Answer
$3.9577$
Work Step by Step
We have
\begin{aligned}
s=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x &=\int_{1}^{2} \sqrt{1+\left(x^{3}\right)^{2}} d x \\
&=\int_{1}^{2} \sqrt{1+x^{6}} d x
\end{aligned}
Use the Trapezoidal Rule, $T_{5},$ to approximate the arc length,
$
T_{5}=\frac{1}{2} \Delta x\left[y_{0}+2 y_{1}+2 y_{2}+2 y_{3}+2 y_{4}+y_{5}\right], \quad y_{j}=f\left(x_{j}\right)
$
where,
$
\Delta x=\frac{b-a}{N}=\frac{2-1}{5}=\frac{1}{5}
$
and,
$
\left[x_{j}\right]_{j=0}^{5}=\left[1, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, 2\right]
$
Then
\begin{aligned}
s &=\int_{1}^{2} \sqrt{1+x^{6}} d x \\
&=\frac{1}{10}[\sqrt{2}+2 \sqrt{1+\left(\frac{6}{5}\right)^{6}}+2 \sqrt{1+\left(\frac{7}{5}\right)^{6}}+2 \sqrt{1+\left(\frac{8}{5}\right)^{6}}+2 \sqrt{1+\left(\frac{9}{5}\right)^{6}}+\sqrt{1+(2)^{6}}]\\
&\approx 3.9577
\end{aligned}
