The diagram shows part of the curve $y = e^{x-1} - 2$. The point $P(2,\ e-2)$ lies on the curve and the normal to the curve at $P$ meets the x-axis at $Q$.

1. Differentiate the equation of the curve to find the gradient function

$y = e^{x-1} - 2$
$\frac{dy}{dx} = e^{x-1}$

2. Find the gradient of the tangent at $P(2, e-2)$

At $x = 2$, $\frac{dy}{dx} = e^{2-1} = e$

3. Find the gradient of the normal at $P$

Gradient of normal $m_n = -\frac{1}{m_t} = -\frac{1}{e}$

4. Form the equation of the normal at $P$

$y - (e-2) = -\frac{1}{e}(x - 2)$

5. Find the x-coordinate of $Q$ by setting $y = 0$

At $Q$, $y = 0$:
$0 - (e-2) = -\frac{1}{e}(x - 2)$
$-(e-2) = -\frac{1}{e}(x - 2)$
$e(e-2) = x - 2$
$x = 2 + e(e-2)$

1. Find the x-intercept of the curve to determine the limits of integration.

At the x-axis, $y = 0$:
$e^{x-1} - 2 = 0$
$e^{x-1} = 2$
$x - 1 = \ln 2$
$x = 1 + \ln 2$

2. Set up the definite integral for the area of the shaded region.

The shaded region is bounded by the curve, the x-axis, and the y-axis. Since the curve is below the x-axis for $0 \leqslant x \leqslant 1 + \ln 2$, the area is:
$\text{Area} = -\int_{0}^{1+\ln 2} (e^{x-1} - 2) \, dx$

3. Integrate the function.

$\int (e^{x-1} - 2) \, dx = e^{x-1} - 2x$

4. Evaluate the definite integral using the limits.

$\text{Area} = - \left[ e^{x-1} - 2x \right]_{0}^{1+\ln 2}$
$= - \left( (e^{1+\ln 2 - 1} - 2(1+\ln 2)) - (e^{-1} - 0) \right)$
$= - \left( e^{\ln 2} - 2 - 2\ln 2 - \frac{1}{e} \right)$
$= - \left( 2 - 2 - 2\ln 2 - \frac{1}{e} \right)$
$= 2\ln 2 + \frac{1}{e}$