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Intervals on an integral

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data


    Screenshot2012-02-04at44658AM.png


    I understand about everything except why the b and a values on the integral change from 0,3 to 9,36
     
  2. jcsd
  3. Feb 4, 2012 #2

    tiny-tim

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    hi bobsmith76! :smile:

    09 dx is an abbreviation for ∫x=0x=9 dx …

    now convert x to u ! :wink:
     
  4. Feb 4, 2012 #3
    I know what it's an abbreviation for but how do you go from 0,3 to 9,36 by what rule is that legal?
     
  5. Feb 4, 2012 #4

    tiny-tim

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    uhh? :confused:

    u = x3 + 9

    so x = 0 -> u = 9,

    so x = 3 -> u = 36 :smile:
     
  6. Feb 4, 2012 #5
    ok, thanks
     
  7. Feb 4, 2012 #6
    Does this rule have a name so that I can look it up in my book? I don't see why u-substitution should be related to the a and b values on an integral.
     
  8. Feb 4, 2012 #7

    Dick

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    I don't think it has any particular name. But if you express your final integral as a function of u, then the limits have to change to the limits for u. If you don't like that, then change the u back into x^3+9 at the end, so you've got 6π(x^3+9)^(1/2). Now use the original limits. It's exactly the same thing.
     
  9. Feb 5, 2012 #8

    tiny-tim

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    (just got up :zzz: …)
    yes … and the name is u-substitution! :biggrin:

    if you substitute u for x, you must do so wherever x occurs,

    including in the the limits!!

    u-substitution does exactly what it says on the tin! :wink:
     
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