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Does dx does not make change in front of the integral?

  1. Aug 10, 2009 #1
    Does "dx" does not make change in front of the integral?

    Hello I got one question. I am confused by seeing dx in front of the integral.

    For ex.

    [tex]\int f(x) dx = F(x) + C[/tex]

    where

    [tex]F'\!(x) =\frac {d}{dx} F(x) = f(x).[/tex]

    As we know F'(x)=f(x), then why F'(x) = f(x) . dx

    Isn't supposed to be F'(x)=f(x) and not F'(x)=f(x) . dx??


    Thanks in advance.
     
  2. jcsd
  3. Aug 10, 2009 #2
    Re: Does "dx" does not make change in front of the integral?

    The former is correct. F'(x) = f(x). Anyone who tells you otherwise is a liar.

    The "dx" is just a formality. It tells you which variable you are integrating or deriving over.
     
  4. Aug 10, 2009 #3
    Re: Does "dx" does not make change in front of the integral?

    The [itex]dx[/itex] inside the integral is a "dummy variable." It could be just as well be any other variable. If you look in you math book for a more precise statement of the fundamental theorem of calculus, it actually says
    [tex]\frac{d}{dx}\int_a^x f(y) dy = f(x) [/tex]

    Note specifically how [itex]x[/itex] appears in the limit of integration and how the differentials don't "cancel out." The other direction looks like this
    [tex] \int_a^x \frac{df}{dy} \,dy = f(x) - f(a) [/tex]

    in this case the differentials can be thought of as "canceling out"
    [tex] \int_a^x \frac{df}{dy} \,dy = \int_a^x df = \Delta f = f(x) - f(a) [/tex]
     
  5. Aug 10, 2009 #4
    Re: Does "dx" does not make change in front of the integral?

    Thanks for the replies.

    Yes but,
    [tex]
    \int f(x) dx = F(x) + C
    [/tex]

    States that F'(x)=f(x).dx and if dx=1 then F'(x)=f(x).

    And is d/dx in the integral above d(f(x))/ dx ?
     
  6. Aug 10, 2009 #5
    Re: Does "dx" does not make change in front of the integral?

    Like I said, the dx is just a formality. It's an artifact of leibniz notation, and it's mathematically rigorous (there's another thread that discusses this at length).

    The prime notation (F(x)) signifies the derivative. If F is an antiderivative of f, then (by definition of an antiderivative) F'(x) = f(x).

    There is no dx involved. In Leibniz notation, it would be spelled

    [tex]\frac{d}{dx}F(x) = f(x)[/tex]
     
  7. Aug 10, 2009 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Дьявол! :smile:
    Sorry, but Tac-Tics :smile: is completely right …

    "F'(x)=f(x).dx" is just plain wrong …

    you can't have a differential (d-something) on one side of an equation but not on the other side.
     
  8. Aug 10, 2009 #7
    Re: Does "dx" does not make change in front of the integral?

    Yes, I know that it is completely wrong, that's what I am asking.
    If F'(x)=f(x) then

    [tex]
    \int f(x) = F(x) + C
    [/tex]

    and not
    [tex]
    \int f(x) dx = F(x) + C
    [/tex]

    That's why I was asking if "dx" changes something.
     
  9. Aug 10, 2009 #8
    Re: Does "dx" does not make change in front of the integral?

    Again, the dx here is just an artifact of leibniz notation. You can safely ignore it. When you include the dx, it has a certain symmetry with the rest of leibniz notation. [tex]\int[/tex] becomes almost-the-opposite of [tex]d[/tex]. So you can make arguments like this:

    [tex]\frac{d}{dx}f(x) = y[/tex]
    [tex]df(x) = y dx[/tex]
    [tex]\int df(x) = \int y dx[/tex]
    [tex]f(x) = \int y dx[/tex]

    But again, this is just a convenient coincidence. Be wary of those "plus C"s.
     
  10. Aug 10, 2009 #9
    Re: Does "dx" does not make change in front of the integral?

    Thanks for the reply. I know that d(f(x))/dx is Leibniz notation for f'(x).
    [tex]
    \frac{d}{dx}F(x) = F'(x)
    [/tex]

    [tex]\int{\frac{dF(x)}{dx}}=\int{F'(x)}[/tex]
    [tex]
    \int{\frac{dF(x)}{dx}}=F(x)
    [/tex]


    Since

    [tex]F'\!(x) =\frac {d}{dx} F(x) = f(x). [/tex]

    then

    [tex]\frac {d}{dx} F(x)=f(x)[/tex]

    So

    [tex]\int{\frac{dF(x)}{dx}}=F(x)[/tex]

    becomes

    [tex]\int{f(x)}=F(x)+C[/tex]

    Exactly. And df(x)=f '(x)dx , so you say that

    [tex]\int{f ' (x)dx}=\int{f ' (x)}[/tex]

    You made df(x) and f'(x) equal, i.e f'(x)dx=f'(x).

    Sorry, for being rebelde, "dx" is kinda irritating.
     
  11. Aug 10, 2009 #10
    Re: Does "dx" does not make change in front of the integral?

    The dx is a pain in the but. My advice: just ignore it, but try to follow the convention your teacher uses. Usually, it will be as simple as "always have a dx under the integral sign" and "put a dx under every d or df".

    The dx's are just a historical thing, carried over from back in the day when mathematicians didn't really know what the hell made calculus work.

    If you go on to take an analysis class in college, all the fundamental rules of calculus will be spelled out in the elegant language of set theory.
     
  12. Aug 10, 2009 #11
    Re: Does "dx" does not make change in front of the integral?

    Thank you for the explanation. I have just finished high school, now my plan is to go to college.

    As I try to solve several integrals using x' (since x'=1 and doesn't change anything to the integral) instead of dx, I see that it is just dummy variable which does not change anything in the integral.

    Regards.
     
  13. Aug 10, 2009 #12
    Re: Does "dx" does not make change in front of the integral?

    I remember when I was in my high school calculus class, I struggled with understanding the "algebra" of calculus in the same way you are. The problem is is that this mixed up "dx" and "plus C" business usually gets you to the correct answer, even though it is not technically sound.

    If you're really ambitious, try picking up an introductory book on analysis. After learning basic analysis, everything made sense. You learn what a limit is. You learn what a real number *really* is (without appealing to "infinite", nonrepeating decimals). You can prove all those handy calculus identities that come out of nowhere, such as the linearity of the integral and derivative and the chain rule.
     
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