Figure 1 shows a sketch of part of the curve C with equation $$y = (x + 1)(x - 5)$$ The curve crosses the x-axis at the points A and B - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3

The area of the region R can be found by integrating the function from the left intersection point A to the right intersection point B:

$A(R) = \int_{-1}^{5} ((x + 1)(x - 5)) \, dx$

First, we expand the function:

$A(R) = \int_{-1}^{5} (x^2 - 4x - 5) \, dx$

Now, we integrate term by term:

$A(R) = \left[ \frac{x^3}{3} - 2x^2 - 5x \right]_{-1}^{5}$

Calculating the definite integral:

1. Evaluate at $x = 5$: $\frac{5^3}{3} - 2(5^2) - 5(5) = \frac{125}{3} - 50 - 25 = \frac{125 - 150 - 75}{3} = \frac{-100}{3}$

2. Evaluate at $x = -1$: $\frac{(-1)^3}{3} - 2(-1)^2 - 5(-1) = \frac{-1}{3} - 2 + 5 = \frac{-1 + 15}{3} = \frac{14}{3}$

Now, subtract the two results:

$A(R) = \left( \frac{-100}{3} - \frac{14}{3} \right) = \frac{-100 - 14}{3} = \frac{-114}{3}$

However, since area is positive, we take the absolute value, and we find that the area of region R is:

$A(R) = 36$