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Thread: Intergration

  1. Nov 9th 2008, 09:57 AM #1
    elitespart
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    Intergration - Help please.

    The area between y^2=x and x=4 is divided into two equal parts by the line x=a. Find a. So I got So I got a=cubed route of 16. right wrong? Thanks.
    Last edited by elitespart; Nov 9th 2008 at 11:47 AM.
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  2. Nov 9th 2008, 01:46 PM #2
    skeeter
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    $\displaystyle a = \sqrt[3]{16} = 2\sqrt[3]{2}$
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