#### Answer

$$\frac{525}{32}$$

#### Work Step by Step

Given
$$ y= y=\left(\frac{x}{4} \right)^{3}+\frac{1}{2x^2}$$
Since \begin{aligned} 1+\left(y^{\prime}\right)^{2} &=1+\left(\frac{x^3}{4} -\frac{1}{x^3}\right)^{2} \\ &=1+\frac{x^{6}}{16}-\frac{1}{2}+x^{-6} \\ &=\frac{x^{6}}{16}+\frac{1}{2}+x^{-6} \\ &=\left(\frac{1}{4} x^{3}+x^{-3}\right)^{2} \end{aligned}
Then the arc length given by
\begin{aligned} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\ &=\int_{1}^{4} \sqrt{\left(\frac{1}{4} x^{3}+x^{-3}\right)^{2}} d x \\ &=\int_{1}^{4}\left(\frac{1}{4} x^{3}+x^{-3}\right) d x \\
&=\frac{1}{16} x^{4}-\frac{1}{2} x^{-2}\bigg|_{1}^{4}\\
&=\frac{525}{32}
\end{aligned}