## Essential University Physics: Volume 1 (4th Edition) Clone

$\frac{2L}{3}$
We know: $\frac{dm}{M} = \frac{2xdx}{L^2}$ We obtain the following integral from this: $x_{cm} = \frac{1}{M} \int xdm = \frac{2}{L^2} \int_0^Lx^2dx =\frac{2L}{3}$