The curve shown in Figure 2 has parametric equations $x = 2 \, ext{sin} \, t, \, y = 1 - 2 \, ext{cos} \, t, \, 0 \, ext{is} \, t \, ext{is} \, 2\pi$ (a) Show that the curve crosses the x-axis where $t = \frac{\pi}{3}$ and $t = \frac{5\pi}{3}$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 7

The area $R$ bounded by the curve and the x-axis can be determined using the integral of the function representing $y$. The expressions for the area is given by:

$\text{Area} = \int y \, dx = \int_{a}^{b} (1 - 2 \text{cos} \, t) \, \frac{dx}{dt} dt$

Here, we note that:

$\frac{dx}{dt} = 2 \text{cos} \, t.$

Thus, the area can indeed be setup as:

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

To compute the area, we evaluate the integral:

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

This integral can be separated into two parts:

1. Evaluate $\int dt = t$ and
2. Evaluate $\int -2 \text{cos} \, t \, dt = -2 \text{sin} \, t.$

Thus, substituting we have: