Figure 3 shows a sketch of part of the curve C with equation $y = x(x + 4)(x - 2)$ The curve C crosses the x-axis at the origin O and at the points A and B - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 4

To find the area of the finite region, we will integrate the function from the left intersection point A to the right intersection point B. Hence, the area can be expressed as:

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

Expanding the integrand:

$= \int_{-4}^{2} (x^3 + 2x^2 - 8x) \, dx$

Now we integrate term by term:

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

Calculating at the bounds:

At $x = 2$: $= \frac{2^4}{4} + \frac{2(2^3)}{3} - 4(2^2) = 4 + \frac{16}{3} - 16 = 4 - 16 + \frac{16}{3} = -12 + \frac{16}{3} = -\frac{36}{3} + \frac{16}{3} = -\frac{20}{3}$

At $x = -4$: $= \frac{(-4)^4}{4} + \frac{2(-4)^3}{3} - 4(-4)^2 = 64 - \frac{128}{3} - 64 = -\frac{128}{3}$

Now subtract the two results:

So,

$Area = \left(-\frac{20}{3} - (-\frac{128}{3})\right) = \left(-\frac{20}{3} + \frac{128}{3}\right) = \frac{108}{3} = 36$

Thus, the total area of the region is 36 square units.