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 curve and the line equal to each other:

$6x - x^2 = 2x$

Rearranging gives:

$x^2 - 4x = 0$

Factoring:

$x(x - 4) = 0$

Thus, the solutions are:

1. $x = 0$
2. $x = 4$

Now, substituting $x = 4$ back into either equation to find y:

$y = 2(4) = 8$

Therefore, the intersections are $(0, 0)$ and $(4, 8)$.

To find the area of region R, we need to integrate the difference between the upper curve C and the lower line L from $x = 0$ to $x = 4$:

$\text{Area} = \int_{0}^{4} (6x - x^2) - (2x) \, dx$

Simplifying the integrand gives:

$\text{Area} = \int_{0}^{4} (4x - x^2) \, dx$

Calculating the integral:

$= \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4}$

Evaluating at the bounds:

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

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