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

1. Jan 12, 2013

Lo.Lee.Ta.

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! :/
Thank you so much! :)

2. Jan 12, 2013

Dick

The integral of x^(3/2) doesn't have anything to do with x^(1/2). What are you doing to get the integral??

3. Jan 12, 2013

Lo.Lee.Ta.

UGH! >_< OF COURSE! DUH!