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(B) The linear mass density is given by $\lambda = \frac{dm}{dx}$,so $dm = \lambda \, dx$.The position of the center of mass $x_{cm}$ is given by:$x_{

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$\frac{12}{7} \ m$

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(B) The linear mass density is given by $\lambda = \frac{dm}{dx}$,so $dm = \lambda \, dx$.The position of the center of mass $x_{cm}$ is given by:$x_{cm} = \frac{\int x \, dm}{\int dm}$Substituting $\lambda = 2 + x$ and the limits from $0$ to $3$:$x_{cm} = \frac{\int_{0}^{3} x(2 + x) \, dx}{\int_{0}^{3} (2 + x) \, dx}$$x_{cm} = \frac{\int_{0}^{3} (2x + x^2) \, dx}{\int_{0}^{3} (2 + x) \, dx}$Evaluating the integrals:Numerator: $[x^2 + \frac{x^3}{3}]_{0}^{3} = (3^2 + \frac{3^3}{3}) = 9 + 9 = 18$Denominator: $[2x + \frac{x^2}{2}]_{0}^{3} = (2(3) + \frac{3^2}{2}) = 6 + 4.5 = 10.5 = \frac{21}{2}$$x_{cm} = \frac{18}{21/2} = \frac{36}{21} = \frac{12}{7} \ m$

The linear mass density of a $3 \ m$ long rod varies as $\lambda = 2 + x$. The position of the center of mass of the rod is at:

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