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Integral from 3 to 9 of (8x^2 + 8)/(sqrt(x))dx. Not getting right answer :/

  1. Jan 12, 2013 #1
    1. Integral from 3 to 9 of (8x^2 + 8)/(√(x))dx

    2. Okay so I thought to rewrite this as:

    ∫(8x^2 + 8)(x^-1/2)dx

    = ∫(8x3/2+ 8x -1/2

    = (2*8x1/2 + (2/3)*8x-3/2) |3 to 9

    = (16)(9^(1/2)) + (2/3)(8)(9^(-3/2)) - [(2)(8)(3^(1/2)) + (2/3)(8)(3^(-3/2))]

    = 48 + .19753 - (27.71 + 1.0264)

    = 19.46113

    This answer is wrong... I can't find where I made my mistake, though! :/
    Please help!
    Thank you so much! :)
     
  2. jcsd
  3. Jan 12, 2013 #2

    Dick

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    The integral of x^(3/2) doesn't have anything to do with x^(1/2). What are you doing to get the integral??
     
  4. Jan 12, 2013 #3
    UGH! >_< OF COURSE! DUH!

    I subtracted 1 instead of adding 1.

    The answer is 748!

    Thanks! ;)
     
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