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 9 - 2008 - Paper 2

To find the intersection points, we set the equations equal to each other:

$6x - x^2 = 2x$

Rearranging gives:

$x^2 - 4x = 0$

Factoring yields:

$x(x - 4) = 0$

Thus, the solutions are:

$x = 0 \quad \text{and} \quad x = 4.$

For $x = 0$, substituting back into the line equation:

$y = 2(0) = 0 \rightarrow (0, 0).$

For $x = 4$:

$y = 2(4) = 8 \rightarrow (4, 8).$

The line L intersects the curve C at the points $(0, 0)$ and $(4, 8)$.

To find the area R between the curve C and the line L, we need to calculate the integral of the top function minus the bottom function from $x = 0$ to $x = 4$:

$\text{Area} = \int_0^4 (6x - x^2 - 2x) \, dx = \int_0^4 (4x - x^2) \, dx.$

Calculating the integral:

$= \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 = \left( 2(4)^2 - \frac{(4)^3}{3} \right) - \left( 2(0)^2 - \frac{(0)^3}{3} \right) = \left( 32 - \frac{64}{3} \right) = \frac{96 - 64}{3} = \frac{32}{3}.$

Therefore, the area of region R is:

$\frac{32}{3} \text{ square units}.$