1. Do you mean y = x^2? You wrote y-x^2. That is not a curve.
If you mean y = x^2, the area-weighted mean value of y is
[integral of y(y/2)dx] divided by 1/3
x = 0 to 1
= 3/10
The area-weighted mean value of x is
integral of y*x dx divided by 1/3
x = 0 to 1
= integral of x^3 dx divided by 1/3
x = 0 to 1
= 3/4
The division by 1/3 is normalizing by the area of the region. I agree with your answers. If you followed the same procedure for problem 2, your answers to that one are also correct.
could some body please check this for me?
1. find the exact coordinates of the centroid of the region bounded by y-x^2, the x axis, the y axis, and the line x=1?
I said the answer was ((3/4), (3/10))
2. Find the exact coordinates of the centroid of the region to the right of the y axis between y=2e^(-2x) and the x axis. Note this region is unbounded but has a finite area.
I said the answer was ((1/2), (1/2))
THANK YOU
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