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Trying to understand the dx after the integrand

  1. Jan 23, 2010 #1
    Every book that I have checked, it says the dx after the integrand is just there to remind us that it is the variable x that is being integrated. I however am trying to see some other meaning. There must be some reason why, when we integrate 2x, it is also multiplied by dx. I have tried real hard but couldn't come up with a reasonable explanation.

    Maybe someone here can help me.

    But first of all, answer this: Is the dx and the integrand being multiplied? For example, in [tex]\int dy = \int 2x dx[/tex], dx is being multiplied by 2x? I believe it is, since from [tex]\frac{dy}{dx} = 2x[/tex] we get [tex]dy=2x \cdot dx[/tex], then we integrate both sides.
     
    Last edited: Jan 23, 2010
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  3. Jan 23, 2010 #2

    HallsofIvy

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    Do you remember the Riemann sums that the integral is derived from? At some point we have to divide whatever space we are working in (line, surface, etc.) into small intervals and determine its "measure" (length, area, volume). Typically, we call that "[itex]\Delta x[/itex]" and dx comes from that in the limit. Basically, "dx" says how we are measuring the underlying space. Notice, also, that the [itex]\Delta x[/itex] multiplies the value of the function in the Riemann sum so, yes, we can think of "dx" as multiplying the function.

    By the way, you have a typo- your last should be "dy= 2x dx", not "dy= 2x".
     
  4. Jan 23, 2010 #3
    The dx is a matter of notation, but like the notation [tex]\frac{dy}{dx}[/tex] can give us some correct results like the chain rule [tex]\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}[/tex] (although not a real prove), the dx notation in the integral can give us results of variable changing [tex]\int dx=\int \frac{dx}{dt} dt[/tex]
    Here I must comment, that while looking at it like a fraction helps to understand the change of variables, it doesn't explain at all what do you do with the integration limits (which as you know, changes to fit values of t).

    The dx is also there as a remainder of the reimann sums definition of integration, to notify we're summing samples of f(x) on realy small intervals.
     
  5. Jan 23, 2010 #4
    From my question, it is obvious I was asking about the definite integral, because then I would have said something like [tex]A = \int x^2 dx[/tex], and not [tex]y = \int 2x \cdot dx[/tex]. My question is about the latter. I really do not understand why dx is being multiplied with the derivative 2x. Is it just a notation--to show that it is the variable x that is being integrated?
     
  6. Jan 24, 2010 #5
    Halls of Ivy already answered your question.
     
  7. Jan 24, 2010 #6
    But in [tex]y = \int 2x \cdot dx[/tex], why dx is being multiplied by the derivative of the original function?
     
  8. Jan 24, 2010 #7

    HallsofIvy

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    What "original" function are you talking about? "dx", here, is simply being multiplied by a function- the function to be integrated. Of course, after the integration, if you differentiate the resulting function you get the integrand again. That is one half of the "Fundamental Theorem of Calculus" and doesn't have anything directly to do with the meaning of "dx".
     
  9. Jan 24, 2010 #8
    Check out about differential forms of first (and perhaps higher) degree (1-forms).

    I think you may just find a meaning of the notation 'dx' there.


    The 'dx' in the Riemann integral does not have any specific meaning as far as I know and it has correctly been traced by HallsofIvy back to the very definition of the Riemann sum.

    Note that if you consider the Lebesgue integral as a concept lying on measure theory you won't find a 'dx' any more, but rather a '[tex]d\mu[/tex]', where [tex]\mu[/tex] is the measure and not a variable!

    I suppose the notation 'dx' is due to Leibniz, but he just 'scored up 10' with it, as it implicitly suggests many other properties and makes calculations much easier.


    regards, marin
     
  10. Jan 24, 2010 #9
    This. If you study measure theory, you see that there are many interesting ways to assign a "length" to an interval, and it's convenient to integrate with respect to these methods of measuring space. In this context, the dx is convenient to indicate that you're using the standard idea of length, i.e. the Lebesgue measure.
     
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