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Calculus terminology/notation question.

  1. Sep 8, 2006 #1
    Ok here's my problem... First, I only took Calculus AB so don't drown me with what I wont even approach to understand.

    Now my problem is the dy/dx part of calculus. In the differential equations etc it makes sense...dy gets integrated into a y and dx is integrated into x etc...

    [tex]\int{dx}=x + C[/tex]

    but look here:

    [tex]\int {x [dx]} = \frac{x^{2}} {2} + C[/tex]

    SO what happens if you do plainly...

    [tex]\int{x} = ?[/tex] + C

    Would it simply...malfunction because you don't have a dx? What does the dx actually do? to my logic the dx should turn into an x somehow...because it does so in my first example. but when dx is by some x it just does nothing. Why is that?
    Last edited: Sep 8, 2006
  2. jcsd
  3. Sep 8, 2006 #2


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    The simple explanation is that dx tells us what is the variable of integration. For example if the integrand was xy, then (assuming y does not depend on x), then integration using dx tells us that the answer is yx2/2 +c.
  4. Sep 8, 2006 #3
    Yes, this expression has to meaning.
  5. Sep 8, 2006 #4


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    It makes no sense because it doesn't tell you what to integrate with respect to. Just like mathman said.

    Unfortunately, dx does not simply mean multiply an x in the equation. This only happens when working with simple functions, like polynomials.

    Not sure if you know this, but what is the derivative of [itex]e^x[/itex]?

    If you remembered or have learned it, you should know that it is [itex]e^x[/itex].

    So, if you integrate [itex]e^x[/itex] with respect to x as follows...

    [tex]\int{e^x} dx = e^x[/tex]

    I never multiplied an x in there. See what's happening? Getting an idea?
  6. Sep 8, 2006 #5
    To say that dx is simply telling us the variable to be integrated makes no sense. If it's just to tell you which variable to integrate with respect to, why use such a distinctive notation? For example, the variable could be written in square brackets at the top of the integral sign.

    Correct me if I'm wrong here (I'm 16 and hence still allowed to make mistakes), but the reason the (Riemann) integral without the dx is meaningless is (geometrically and extremely loosely) because that would be summing an infinite number of heights without multiplying each by an infinitesimal width, and so it has no meaning.

    An interesting historical note, I think that when the long-S notation was first used, the dx wasn't written.
    Last edited: Sep 8, 2006
  7. Sep 8, 2006 #6
    Do you know how the dx is derived (derived is probably a bad word to use).

    If you think about [itex] \int_a^b f(x) \, dx [/itex] as the area under the curve. It is saying take a bunch of rectangles of width [itex] dx [/itex] and height [itex] f(x) [/itex] and add them together to get the area.

    Lets say you have a line, say of height 5. So,
    [tex] f(x) = 5 [/itex].

    Now you want to take the integral of [itex] f(x) [/itex] from 0 to 10.

    This would look like:
    [tex] \int_0^{10} f(x)\, dx [/tex]

    This is asking for the area under the curve from 0 to 10. Now if you think about this for a second, the area is just a rectangle.
    So the integral above would be [itex] 10 \times 5 [/itex].

    Now what if you broke this up into two rectangles?
    You could say that the integral equals [itex] (5 \times 5) + (5 \times 5) [/itex] right?

    Now what if you want to break this up into 5 rectangles?
    You could say the integral is equal to:
    [tex] 5\times (2 \times 5) [/tex].

    Notice how the width is shrinking each time?
    Well what if you made the width infinitely small and called it dx? So you have,
    height * dx

    If you added all these height*dx rectangles together you would get the area under the curve. Does that makes sense?

    Sorry for the crappy explanation... this was actually a lot harder to explain then I thought it would be . This discussion (at least how I would know how to explain it) requires a white board or chalk board or something.

    I don't know if this would help you at all,

    That would have definitely been over my head when taking calc I though.
    Last edited: Sep 8, 2006
  8. Sep 8, 2006 #7


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    And so integrating with respect to x implies that you are multiplying each by an infinitesimal width dx. That's integration.

    Then what is it telling us?
  9. Sep 8, 2006 #8
    That notation would be fine. Only who would know what you mean?

    I could be wrong here, someone please correct me. But dx means the d part is the differential length of x.

    where the d is somewhat equivalent to the [itex] \Delta [/itex] in the Riemann integral when [itex] \Delta x_i [/itex] shrinks to 0.
  10. Sep 9, 2006 #9
    Exactly -- and that's why leaving out the dx makes no sense. That's what I said. The point is that the lack of a dx doesn't only mean that you don't know with what to integrate with respect to. Even if it was, it could just be implied for simpler integrations, like in the f-prime notation for derivitives. There's a good reason it's there.

    To multiply each height by an infinitesimal length dx! To leave it out would be to say [itex]\int_a^b f(x)=\lim_{\Delta x\to\infty} \sum_1^n f(x_k*)[/itex], a useless, often divergent sum.

    Now, with the theory of differential forms, the problem is solved by defining dx as a 1-form. Then the integral sign is a function with a differential form, mapping an interval to a real number. Thus the interval sign has no meaning when it's followed with a non-differential form.
    Last edited: Sep 9, 2006
  11. Sep 9, 2006 #10
    [tex]\int_a^b f(x)=\lim_{\Delta x\to\infty} \sum_1^n f(x_k*)[/tex]

    Oh...I sould have known this...I seen this before. My teacher expalined it, how the x-part is the height and the dx part is the width between sectors...I understand it now. i remember something about a [tex]||p||[/tex] in the limit bottom. Oh well.

    [tex]\int{dx}[/tex] is the same as [tex]\int{1}dx[/tex] where the 1 turns into the x not the dx.

    Yes I know the [tex]e^{x}[/tex] has a derivative and an integral of [tex]e^{x}[/tex]. SO by looking at the tiny little rectangles...x tells you basically how big they are (height) and the dx tells you how many spaces are in between them. Like applying a LRAM or RRAM or MRAM for many, tiny rectangles.
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