Those really aren't relevant. What is relevant is where the two curves intersect- where y= 4x^2= 2x^3. Obviously, one root is x= 0. If x is not 0 we can divide by 2x^2 to get x= 2.

Yes, the boundary is 0 and 2. But where did you get 2x^3- 2x^2? Your two functions are y= 4x^2 and 2x^3. Further, for all x between 0 and 2, 4x^2> 2x^3 so you want to integrate 4x^2- 2x^3, not 2x^3- 2x^2.Attempt:

A = (int) 2x^3 - 2x^2

boundary 0 and 2

Be careful about your notation. Many people would interpret "1/2x^4" as 1/(2x^4) not (1/2)x^4.1/2x^4 - 2/3x^3 =

In any case, the integral of 4x^2- 2x^3 is (4/3)x^3- (1/2)x^4 which, evaluated at 2 and 4 is 32/3- 8= 32/3- 24/3= 8/3.

What? The integral should be just x times the area integrand: x(2x^3- 4x^2). I have no idea where those squares came from1/2(2^4) - 2/3(2^3) = 8 - 2.666 = 5 1/3

X centroid:

1/2 * 1/(5.333) (int) [(2x^3)^2 - (4x^2)^2]

boundaries 2 and -2

[tex]= .09375 (int) 4/7x^7 - 16/5x^5

plugging in 2

= 4/7(2^7) - 16/5(2^5) = -29.26

completely wrong... so where'd I go wrong? Thanks[/QUOTE]

Pretty much everything is wrong! You need to review "centroid".