the linear density $\rho$ in a rod 7 m long is $\frac{15}{sqrt{x + 9}}$ kg/m, where x is measured in meters…

Explanation:

Step1: Recall the formula for average value

The formula for the average value of a function (y = f(x)) over the interval ([a,b]) is (\rho_{ave}=\frac{1}{b - a}\int_{a}^{b}f(x)dx). Here, (a = 0), (b=7), and (f(x)=\frac{15}{\sqrt{x + 9}}), so (\rho_{ave}=\frac{1}{7-0}\int_{0}^{7}\frac{15}{\sqrt{x + 9}}dx=\frac{15}{7}\int_{0}^{7}(x + 9)^{-\frac{1}{2}}dx).

Step2: Integrate the function

Use the power - rule for integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). Let (u=x + 9), then (du=dx). When (x = 0), (u = 9); when (x = 7), (u=16). So (\int(x + 9)^{-\frac{1}{2}}dx=\int u^{-\frac{1}{2}}du). By the power rule, (\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C = 2u^{\frac{1}{2}}+C=2\sqrt{u}+C). Then (\frac{15}{7}\int_{0}^{7}(x + 9)^{-\frac{1}{2}}dx=\frac{15}{7}\left[2\sqrt{x + 9}\right]_{0}^{7}).

Step3: Evaluate the definite integral

(\frac{15}{7}\left(2\sqrt{7 + 9}-2\sqrt{0 + 9}\right)=\frac{15}{7}(2\times4-2\times3)).

Step4: Simplify the expression

(\frac{15}{7}(8 - 6)=\frac{15\times2}{7}=\frac{30}{7}\approx4.29).

Answer:

(\frac{30}{7}\text{ kg/m})