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 6 - 2011 - Paper 3

The area of the region R can be calculated using the definite integral of the curve from $x = -1$ to $x = 5$:

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

First, we expand the integrand:

$(x + 1)(x - 5) = x^2 - 4x - 5$

Now we compute the integral:

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

Calculating the antiderivative:

$\int (x^2 - 4x - 5) \, dx = \frac{x^3}{3} - 2x^2 - 5x + c$

Now, we evaluate from $-1$ to $5$:

1. 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. At $x = -1$: $= \frac{(-1)^3}{3} - 2(-1)^2 - 5(-1) = -\frac{1}{3} - 2 + 5 = -\frac{1}{3} + 3 = \frac{8}{3}$

Now, we find area:

$A = \left( \frac{-100}{3} - \frac{8}{3} \right) = -\frac{108}{3} = -36$

Since we are looking for the area, we take the absolute value:

Thus, the area of region R is $36$.