calculus
posted by travis .
find the centroid of the region bounded by x^212x and the xaxis

The expression x²12x crosses the xaxis twice, at x=0 and x=12.
The curve stays below the xaxis on the interval [0,12].
See:
http://img207.imageshack.us/img207/3690/1291001724centroid.png
The area is therefore:
I1=∫(x^212x)dx from x=0 to x=12.
By symmetry, the centroid lies on the line x=6.
To find the ydistance, evaluate the integral in which each slice is multiplied by y/2=(x^212x), equal to the centroid of each slice:
I2=∫(1/2)(x^212x)²dx
The yposition of the centroid is then:
yc=I2/I1
I get yc=14.4 (below the xaxis)