The linear mass density of a thin rod AB of length L varies from A to B as $\lambda \left(x\right)={\lambda }_0\left(1+\frac{x}{L}\right),$ where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :

Mass of the small element of the rod
$dm=\lambda \cdot dx$
Moment of inertia of small element,
$dI=dm\cdot x^2={\lambda }_0\left(1+\frac{x}{L}\right)\cdot x^{2}dx$
Moment of inertia of the complete rod can be obtained by integration 
$\begin{array}{l}I={\lambda }_0{\int }_0^L \left(x^2+\frac{x^3}{L}\right)dx={\lambda }_0{\left|\frac{x^3}{3}+\frac{x^4}{4L}\right|}_0^L={\lambda }_0\left[\frac{L^3}{3}+\frac{L^3}{4}\right]\\ \Rightarrow I=\frac{7{\lambda }_0{L}^3}{12}\end{array}$
Mass of the thin rod ,
$\begin{array}{l}M={\int }_0^L \lambda dx={\int }_0^L {\lambda }_0\left(1+\frac{x}{L}\right)dx=\frac{3{\lambda }_{0}L}{2}\therefore {\lambda }_0=\frac{2M}{3L}\\ \therefore I=\frac{7}{12}\left(\frac{2M}{3L}\right)L^3\Rightarrow I=\frac{7}{18}M{L}^2\end{array}$