#### Answer

$$2\sinh(1)$$

#### Work Step by Step

Given $$y=e^{x / 2}+e^{-x / 2},\ \ \ \ [0,2] $$
Since
$$y^{\prime}=\frac{e^{x / 2}}{2}-\frac{e^{-x / 2}}{2} $$
Then the arc length is given by
\begin{aligned}
s &=\int_{a}^{b} \sqrt{1+y'^{2}} d x \\
&= \int_{0}^{2} \sqrt{1+\left(e^{x / 2}+e^{-x / 2}\right)^2} d x \\
&= \int_{0}^{2} \sqrt{1+\frac{e^{x}}{4}-\frac{1}{2}+\frac{e^{-x}}{4}} d x \\
&= \int_{0}^{2} \sqrt{\frac{e^{x}}{4}+\frac{1}{2}+\frac{e^{-x}}{4}} d x \\
&=\int_{0}^{2} \sqrt{\left(\frac{e^{x / 2}}{2}+\frac{e^{-x / 2}}{2}\right)^{2}} d x \\
&=\int_{0}^{2}\left(\frac{e^{x / 2}}{2}+\frac{e^{-x / 2}}{2}\right) d x\\
&= \int_{0}^{2} \cosh \left(\frac{x}{2}\right) d x\\
&= 2 \sinh \left(\frac{x}{2}\right)\\
&= \left[2 \sinh \left(\frac{x}{2}\right)\right]_{0}^{2}\\
&=2\sinh(1)
\end{aligned}