The linear density of a $48$ m long rod is $\frac{72}{\sqrt{x+1}} \mathrm{~kg} / \mathrm{m}$ where x is measured in meters from one end of the rod. Find the average density of the rod.
A) $\rho_{\text {ave }}=21 \mathrm{~kg} / \mathrm{m}$
B) $\rho_{\text {ave }}=11 \mathrm{~kg} / \mathrm{m}$
C) $\rho_{\text {ave }}=41 \mathrm{~kg} / \mathrm{m}$
D) $\rho_{\text {ave }}=26 \mathrm{~kg} / \mathrm{m}$
E) $\rho_{\text {ave }}=23 \mathrm{~kg} / \mathrm{m}$

Question

Quizplus · Accepted Answer

The average density is given by:$$ho_{	ext {ave }}=\frac{M}{L}=\frac{\int_{0}^{48} \frac{72}{\sqrt{x+1}} d x}{48}=\frac{144 \sqrt{49}-144 \sqrt{1}}{48}=21 \mathrm{~kg} / \mathrm{m}$$

The Linear Density of A 484848 M Long Rod Is 72x+1 kg/m\frac{72}{\sqrt{x+1}} \mathrm{~kg} / \mathrm{m}x+1​72​ kg/m

The Linear Density of A $48$ M Long Rod Is $\frac{72}{\sqrt{x+1}} \mathrm{~kg} / \mathrm{m}$