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Basic intergration

  1. May 17, 2010 #1
    Hi

    could someone explain to me why, if I integrate dv/v between V2 and V1 the result is nastural log of V2/V1?

    Thanks
     
  2. jcsd
  3. May 17, 2010 #2

    Cyosis

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    Is your question why the primitive of 1/v is log v or are you wondering why you get log v2/v1 instead of log v2 - log v1? Anyhow [itex]\log v_2-\log v_1=\log v_2/v_1[/itex].
     
  4. May 17, 2010 #3
    Yes thanks Cyosis!

    Missed the fact that logV2-logV1=LogV2/V1

    Thank you very much for your insight!
    james
     
  5. May 17, 2010 #4
    Also is not that 1/v.dv between v2 and v1 = [log|v2|-log|v1|] between v2 and v1? and that the dv is ignored?
     
  6. May 18, 2010 #5

    Redbelly98

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    The dv is not ignored, it is necessary in an integral to specify the variable of integration.
     
  7. May 18, 2010 #6

    lavinia

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    can you restate this more clearly? I don't get what you are asking.
     
  8. May 18, 2010 #7
    [tex]\text{By the above phrase I think most people would understand }\int_{v_2}^{v_1}\frac{dv}{v}=log_e(v_1/v_2)\text{ not }log_e(v_2/v_1)\text{.}[/tex]
    [tex]\text{Does the use of }|v_1|,|v_2|\text{ here mean that you're interested in values of }v_1\text{ and }v_2\text{ other than positive and real?}[/tex]
     
    Last edited: May 18, 2010
  9. May 18, 2010 #8
    Hi, I think what I'm getting confused about, is that for entropy the change in entropy is ds=dQ/T. Integrating this gives s2-s1=the integral of dQ/T between 2 and 1. So are you saying that the change in Q is now just there to specify the variable of integration? Sorry id this all sounds confusing.
     
  10. May 18, 2010 #9
    It does, apology accepted :)
     
  11. May 24, 2010 #10
    integral (dv / v) = integral (1/v)*dv = integral (1/v) dv = ln|v|

    since our integral is between V2 and V1, we do ln|V2| - ln|V1| = ln|V2/V1|
     
  12. May 25, 2010 #11
    (As before.)
     
  13. May 25, 2010 #12
    Thanks everyone for there help.
     
  14. May 25, 2010 #13

    Char. Limit

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    If you integrate from v2 to v1, yes, that is correct. I was under the impression that e was integrating from v1 to v2. The word "between" creates an ambiguity there, I suppose.
     
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