Disk Integration to find Moment of Inertia?

Your answer is correct beyond doubt.

Consider a thin ring of width dx at a distance x from the center of the ring, r_i < x < r_o.

The moment of inertia of this thin ring is

dI = (2πx*t*ρ)*x^2 dx, where, t = thickness and ρ = density

Integrating,

Moment of inertia, I

= (2π*t*ρ)* [x^4/4] (x = r_i to r_o)

= 0.5πρt(r_o^4 - r_i^4).

If you need the answer in terms of mass, m

m = πρt(r_o^2 - r_i^2).

=> moment of inertia

I = 0.5 m (r_o^2 + r_i^2).

Alternate method:

Using the formula for moment of inertia of the disc,

I = (1/2)m1*r_0^2 - (1/2)m2*r_i^2

where

m1 = mass of the disc having radius r_o = 0.5πρt*r_o^2

and

m2 = mass of the disc having radius r_i = 0.5πρt*r_i^2

=> I = 0.5 m (r_o^2 + r_i^2), same as what you had got.

This is then coverted in terms of mass of the annular ring, m as

I = 0.5 m (r_o^2 + r_i^2).

Moment Of Inertia By Integration

Suppose you had thre masses "m" located at r1=1 ,r2 = 2 & r3 = 3. The moment of inertia is; I = m(1) + m(4) + m(9) = 14m Now suppose the same three masses were all located at r = 3. The moment of inertia would be; I = (3m)(9) = 27m As you can see the top one is approximately half the bottom one. Its just how the summation works out when the mass is distributed between 0 & R as compared to all "bunched" up at R.