In Figure 2 the curve C has equation $y = 6x - x^2$ and the line L has equation $y = 2x$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

To find the intersection points, we set the equations of the line L and the curve C equal to each other:

$6x - x^2 = 2x$

Rearranging gives:

$x^2 - 4x = 0$

Factoring out $x$:

$x(x - 4) = 0$

Thus, $x = 0$ or $x = 4$. For $x = 0$, substituting back, we find the point $(0, 0)$.

For $x = 4$:

$y = 2(4) = 8$

Therefore, the points of intersection are $(0, 0)$ and $(4, 8)$.

To find the area of region R, we set up an integral between the intersection points.

The area $A$ can be computed as:

$A = \int_0^4 (6x - x^2) - (2x) \, dx$

Which simplifies to:

$A = \int_0^4 (4x - x^2) \, dx$

Calculating the integral:

$A = \left[2x^2 - \frac{x^3}{3}\right]_0^4$

Evaluating at the bounds:

$A = \left(2(4^2) - \frac{(4)^3}{3}\right) - 0 = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}$

Thus, the area of region R is rac{32}{3} square units.