Goal: evaluate the moment of inertia integral I = ∫ r^{2} dm
For a uniform rod of length L rotating about an axis passing perpendicularly through one end of the rod, align the rod along the xaxis. Split the rod into pieces of size dx.
The mass of each piece is: dm = λ dx, where λ = M/L is the mass per unit length of the rod.
The integral becomes: I = ∫_{0}^{L} x^{2} λ dx = λ x^{3}/3 _{0}^{L} = λ L^{3}/3
Substituting M = λL
gives:
I  = 

ML^{2}  for a uniform rod rotating about one end. 