Figure 2 shows a sketch of the curve C with parametric equations $x = 5t^2 - 4$, y = (9 - t^2)$ The curve C cuts the x-axis at the points A and B - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 7

To find the area of region R, we will integrate the function defined by the parametric equations. We know:

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

First, we calculate:

$\frac{dy}{dt} = -2t$

$\frac{dx}{dt} = 10t$

Then we find:

$\frac{dy}{dx} = \frac{-2t}{10t} = -\frac{1}{5}$

To find the area, we can use the formula:

$A = \int_{a}^{b} y \frac{dx}{dt} dt$

Here, $y = 9 - t^2$ and the limits of integration will be from $-3$ to $3$:

$A = \int_{-3}^{3} (9 - t^2) \cdot 10 dt$

This simplifies to:

$A = 10 \int_{-3}^{3} (9 - t^2) dt$

Calculating the definite integral, we evaluate:

$\int (9 - t^2) dt = 9t - \frac{t^3}{3}$

Thus, compute from -3 to 3:

$A = 10 [ (9(3) - \frac{(3)^3}{3}) - (9(-3) - \frac{(-3)^3}{3})]$

Calculating this gives us:

$A = 10 [ (27 - 9) - (-27 + 9) ] = 10 [ 18 + 18 ] = 10 \cdot 36 = 360 \text{ units}^2$