#### Answer

$$4\sqrt{17}$$

#### Work Step by Step

Given $$y=4 x-2, \quad[-2,2] $$
Since
$$y' = 4 $$
Then the arc length is given by
\begin{aligned}
s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x \\
&=\int_{-2}^{2} \sqrt{1+16} d x \\
&=\int_{-2}^{2} \sqrt{17} d x\\
&=\sqrt{17}x\bigg|_{-2}^{2}\\
&= 4\sqrt{17}
\end{aligned}