Area Bounded by Exponential and Absolute Functions in First Quadrant

To find the area of the region lying in the first quadrant enclosed by the curves $y = f(x)$, $y = g(x)$, and the line $x = 0$, we first simplify the definition of the function $f(x)$.

The given functions are: $f(x) = e^{x-1} - e^{-|x-1|}$ $g(x) = \frac{1}{2}\left(e^{x-1} + e^{1-x}\right)$

Step 1: Analyze $f(x)$ for different intervals of $x$

1. For $x \ge 1$: $|x - 1| = x - 1$ $f(x) = e^{x-1} - e^{-(x-1)} = e^{x-1} - e^{1-x}$

2. For $x < 1$: $|x - 1| = -(x - 1) = 1 - x$ $f(x) = e^{x-1} - e^{-(1-x)} = e^{x-1} - e^{x-1} = 0$

Thus, the piecewise definition of $f(x)$ is:

$$f(x) = \begin{cases} 0, & \text{if } x < 1 \\ e^{x-1} - e^{1-x}, & \text{if } x \ge 1 \end{cases}$$

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Step 2: Find the point of intersection of $y = f(x)$ and $y = g(x)$

Since $g(x) > 0$ for all real $x$ and $f(x) = 0$ for $x < 1$, the curves can only intersect in the region $x \ge 1$.

Setting $f(x) = g(x)$ for $x \ge 1$: $e^{x-1} - e^{1-x} = \frac{1}{2}\left(e^{x-1} + e^{1-x}\right)$

Let $t = e^{x-1} \ge 1$. The equation becomes: $t - \frac{1}{t} = \frac{1}{2}\left(t + \frac{1}{t}\right)$

Multiplying both sides by $2t$: $2t^2 - 2 = t^2 + 1$ $t^2 = 3 \implies t = \sqrt{3} \quad (\text{since } t \ge 1)$

Now, solving for $x$: $e^{x-1} = \sqrt{3} \implies x - 1 = \ln \sqrt{3} \implies x = 1 + \ln \sqrt{3}$

Let $x_0 = 1 + \ln \sqrt{3}$.

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Step 3: Calculate the area of the bounded region

The region is bounded on the left by $x = 0$ and on the right by the intersection point $x = x_0 = 1 + \ln \sqrt{3}$. Since $g(x) \ge f(x)$ on the interval $[0, x_0]$, the area $A$ is given by: $A = \int_{0}^{x_0} [g(x) - f(x)] \, dx$

We split this integral at $x = 1$:

$A = \int_{0}^{1} [g(x) - f(x)] \, dx + \int_{1}^{x_0} [g(x) - f(x)] \, dx$

Part 1: Integral from $x = 0$ to $x = 1$

Here, $f(x) = 0$, so: $A_1 = \int_{0}^{1} \frac{1}{2}\left(e^{x-1} + e^{1-x}\right) dx$ $A_1 = \frac{1}{2} \left[ e^{x-1} - e^{1-x} \right]_{0}^{1}$ $A_1 = \frac{1}{2} \left[ (e^0 - e^0) - (e^{-1} - e^1) \right]$ $A_1 = \frac{1}{2} (e - e^{-1})$

Part 2: Integral from $x = 1$ to $x = 1 + \ln \sqrt{3}$

Here, $g(x) - f(x) = \frac{1}{2}(e^{x-1} + e^{1-x}) - (e^{x-1} - e^{1-x}) = \frac{3}{2} e^{1-x} - \frac{1}{2} e^{x-1}$.

$A_2 = \int_{1}^{1 + \ln \sqrt{3}} \left( \frac{3}{2} e^{1-x} - \frac{1}{2} e^{x-1} \right) dx$ $A_2 = \left[ -\frac{3}{2} e^{1-x} - \frac{1}{2} e^{x-1} \right]_{1}^{1 + \ln \sqrt{3}}$

Evaluating at the upper limit $x = 1 + \ln \sqrt{3}$: $-\frac{3}{2} e^{-\ln \sqrt{3}} - \frac{1}{2} e^{\ln \sqrt{3}} = -\frac{3}{2\sqrt{3}} - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3}$

Evaluating at the lower limit $x = 1$: $-\frac{3}{2} e^0 - \frac{1}{2} e^0 = -\frac{3}{2} - \frac{1}{2} = -2$

Thus, $A_2 = (-\sqrt{3}) - (-2) = 2 - \sqrt{3}$

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Step 4: Total Area

Adding $A_1$ and $A_2$: $A = A_1 + A_2 = (2 - \sqrt{3}) + \frac{1}{2}(e - e^{-1})$

Hence, the correct option is (A).