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Integrating dx/dv^2 ?

  1. May 9, 2005 #1
    Integrating dx/dv^2 ??

    i'm trying to figure out an example in my physics book but i don't quite understand the maths.

    \int \frac {dv} {v^2} = - \frac {1} {v}

    how does this happen??

    looking at the basic antiderivative formulas section in my maths book, it says that:

    \int \frac {dv} {v} = ln v

    but nowwhere do i find info on how to do the problem in my physics book.

    would reallly appreciate the help! :)


    sorry, very new to the forums.. why doesn't the latex work? :confused:

    just in case others can't view it as well.. what i meant was..

    how do you integrate dv/v^2? i know the answer is -1/v but don't know the rules for this. the closest rule i can find is that integrating dv/v is ln v..

    incidentally, i don't think i've activated this account properly.. how can i activate my account? (i deleted the activate account e-mail i think)
    Last edited: May 9, 2005
  2. jcsd
  3. May 9, 2005 #2
    the closing tag should be [/tex]
  4. May 9, 2005 #3
    I can't use LaTeX properly, so I'll just write it out again.

    The integral of x^n dx = [x^(n + 1)]/(n + 1) + C for all n (positive or negative), except n = -1, in which case, it is ln x + C.

    So, in your case you have the integral of dx/x^2 = x^-2 dx and just apply the rule above.
  5. May 9, 2005 #4
    [tex] \int \frac{dv}{v^2} = \frac{-1}{v} [/tex]

    [tex] \int \frac{dv}{v} = ln|v| [/tex]
  6. May 9, 2005 #5


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    Does it make sense to have [itex] \frac{1}{v^{2}} [/itex] in one member and [itex] \frac{1}{v} [/itex] in the other...?I think not.

  7. May 10, 2005 #6
    ah thanks a lot! thank you inha for that [/tex]! and thank you nylex for that explanation!! i never knew that it didn't apply when n = -1.. now that i look in my textbook, it does specify that.. though wrongly.. it says when n = 1 (checking in another book, shows that it is indeed when n = -1)

    but what do you mean to have [tex] \frac {1} {v^2} [/tex] in one member and [tex] \frac {1} {v} [/tex] in another? i don't understand at all. (what's a member, first of all?)

    and thanks whozum for the proper rendition in latex! but why is it [tex] \int \frac{dv}{v^2} = \frac{-1}{v} [/tex] and not [tex] \int \frac{dv}{v^2} = - \frac{1}{v} [/tex]? i know that both actually mean the same thing but when i do the calculation i reach [tex] \int \frac{dv}{v^2} = \frac{1}{-v} [/tex].. which means that i may doing something wrong so, just in case.. i'll write out how i do it and if you could be so kind, could you point out where i went wrong?

    [tex] \int \frac{dv}{v^2}
    = \int \frac{1}{v^2} dv
    = \int (v^-2) dv
    = \frac{v^-1}{-1}
    = \frac {\frac{1}{v}} {\frac{-1}{1}}
    = \frac{-1}{v} [/tex]

    i guess this is where the difference is? [tex] \frac{1}{-1} [/tex] instead of [tex] \frac{-1}{1} [/tex]? does it matter? i haven't done any maths for a long long time and i have no idea if it does or not..

    ** by the way, why is it that the "1" is not in superscript in [tex] \frac{v^-1}{-1}[/tex]?
    Last edited: May 10, 2005
  8. May 10, 2005 #7
    The answers you gave are identical and both correct. To superscript it, put { } about the exponential argument

    a^{apples} = [tex] a^{apples} [/tex]
  9. May 10, 2005 #8
    thanks! :)
  10. May 10, 2005 #9


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    The general formula, by the way is that the anti-derivative of [tex]x^n[/tex] is
    [tex]\frac{1}{n+1}x^{n+1}+ C[/tex] unless n+1= 0 (since we can't have 1/0) . That happens, of course, when n= -1. The anti-derivative of [tex]x^{-1}= \frac{1}{x}[/tex] is ln|x|. There are a variety of ways to show that- the simplest is to define ln|x| in that way.
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