The distance of centre of mass from end A of a one dimensional rod...

Given here:

Mass density, $ρ = {ρ}_{0}\left(1 - \dfrac{{x}^{2}}{{L}^{2}}\right)kg{m}^{- 1}$

Let us suppose there is an element of length $d x$ at a distance $x$ from left must end.

Mass of this element is $d m = ρ d x = {ρ}_{0}\left(1 - \dfrac{{x}^{2}}{{L}^{2}}\right)$

The centre of mass of this element is ${X}_{cm} = \dfrac{∫ xdm}{∫ dm}$

$= \dfrac{{ρ}_{0}{∫}_{0}^{l}\left(x - \dfrac{{x}^{2}}{{L}^{2}}\right)x d x}{{∫}_{0}^{l}{ρ}_{0}\left(1 - \dfrac{{x}^{2}}{{L}^{2}}\right) d x}$

$= \dfrac{\dfrac{{L}^{2}}{2} - \dfrac{{L}^{4}}{4{L}^{2}}}{L - \dfrac{{L}^{3}}{3{L}^{2}}} = \dfrac{\dfrac{{L}^{2}}{4}}{\dfrac{2L}{3}} = \dfrac{3L}{8}$

Hence, the value of $\alpha = 8$.