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September 2, 2014

September 2, 2014

Posted by **Laura** on Thursday, September 27, 2012 at 5:35pm.

Please show all work

- Calculus 2 -
**bobpursley**, Thursday, September 27, 2012 at 6:05pmI would approach this as a centroid problem.

first, find the area under the curve

area=int y dx from x=0 to 2

and you put in for y, -x^3+8

centroid x= (INT x*y dx )/area

centroid y= (INT y*y dx)/area

where in both of these y=8-x^3

and the center will be the x,y you get

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