Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 13

For an irregularly shaped planar lamina of uniform density (rho) , bounded by graphs y=f(x),y=g(x) and a<=x<=b ,the mass (m) of this region is given by,
m=rhoint_a^b[f(x)-g(x)]dx
m=rhoA ,where A is the area of the region.
The moments about the x and y-axes are given by the formula,
M_x=rhoint_a^b1/2([f(x)]^2-[g(x)]^2)dx
M_y=rhoint_a^bx(f(x)-g(x))dx
The coordinates of the center of mass (barx,bary) are given by,
barx=M_y/m
bary=M_x/m
We are given y=1/2x,y=0,x=2
The attached image shows the region bounded by the functions and the limits of integration,
Let's evaluate the area of the region,
A=int_0^2x/2dx
Evaluate the integral by applying power rule,
A=[1/2(x^2/2)]_0^2
A=1/4[2^2-0^2]
A=1/4(4)
A=1
Now let's evaluate the moments about the x and y-axes,
M_x=rhoint_0^2 1/2(x/2)^2dx
M_x=rhoint_0^2 1/2(x^2/4)dx
M_x=rho/8int_0^2x^2dx
Apply the power rule,
M_x=rho/8[x^3/3]_0^2
M_x=rho/8[1/3(2)^3]
M_x=rho/8(8/3)
M_x=rho/3
M_y=rhoint_0^2x(x/2)dx
M_y=rho/2int_0^2x^2dx
Apply power rule,
M_y=rho/2[x^3/3]_0^2
M_y=rho/2[1/3(2)^3]
M_y=rho/2[8/3]
M_y=4/3rho
Now let's find the coordinates of the center of mass,
barx=M_y/m=M_y/(rhoA)
barx=(4/3rho)/(rho(1))
barx=4/3
bary=M_x/m=M_x/(rhoA)
bary=(rho/3)/(rho(1))
bary=1/3
The coordinates of the center of mass are (4/3,1/3)