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Math Help - Notation problems

  1. #1
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    Notation problems

    I've recently been told that dy and dx can be treated as very small numbers since they're a ratio.

    If this is true then how is the integral of dx/x = lnx ?

    it would mean that x' = dx
    Which means that dx=1

    Could anyone explain?
    Thanks!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Notation problems

    If you want to calculate:
    \int \frac{dx}{x}, the primitive is indeed \ln(x)+C

    What do you mean with x'=dx and so dx=1?

    Do you know what dx is? if you don't I think it can be useful to look at the meaning of the limit definition of the definite integral. Are you familiar with that? ...
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  3. #3
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    Re: Notation problems

    dx represents the change in x,
    so it's just a number right?
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  4. #4
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    Re: Notation problems

    Quote Originally Posted by elieh View Post
    dx represents the change in x,
    so it's just a number right?
    That is certainly historically correct. It maybe that dx is just an artifact of history. I knew an American analyst who refused to use it.
    He would simply write  \int_a^b f whereas most would write  \int_a^b {f(x)dx} .

    I have been guilty of telling students that artifact simply reminds what the variable of integration is.

    So in your question \int_a^b {dx}=\int_a^b {1dx}=\left. x \right|_{x = a}^b
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  5. #5
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    Re: Notation problems

    Makes sense, so I'm assuming that dx is negligible.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Notation problems

    What I want to say was that dx has definitely a meaning if you take a closer look to the geometric aspect of definite integrals. But in the case of indefinite integrals I agree with what Plato said.

    Meaning in case of definite integrals:
    Why do you calculate integrals? Probably you know that (calculating the area under a curve). If you have given a curve of the function f and you want to calculate the area in an interval [a,b] then you've to calculate:
    \int_{a}^{b}f(x)dx
    If you're going to take a closer look tot the geometric aspect you're going to divide the interval [a,b] in n equal part intervals with length \Delta (x_i)=\frac{b-a}{n} and height f(x'_i), if you want to calculate the area then you've to make the sum of the area's of all the part intervals that means:
    \sum f(x'_i)\cdot \Delta(x_i)
    But you want a very good approximation of the area under the curve so limits will be a solution for that:
    \lim_{n\to \infty} \left(\sum_{i=1}^{n} f(x'_i)\cdot \Delta(x_i)\right)
    Because the approximation will be better if the length of the part intervals is very small that means you want \Delta(x_i) as small as possible (so you'll get more part intervals and so the approximation of the area under the curve will be better)
    Because if  n \to \infty then  \Delta(x_i)=\frac{b-a}{n} \to 0 . So dx is the length of a very small part interval.
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