Answer
$\frac{e^{2}-1}{2e}$
Work Step by Step
$\frac{dy}{dx}$ = $\frac{1}{2}(e^{x}-e^{-x})$
$L$ = $\int_{{\,0}}^{{\,1}}$$\sqrt {1+(\frac{1}{2}(e^{x}-e^{-x}))^{2}}$ $dx$
$L$ = $\int_{{\,0}}^{{\,1}}$$\sqrt {1+\frac{1}{4}e^{2x}-\frac{1}{2}+\frac{1}{4}e^{-2x}))}$ $dx$
$L$ = $\int_{{\,0}}^{{\,1}}$$\sqrt {\frac{1}{4}e^{2x}+\frac{1}{2}+\frac{1}{4}e^{-2x}))}$ $dx$
$L$ = $\int_{{\,0}}^{{\,1}}$$\sqrt {(\frac{1}{2}(e^{x}+e^{-x}))^{2}}$ $dx$
$L$ = $\int_{{\,0}}^{{\,1}}$${\frac{1}{2}(e^{x}+e^{-x})}$ $dx$
$L$ = ${\frac{1}{2}(e^{x}-e^{-x})}$$|_{{\,0}}^{{\,1}}$
$L$ = ${\frac{1}{2}[(e-e^{-1})-(1-1)]}$
$L$ = $\frac{1}{2}(e-\frac{1}{e})$
$L$ = $\frac{e^{2}-1}{2e}$
