Let the functions f: R → R and g: R → R be defined by f(x) = e ^ x...

f\left(x\right) = {e}^{x - 1} - {e}^{- \left|x - 1\right|} = \left\{$\begin{cases}0, & x \le 1 \\ {e}^{x - 1} - {e}^{1 - x}, & x > 1\end{cases}$\right) and

$g\left(x\right) = \dfrac{1}{2}\left({e}^{x - 1} + {e}^{1 - x}\right)$

At x = 0, f\left(0\right) = 0 & g\left(0\right) = \dfrac{e + {e}^{- 1}}{2}

At x = 1, f\left(1\right) = 0 & g\left(1\right) = \dfrac{1}{2}\left({e}^{1 - 1} + {e}^{1 - 1}\right) = 1

For point of intersection of the two curves for $x > 1$, $f\left(x\right) = g\left(x\right)$

$⇒ {e}^{x - 1} - {e}^{1 - x} = \dfrac{1}{2}\left({e}^{x - 1} + {e}^{1 - x}\right)$

$⇒ 2{e}^{x - 1} - 2{e}^{1 - x} = {e}^{x - 1} + {e}^{1 - x}$

$⇒ {e}^{x - 1} = 3{e}^{1 - x}$

$⇒ {\left({e}^{x - 1}\right)}^{2} = 3$

$⇒ x = 1 + ln\sqrt{3}$

$∴$Required area = Area of the shaded region

$= {∫}_{0}^{1}g\left(x\right)dx + {∫}_{1}^{1 + ln\sqrt{3}}\left\{g\left(x\right) - f\left(x\right)\right\}dx$

$= {∫}_{0}^{1}\dfrac{1}{2}\left({e}^{x - 1} + {e}^{1 - x}\right)dx + {∫}_{1}^{1 + ln\sqrt{3}}\left\{\dfrac{1}{2}\left({e}^{x - 1} + {e}^{1 - x}\right) - \left({e}^{x - 1} - {e}^{1 - x}\right)\right\}dx$

$= \dfrac{1}{2}{∫}_{0}^{1}\left({e}^{x - 1} + {e}^{1 - x}\right)dx + \dfrac{1}{2}{∫}_{1}^{1 + ln\sqrt{3}}\left(3{e}^{1 - x} - {e}^{x - 1}\right)dx$

$= \dfrac{1}{2}{\left[{e}^{x - 1} - {e}^{1 - x}\right]}_{0}^{1} - \dfrac{1}{2}{\left[3{e}^{1 - x} + {e}^{x - 1}\right]}_{1}^{1 + ln\sqrt{3}}$

$= \dfrac{1}{2}\left(e - {e}^{- 1}\right) - \dfrac{1}{2}\left[2\sqrt{3} - 4\right]$

$= \dfrac{e - {e}^{- 1}}{2} + \left(2 - \sqrt{3}\right)$