Let the functions f: R rightarrow R and g: R | vedclass

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(A) Given $f(x) = e^{x-1} - e^{-|x-1|}$ and $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$.For $x \leq 1$,$|x-1| = 1-x$,so $f(x) = e^{x-1} - e^{x-1} = 0$.For

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

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(A) Given $f(x) = e^{x-1} - e^{-|x-1|}$ and $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$.For $x \leq 1$,$|x-1| = 1-x$,so $f(x) = e^{x-1} - e^{x-1} = 0$.For $x \geq 1$,$|x-1| = x-1$,so $f(x) = e^{x-1} - e^{1-x}$.To find the intersection point of $f(x)$ and $g(x)$ for $x \geq 1$:$e^{x-1} - e^{1-x} = \frac{1}{2}(e^{x-1} + e^{1-x})$$2e^{x-1} - 2e^{1-x} = e^{x-1} + e^{1-x}$$e^{x-1} = 3e^{1-x} \Rightarrow e^{2(x-1)} = 3 \Rightarrow 2(x-1) = \ln 3 \Rightarrow x = 1 + \frac{1}{2}\ln 3 = 1 + \ln \sqrt{3}$.The area $A$ in the first quadrant bounded by $y=f(x)$,$y=g(x)$,and $x=0$ is:$A = \int_0^1 (g(x) - 0) dx + \int_1^{1+\ln \sqrt{3}} (g(x) - f(x)) dx$$A = \int_0^1 \frac{1}{2}(e^{x-1} + e^{1-x}) dx + \int_1^{1+\ln \sqrt{3}} (\frac{1}{2}(e^{x-1} + e^{1-x}) - (e^{x-1} - e^{1-x})) dx$$A = \frac{1}{2}[e^{x-1} - e^{1-x}]_0^1 + \int_1^{1+\ln \sqrt{3}} (\frac{3}{2}e^{1-x} - \frac{1}{2}e^{x-1}) dx$$A = \frac{1}{2}[(e^0 - e^0) - (e^{-1} - e^1)] + [-\frac{3}{2}e^{1-x} - \frac{1}{2}e^{x-1}]_1^{1+\ln \sqrt{3}}$$A = \frac{1}{2}(e - e^{-1}) + [(-\frac{3}{2}e^{-\ln \sqrt{3}} - \frac{1}{2}e^{\ln \sqrt{3}}) - (-\frac{3}{2} - \frac{1}{2})]$$A = \frac{1}{2}(e - e^{-1}) + [-\frac{3}{2}(\frac{1}{\sqrt{3}}) - \frac{1}{2}(\sqrt{3}) + 2]$$A = \frac{1}{2}(e - e^{-1}) + [-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + 2] = 2 - \sqrt{3} + \frac{1}{2}(e - e^{-1})$.

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

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