The curve shown in Figure 2 has parametric equations $x = 2 \, ext{sin} \, t, \quad y = 1 - 2 \, ext{cos} \, t, \quad 0 \leq t \leq 2\pi$ - Edexcel - A-Level Maths Pure - Question 1 - 2005 - Paper 7

The area under the curve can be computed using the formula for the area between a curve and the x-axis from the point of intersection. We first express the area $R$ as:

$\int_a^b y \, dx$.

Here, the limits of integration are from $t = 0$ to $t = \frac{\pi}{3}$, and therefore:

$A = \int_0^{\frac{\pi}{3}} (1 - 2 \, \text{cos} \, t) \, \frac{dx}{dt} \, dt = \int_0^{\frac{\pi}{3}} (1 - 2 \, \text{cos} \, t) \, dt.$

To find the exact value of the shaded area, we evaluate the integral:

$A = \int_0^{\frac{\pi}{3}} (1 - 2 \, \text{cos} \, t) \, dt$

This can be separated into two simpler integrals:

$A = \int_0^{\frac{\pi}{3}} 1 \, dt - 2 \int_0^{\frac{\pi}{3}} \text{cos} \, t \, dt.$

Calculating the first integral:

$\int_0^{\frac{\pi}{3}} 1 \, dt = \left[t\right]_0^{\frac{\pi}{3}} = \frac{\pi}{3}.$

For the second integral, we have:

$\int_0^{\frac{\pi}{3}} \text{cos} \, t \, dt = \left[\text{sin} \, t\right]_0^{\frac{\pi}{3}} = \text{sin} \left(\frac{\pi}{3}\right) - \text{sin} (0) = \frac{\sqrt{3}}{2}.$

Now substituting these values back into the area calculation:

$A = \frac{\pi}{3} - 2 \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \sqrt{3}.$

Thus, the exact value of the shaded area is:

$\frac{\pi}{3} - \sqrt{3}.$