A rod of length ‘L’ has non – uniform linear mass density given by $\rho \left(X\right)=\text{a+b}\left(\frac{\text{X}}{\text{L}}\right)$ when a,b are constants and $0\le x\le L;$ the value of ‘x’ for the centre of mass of the rod is at

Position of com is at ${\text{X}}_{\text{cm}}\text{=}\frac{{\displaystyle {\int }_{\text{0}}^{\text{L}}\text{xdm}}}{{\displaystyle {\int }_{\text{0}}^{\text{L}}\text{dm}}}\text{=}\frac{{\displaystyle {\int }_{\text{0}}^{\text{L}}\text{x}\rho \left(\text{x}\right)\text{dx}}}{{\displaystyle {\int }_{\text{0}}^{\text{L}}\rho \left(\text{x}\right)\text{dx}}}\text{=}\frac{{\displaystyle {\int }_{\text{0}}^{\text{L}}\text{x}\left(\text{a+b}\frac{\text{X}}{\text{L}}\right)\text{dx}}}{{\displaystyle {\int }_{\text{0}}^{\text{L}}\left(\text{a+b}\frac{\text{X}}{\text{L}}\right)\text{dx}}}$

$\text{=}\frac{{\displaystyle {\int }_{\text{0}}^{\text{L}}\left(\text{ax+}\frac{{\text{bx}}^{\text{2}}}{\text{L}}\right)\text{dx}}}{{\displaystyle {\int }_{\text{0}}^{\text{L}}\left(\text{a+}\frac{\text{bx}}{\text{L}}\right)\text{dx}}}=\frac{(a\frac{L^2}{2}+\frac{b}{L}.\frac{L^3}{3})}{\text{aL+}\frac{\text{b}}{\text{L}}\text{.}\frac{{\text{L}}^{\text{2}}}{\text{2}}}\text{=}\frac{\frac{{\text{aL}}^{\text{2}}}{\text{2}}\text{+}\frac{{\text{bL}}^{\text{2}}}{\text{3}}}{\text{aL+}\frac{\text{bL}}{\text{2}}}\frac{\text{=}\frac{\text{aL}}{\text{2}}\text{+}\frac{\text{bL}}{\text{3}}}{\text{a+}\frac{\text{b}}{\text{2}}}\text{=}\frac{\text{2}\left(\frac{\text{3aL+2bL}}{\text{6}}\right)}{\left(\text{2a+b}\right)}$