Answer:
[tex]I = \frac{\lambda_o L^3}{12}[/tex]
Explanation:
Rotational inertia of the rod about its one end is given as
[tex]I = \int dm x^2[/tex]
here we know that
[tex]dm = \lambda dx[/tex]
so we will have
[tex]I = \int (\lambda dx) x^2[/tex]
[tex]I = \int \lambda_o(1 - x/L) x^2 dx[/tex]
so we have
[tex]I = \lambda_o(\frac{L^3}{3} - \frac{L^3}{4})[/tex]
[tex]I = \frac{\lambda_o L^3}{12}[/tex]