# Intergration

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• November 9th 2008, 09:57 AM
elitespart
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.
• November 9th 2008, 01:46 PM
skeeter
$a = \sqrt[3]{16} = 2\sqrt[3]{2}$