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Question on integral notation (dt, dx, etc.)

  1. May 9, 2010 #1
    Say the limit of the integral is from 0 to t and the integral is ended with a dt. Is this okay?

    Generally, all the integrals I see with a variable limit end with a d-letter that is not the same as the variable in the limit. ie: limit is from 0 to t, ends with du
  2. jcsd
  3. May 9, 2010 #2


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    Welcome to PF!

    Hi hulgy! Welcome to PF! :smile:
    Noooo …

    If you have an ∫ f(t) dt, that has to be between limits of t …

    t can't depend on itself. :wink:

    (once you've integrated over a variable like t, that variable should disappear, not come back to haunt you!)
  4. May 9, 2010 #3
    Re: Welcome to PF!

    Hmmm... what if t stood for a value like 1<t<3 which you would have to place into the 0 to t limit (this would technically eliminate the variable...right?). Would the integral as a whole still be wrong then?
  5. May 9, 2010 #4


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    Not following you :confused:

    can you write the integral out in full? :smile:

    (you can write integrals like this: [noparse]∫ab[/noparse] … ∫ab :wink:)
  6. May 9, 2010 #5
    0t (50)dt for 1<t<3
  7. May 9, 2010 #6


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    hmm …

    If you mean ∫0s (50)dt for 1<s<3, then that's fine.

    But no, you can't use t for two different things.
  8. May 9, 2010 #7

    D H

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    That is a bit of abuse of notation -- not that you won't see such abuse.

    The dt inside the integral is a dummy variable. It might as well be tau, or x, or anything. It has nothing to do with the upper limit of the integration.

    This is at best confusing:

    [tex]f(t) = \int_0^t \cos t\, dt[/tex]

    What is [itex]df/dt[/itex]?

    If the above integral is expressed as

    [tex]f(t) = \int_0^t \cos \tau\,d\tau[/tex]

    then it is pretty clear that [itex]f(t)=\sin t[/itex] and that [itex]df/dt = \cos t[/itex].

    Occasionally you will run across things like

    [tex]f(t) = \int_0^t \cos (t-\tau)\,d\tau[/tex]

    The t inside the above integral is not a dummy variable. As far as the integration is concerned that t inside the integral is a constant.
  9. May 9, 2010 #8
    I guess I'm gonna get points off for notation on my AP calc test then. Well thanks for all the help, really cleared things up.
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