<<y=(1/A)*integral from a to b of: (1/2)[f(x)]^2 dx >>

If that is the y value of the center of mass, I don't know why the factor (1/2) is there

I also don't agree with your calculation of the x value, which should be 1/(ln 2).

I agree with you that the area is ln 2

find the center of mass of the region bounded by y=(1/x), y=0, x=1, and x=2.

please check what i've done:

x=(1/A)*integral from a to b of: xf(x) dx
y=(1/A)*integral from a to b of: (1/2)[f(x)]^2 dx

x = (1/ln(2))*interal from 1 to 2 of: x[1/x] dx
x = (1/ln(2))*[x] from 1 to 2 = 2.885390082-???

y = (1/ln(2))*integral from 1 to 2 of: (1/2)[1/x]^2 dx
y = (1/ln(2))*(1/2)*[-(1/x)] from 1 to 2 = -0,3606737-???

area = integral of f(x)dx
= integral of (dx/x)
= ln(x)] from 1 to 2
= ln(2)-ln(1)
= ln(2) = area

did i do this correctly? please check. thank you so much! :)

To find the center of mass of the region bounded by y = 1/x, y = 0, x = 1, and x = 2, you have correctly set up the formulas for x and y.

The formula for the x-coordinate of the center of mass is given by:

x = (1/A) * integral from a to b of: x * f(x) dx

And the formula for the y-coordinate of the center of mass is given by:

y = (1/A) * integral from a to b of: (1/2) * [f(x)]^2 dx

Let's start by calculating the x-coordinate:

x = (1/ln(2)) * integral from 1 to 2 of: x * [1/x] dx

Simplifying the integrand, we get:

x = (1/ln(2)) * integral from 1 to 2 of: 1 dx

Evaluating the integral, we have:

x = (1/ln(2)) * [x] from 1 to 2

x = (1/ln(2)) * (2-1)

x = 1/ln(2)

So, the x-coordinate of the center of mass is indeed 1/ln(2).

Now let's calculate the y-coordinate:

y = (1/ln(2)) * integral from 1 to 2 of: (1/2) * [1/x]^2 dx

Simplifying the integrand, we get:

y = (1/ln(2)) * integral from 1 to 2 of: 1/(2x^2) dx

Evaluating the integral, we have:

y = (1/ln(2)) * (-(1/(2x))) from 1 to 2

y = (1/ln(2)) * (-(1/(2*2)) + (1/(2*1)))

y = -1/(4ln(2)) + 1/(2ln(2))

Combining the terms, we get:

y = (2-ln(2))/(4ln(2))

So, the y-coordinate of the center of mass is (2-ln(2))/(4ln(2)).

You have correctly calculated the area as ln(2).

Overall, your calculations are correct. The factor of (1/2) in the y-coordinate formula is there because it is part of the formula for the moment of inertia, which is used to calculate the y-coordinate of the center of mass.