Answer
$$2.29896$$
Work Step by Step
Since
\begin{aligned}
s &=\int_{a}^{b}\sqrt{1+f'^2}dx\\
&=\int_{1}^{3} \sqrt{1+\left[\frac{1}{x}\right]^{2}} d x \\
&=\int_{1}^{3} \sqrt{1+\frac{1}{x^{2}}} d x
\end{aligned}
Since
$$ \Delta x= \frac{3-1}{6}=\frac{1}{3}$$
Then
\begin{aligned}
M_{6}=& \frac{1}{3}(f(7 / 6)+f(3 / 2)+f(11 / 6)+f(13 / 6)+f(5 / 2)+f(17 / 6)) \\
=& \frac{1}{3}[\sqrt{1+\frac{1}{(7 / 6)^{2}}}+\sqrt{1+\frac{1}{(3 / 2)^{2}}}+\sqrt{1+\frac{1}{(11 / 6)^{2}}}+\sqrt{1+\frac{1}{(13 / 6)^{2}}}+\sqrt{1+\frac{1}{(5 / 2)^{2}}}\\
&+\sqrt{1+\frac{1}{(17 / 6)^{2}}}] \\
& \approx 2.29896
\end{aligned}
