Answer:
The moment of inertia is   [tex]I= \frac{1}{3} ML^2[/tex]
Explanation:
From the question we a given an equation
            [tex]I = \int\limits{r^2} \, dm[/tex]
Now let look at a small length da of this uniform rod that a meters form the x-axis
 The mass of this small section can be mathematically evaluated as
            [tex]dM = da *\frac{M}{L}[/tex]
For this small portion for the rod the moment of inertia is
          [tex]dI = (dM)a^2[/tex]
Substituting for dM
         [tex]dI = da [\frac{M}{L} ] a^2[/tex]
to get I we integrate both sides
         [tex]I = \int\limits^L_0 {\frac{M}{L} * a^2} \, da[/tex]
          [tex]= [\frac{M}{L} ][\frac{a^3}{3} ]\left L} \atop {0}} \right.[/tex]
          [tex]I= \frac{1}{3} ML^2[/tex]
