#### Answer

$$\frac{8}{27}\left[10 \sqrt{10}-\frac{13 \sqrt{13}}{8}\right]$$

#### Work Step by Step

Given
$$y= x^{2/3} ,\ \ \ \ \ \ \ [1,8]$$
Since
\begin{align*}
y'&= \frac{2}{3}x^{-1/3}\\
1+y'^2&= 1+ \frac{4}{9}x^{-2/3}
\end{align*}
Then the arc length is given by
\begin{aligned}
\text{Arc Length }&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x \\
&=\int_{1}^{8} \sqrt{1+\frac{4 x^{-2 / 3}}{9}} d x \\
&=\int_{1}^{8} \sqrt{\frac{4 x^{-2 / 3}}{9}\left(1+\frac{9 x^{2 / 3}}{4}\right)} d x \\
&=\int_{1}^{8} \sqrt{\frac{4 x^{-2 / 3}}{9}} \cdot \sqrt{\left(1+\frac{9 x^{2 / 3}}{4}\right)} d x \\
&=\int_{1}^{8} \frac{2 x^{-1 / 3}}{3} \sqrt{1+\frac{9 x^{2 / 3}}{4}} d x \\
&=\frac{8}{27}\left(\frac{9 x^{2 / 3}}{4}+1\right)^{3 / 2}\bigg|_{1}^{8}\\
&= \frac{8}{27}\left[10 \sqrt{10}-\frac{13 \sqrt{13}}{8}\right]
\end{aligned}