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Homework Help: Definite Integ

  1. Nov 23, 2003 #1
    INT[-8 to -1] x - x^2 / 2*x^(1/3) dx

    (x - x^2)*(2x^(-1/3))

    Distributed:
    2x^(2/3) - 2x^(5/3)

    [6x^(5/3) / 5] -[ 3x^(8/3) / 4]

    Plug in -1, and -8

    -1.2 - -.75 = -.45

    -38.4 - - 192 = 153.6

    -.45 - 153.6 = -154.05

    When I put this into the calculator straight to check my work, I get 57.112 as the answer. What did I do wrong here?
     
  2. jcsd
  3. Nov 23, 2003 #2

    ShawnD

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    There is the error. In your original equation, the 2 is in the denominator. In your second equation, you brought the 2 to the top without putting a negative exponent on it; you only put a negative exponent on the x.
     
  4. Nov 23, 2003 #3

    Hurkyl

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    [tex]
    \frac{x-x^2}{2x^{1/3}}
    = (x-x^2) (\frac{1}{2} x^{-1/3})
    [/tex]

    [tex]
    (-1)^{8/3} = 1
    [/tex]
     
  5. Nov 23, 2003 #4
    Ah, I thought it was just a constant multiplier that stuck with the x.

    Hmm...

    So that would make it:
    [ 2^-1 * x^(2/3)] - [2^-1 * x^(5/3)]

    5x^(5/3)/6 - 3x^(8/3)/16

    For -8, I get -26.7 - - 48 = 21.3

    -1, -.833 - -.1875 = -.6455

    Ay... I must have done something else wrong?
     
  6. Nov 23, 2003 #5

    Hurkyl

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    [tex](-1)^{8/3} = 1[/tex]

    (and this time you multiplied by five-thirds instead of divided)
     
  7. Nov 23, 2003 #6
    Urgh, hopefully I won't make these simple mistakes on the test.
     
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