#### Answer

$$1.91009 $$

#### Work Step by Step

We have
\begin{align*}
s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\
&=\int_{0}^{\frac{\pi}{2}} \sqrt{1+\cos ^{2}(x)} d x
\end{align*}
Use the midpoint rule with $n=8$ and $f(x)=\sqrt{1+\cos ^{2}(x)}$ to approximate the arc length,
$
\Delta x=\frac{b-a}{n}=\frac{\left(\frac{\pi}{2}-0\right)}{8}=\frac{\pi}{16}
$
Then
\begin{aligned}
s&=\int_{0}^{\frac{\pi}{2}} \sqrt{1+\cos ^{2}(x)} d x\\
&=\Delta x \sum_{i=a}^{8} f\left(x_{i}^{*}\right) \\
& \approx \frac{\pi}{16}[1.411+1.384+1.333+1.264+1.184+1.106+1.041+1.005] \\
& \approx \frac{\pi}{16}[9.728] \\
& \approx 1.91009
\end{aligned}