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 5 - 2013 - Paper 4

To calculate the total area between the curve and the x-axis from $x = -4$ to $x = 2$, we evaluate the integral:

ext{Area} = igg| ext{integral from } -4 ext{ to } 2 ext{ of } x(x + 4)(x - 2) \, dx \bigg|

First, we need to expand the function:

$x(x + 4)(x - 2) = x(x^2 + 2x - 8) = x^3 + 2x^2 - 8x$

Now, we can integrate:

ext{Area} = igg| igg[ \frac{x^4}{4} + \frac{2x^3}{3} - 4x^2 \bigg]_{-4}^{2} \bigg|

Calculating the upper limit ($x = 2$):

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

Calculating the lower limit ($x = -4$):

$\frac{(-4)^4}{4} + \frac{2(-4)^3}{3} - 4(-4)^2 = 64 - \frac{128}{3} - 64 = -\frac{128}{3} + 64 = \frac{192 - 128}{3} = \frac{64}{3}$

Now substituting into the area formula:

$\text{Area} = \bigg| -\frac{20}{3} - \frac{64}{3} \bigg| = \bigg| -\frac{84}{3} \bigg| = \frac{84}{3} = 28$

Hence, the total area of the finite region is 28 square units.