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Integral Syntax Question

  1. Dec 29, 2007 #1
    Is the following syntax correct?

    [tex] dx = v\ dt [/tex]
    [tex] x = \int v\ dt [/tex]

    or should it be:

    [tex] dx = v \ dt[/tex]
    [tex] dx = \int v \ dt [/tex]
  2. jcsd
  3. Dec 29, 2007 #2

    Doc Al

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    Staff: Mentor

    This is OK. Realize that this is just saying that:
    [tex] \int \ dx = x [/tex]

    That's no good--you must integrate both sides.
  4. Dec 29, 2007 #3


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    The first. Obviously, the two right sides of the second are not the same and cannot both be equal to dx.

    What you are doing is starting with dx= v dt and integrating both sides:
    [itex]\int x= \int v dt[/itex]. Since [itex]\int dx= x[/itex] the result is [itex]x= \int v dt[/itex].

    (The integral is not "well defined" so that should be [itex]x= \int v dt+ C[/itex].)
    (Once again, Doc Al comes in 2 seconds ahead of me!)
  5. Dec 29, 2007 #4

    Gib Z

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    We don't really need to include the additional C, since indefinite integrals are only unique up to a additive constant anyway.
  6. Dec 30, 2007 #5


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    Yes, of course. The anti-derivative of [itex]x^2[/itex] is [itex]\int x^2 dx[/itex] which is, itself, equal to [itex](1/3)x^3+ C[/itex]. It is only in the last that we need the "C".
  7. Dec 30, 2007 #6


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    Actually, I prefer to think of
    [tex]\int v \, dt[/tex]
    as notation where the [itex]\int dt [/itex] is a single symbol.
    The equation
    [tex]dx = v dt[/tex]
    wouldn't make sense then but can be written
    [tex]dx/dt = v[/tex]
    or considered as a limit.

    (Of course, I also use them as mnemonics and manipulate them as ordinary fractions, but sometimes it's good to keep things clear and separate legal operations from convenient notation).
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