#### Answer

$$2.5065$$

#### Work Step by Step

Since
\begin{aligned}
s &=\int_{a}^{b} \sqrt{1+f'^{2}(x)} d x \\\\
&=\int_{0}^{2} \sqrt{1+[-\sin x]^{2}} d x \\
&=\int_{0}^{2} \sqrt{1+\sin ^{2} x} d x
\end{aligned}
Since
$$ \Delta x= \frac{2-0}{8}=\frac{1}{4}$$
Then
\begin{aligned}
T_{8} &=\frac{1}{2} \cdot \frac{1}{4}\left[f(0)+2 \sum_{j=1}^{7} f\left(0+\frac{1}{4} \cdot j\right)+f(2)\right] \\
&=\frac{1}{8}[\sqrt{1+\sin ^{2} 0}+2(\sqrt{1+\sin ^{2} \frac{1}{4}}+\sqrt{1+\sin ^{2} 2 \cdot \frac{1}{4}}+\sqrt{1+\sin ^{2} 3 \cdot \frac{1}{4}}+\sqrt{1+\sin ^{2} 4 \cdot \frac{1}{4}}+\sqrt{1+\sin ^{2} 5 \cdot \frac{1}{4}}\\
&+\sqrt{1+\sin ^{2} 6 \cdot \frac{1}{4}}+\sqrt{1+\sin ^{2} 7 \cdot \frac{1}{4}})+\sqrt{1+\sin ^{2} 2}] \\
& \approx 2.5065
\end{aligned}