Answer
$$ 2.28$$
Work Step by Step
Since
\begin{aligned}
s&=\int_{a}^{b} \sqrt{1+\left[f'(x)\right]^{2}} d x \\ \\
&=\int_{0}^{2} \sqrt{1+\left[-2 x e^{-x^{2}}\right]^{2}} d x \\
&=\int_{0}^{2} \sqrt{1+4 x^{2} e^{-2 x^{2}}} d x
\end{aligned}
and
$$\Delta x= \frac{2-}{8}=\frac{1}{4} $$
Then
\begin{aligned}
S_{8}=& \frac{1}{3} \Delta x[f(a)+4 f(a+\Delta x)+2 f(a+2 \Delta x)+4 f(a+3 \Delta x)+2 f(a+4 \Delta x)+4 f(a+5 \Delta x)\\
&+2 f(a+6 \Delta x)+4 f(a+7 \Delta x)+f(b)] \\
=& \frac{1}{3} \cdot \frac{1}{4}[f(0)+4 f(1 / 4)+2 f(2 / 4)+4 f(3 / 4)+2 f(4 / 4)+4 f(5 / 4)+2 f(6 / 4)+4 f(6 / 4)+f(2)] \\
=& \frac{1}{12}[\sqrt{1+0}+4 \sqrt{1+4\left(\frac{1}{4}\right)^{2} e^{-2(1 / 4)^{2}}}+2 \sqrt{1+4\left(\frac{2}{4}\right)^{2} e^{-2(2 / 1)^{2}}}+4 \sqrt{1+4\left(\frac{3}{4}\right)^{2} e^{-2(3 / 4)^{2}}}+2 \sqrt{1+4 e^{-2}}\\
&+4 \sqrt{1+4\left(\frac{5}{4}\right)^{2} e^{-2(5 / 4)^{2}}}+2 \sqrt{1+4\left(\frac{6}{4}\right)^{2} e^{-2(6 / 4)^{2}}}+4 \sqrt{1+4\left(\frac{7}{4}\right)^{2} e^{-2(7 / 4)^{2}}}+\sqrt{1+4(2)^{2} e^{-2(2)^{2}}}]\\
&\approx 2.28
\end{aligned}
