Explanation

(b) Let rod is placed along $x$-axis. Mass of element $PQ$ of length $dx$ situated at $x = x$ is
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$dm = \lambda\ dx = (2 + x)\ dx$
The CM of the element has coordinates ($x$, 0, 0).
Therefore, $x$-coordinates of CM of the rod will be
$x_{CM} = \frac{\int_{0}^{3}{xdm}}{\int_{0}^{3}{dm}}$
$= \frac{\int_{0}^{3}{x(2 + x)dx}}{\int_{0}^{3}{(2 + x)dx}}$
$= \frac{\int_{0}^{3}{\left( 2x + x^{2} \right)dx}}{\int_{0}^{3}{(2 + x)dx}}$
$= \frac{\left\lbrack \frac{2x^{2}}{2} + \frac{x^{3}}{3} \right\rbrack_{0}^{3}}{\left\lbrack 2x + \frac{x^{2}}{2} \right\rbrack_{0}^{3}}$
$= \frac{\left\lbrack (3)^{2} + \frac{(3)^{3}}{3} \right\rbrack}{\left\lbrack 2 \times 3 + \frac{(3)^{2}}{2} \right\rbrack} = \frac{9 + 9}{6 + \frac{9}{2}}$
$= \frac{18 \times 2}{21} = \frac{12}{7}$m