A rod of length L has non-uniform linear mass
density given by $\rho$(x) = $a + b{\left( {{x \over L}} \right)^2}$ , where a
and b are constants and 0 $\le$ x $\le$ L. The value
of x for the centre of mass of the rod is at :

$\rho$ = a + b${\left( {{x \over L}} \right)^2}$ <br><br>dm = $\rho$dx = $\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx$ <br><br>M = $\int {dm}$ = $\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx}$ <br><br>X_com = ${{\int {xdm} } \over {\int {dm} }}$ <br><br>= $${{\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)xdx} } \over {\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx} }}$$ <br><br>= $${{{{a{L^2}} \over 2} + {b \over {{L^2}}}.{{{L^4}} \over 4}} \over {aL + {b \over {{L^2}}}.{{{L^3}} \over 3}}}$$ <br><br>= $${{\left( {{{4a + 2a} \over 8}} \right)L} \over {\left( {{{3a + b} \over 3}} \right)}}$$ <br><br>= ${3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L$