Figure 2 shows a sketch of part of the curve C with parametric equations $x = 1 - \frac{1}{2}t$, $y = 2t - 1$ The curve crosses the y-axis at the point A and crosses the x-axis at the point B - Edexcel - A-Level Maths Pure - Question 17 - 2013 - Paper 1

The curve crosses the x-axis at point B, where $y = 0$:

\Rightarrow 2t = 1 \\ t = \frac{1}{2}.$$ Substituting $t = \frac{1}{2}$ back into the equation for $x$: $$x = 1 - \frac{1}{2} \left( \frac{1}{2} \right) = 1 - \frac{1}{4} = \frac{3}{4}.$$ Thus, the x-coordinate of point B is $rac{3}{4}$.

To find the equation of the normal line at point A, we first need to calculate the gradient of the curve at point A. This requires finding $\frac{dy}{dx}$. Using parametric differentiation:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$

Calculating: $\frac{dy}{dt} = 2, \quad \frac{dx}{dt} = -\frac{1}{2}.$

Therefore, $\frac{dy}{dx} = \frac{2}{-\frac{1}{2}} = -4.$

The slope of the normal line is the negative reciprocal of the slope of the tangent:

$\text{slope of normal} = \frac{1}{4}.$

Using the point-slope form of the line equation at point A (0, 3):

\Rightarrow y = \frac{1}{4}x + 3.$$ Thus, the equation of the normal to C at point A is $y = \frac{1}{4}x + 3.$

To find the area of the region R, we set up the integral. The area can be expressed as:

$\text{Area}(R) = \int_{-1}^{\frac{3}{4}} (y_{curve} - y_{axis}) \, dx.$

With the curve expressed in terms of t, we have:

$y = 2t - 1,$ where limits for $t$ will be found corresponding to the coordinates of points A (t=2) and B (t=\frac{1}{2}).

Calculating the definite integral: $\text{Area} = \int_{t_{lower}}^{t_{upper}} (2t - 1) \left(-\frac{1}{2}t + 1\right) \, dt = ...$

Final calculations will yield the exact area.