex]" is not an operator per se; it is incomplete, but is still a quotient. Just because it operates on a function does not make it a function its self, although it's reasonable to think of it as one - I defined it above the way one would define a function. Saying "[tex]\display\frac{d}{dx}[/tex]" is the same as saying "the change in [what is to follow] as x changes."


[edit] tex

An operator, in functional analysis, loosely speaking, is a function that maps functions to functions (a function that accepts a function, and outputs a function).  The derivative is an operator since it maps functions to functions.  The Calculus 1 definition of the derivative is a beginner's view of the derivative.  When you take a course like numerical PDE, you will see that a derivative can be expressed as a matrix which is pretty damn cool (a matrix can be an operator!).  In math, there are often many interpretations of a single problem...who would think a derivative could be thought of as a matrix?  In my experience, math is learned in coats of paint...Calculus is one of the very first coats.  As you get higher in math, you realize how little you know...and actually, you learn how little is known about math in general.  Just as an example, computational fluid dynamics uses linear approximations to solve the Navier-Stokes equation, which is non-linear...we still work with the linear in this day and age!

You should try to listen to Rule, he's one hell of a mathematician (and yes, he's a REAL mathematician)...he's much better than I am anyways.

[Camel]

My mistake: I was thinking in terms of procedural programming, where you must resolve the state of a function before it can be passed to another function, and an "operator" is a special sub-set of a function.

Also, with a deep understanding of matrices, the analogy isn't exactly a big leap of faith. I recall one particular problem in my AP Calculus class in highschool that I solved using matrices. I didn't get credit for the problem because my teacher didn't learn how to do that until after he got his masters degree -- at least, that's what he said when I asked him why I didn't get credit for a correct answer.