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)`