Let the functions $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ be defined by $f(x) = e^{x-1} - e^{-|x-1|}$ and $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$. Then the area of the region in the first quadrant bounded by the curves $y = f(x)$, $y = g(x)$ and $x = 0$ is

$f(x) = e^{x-1} - e^{-|x-1|},\qquad g(x) = \frac{1}{2}\left(e^{x-1} + e^{1-x}\right)$ $\text{Region in first quadrant bounded by } y=f(x),\ y=g(x),\ x=0$ $\text{To find: area of the bounded region}$ $\text{Plan: write } f \text{ piecewise; note } g(x) = \cosh(x-1)$ $\text{find intersection of } f, g; \text{ split area into two integrals}$ $f(x) = \begin{cases} 0, & x \leq 1 \\ e^{x-1} - e^{1-x} = 2\sinh(x-1), & x \geq 1 \end{cases}$ $g(x) = \dfrac{e^{x-1}+e^{1-x}}{2} = \cosh(x-1)$ $u = x-1:\qquad f = 2\sinh u,\qquad g = \cosh u$ $2\sinh u = \cosh u \Rightarrow e^{2u} = 3 \Rightarrow u = \tfrac{1}{2}\ln 3$ $x^{*} = 1 + \tfrac{1}{2}\ln 3$ $A = \int_{0}^{1} g(x)\,dx + \int_{1}^{x^{*}} \bigl(g(x)-f(x)\bigr)\,dx$ $\int_{0}^{1} g(x)\,dx = \bigl[\sinh(x-1)\bigr]_{0}^{1} = \sinh 1 = \dfrac{1}{2}(e-e^{-1})$ $g-f = \cosh u - 2\sinh u = -\tfrac{1}{2}e^{u} + \tfrac{3}{2}e^{-u}$ $\int_{0}^{\frac{1}{2}\ln 3}\!\left(-\tfrac{1}{2}e^{u}+\tfrac{3}{2}e^{-u}\right)du = [-\tfrac{1}{2}e^{u}-\tfrac{3}{2}e^{-u}]_{0}^{\frac{1}{2}\ln 3} = -\sqrt{3}-(-2) = 2-\sqrt{3}$ $A = (2-\sqrt{3}) + \dfrac{1}{2}(e-e^{-1})$ $\boxed{\text{option (1)}}$