The curve shown in Figure 2 has parametric equations $x = 2 \, ext{sin} \, t,$ y = 1 - 2 \, ext{cos} \, t, \, 0 \leq t \leq 2 \pi$ (a) Show that the curve crosses the x-axis where $t = \frac{\pi}{3}$ and $t = \frac{5\pi}{3}.$ (b) The finite region $R$ is enclosed by the curve and the x-axis, as shown shaded in Figure 2 - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 5

The area under the curve and above the x-axis can be calculated using the definite integral of the function. The area $R$ is given by:

$\int_{a}^{b} y \, dx$
Since we are using the parametric equations, we derive:

$\frac{dx}{dt} = 2\cos(t).$
Thus, the area can be established as:

$\int_{0}^{2\pi} (y) \frac{dx}{dt} \, dt = \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (1 - 2\cos(t)) \, dt.$

To find the area $R$, we evaluate the integral:

$\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (1 - 2\cos(t)) \, dt.$
This simplifies to:

$\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} 1 \, dt - 2\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \cos(t) \, dt.$
Calculating the first integral gives:

$\left[t\right]_{\frac{\pi}{3}}^{\frac{5\pi}{3}} = \frac{5\pi}{3} - \frac{\pi}{3} = \frac{4\pi}{3}.$
For the second integral, with bounds evaluated, we have:

$\left[\sin(t)\right]_{\frac{\pi}{3}}^{\frac{5\pi}{3}} = \sin(\frac{5\pi}{3}) - \sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3}.$
Thus, substituting back, we arrive at:

$\frac{4\pi}{3} - 2(-\sqrt{3}) = \frac{4\pi}{3} + 2\sqrt{3}.$
Therefore, the exact value of the shaded area is:

$\frac{4\pi}{3} + 2\sqrt{3}.$