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 7 - 2013 - Paper 4

To find the area under the curve from point A ($x = -4$) to point B ($x = 2$), we set up the integral:

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

First, we expand the integrand:

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

Now we integrate term by term:

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

Evaluating this from $-4$ to $2$:

$\left[ \frac{(2)^4}{4} + \frac{2(2)^3}{3} - 4(2)^2 \right] - \left[ \frac{(-4)^4}{4} + \frac{2(-4)^3}{3} - 4(-4)^2 \right]$

Calculating this gives:

$= \left[ 4 + \frac{16}{3} - 16 \right] - \left[ 64 - \frac{128}{3} - 64 \right]$

Simplifying:

$= \left[ -12 + \frac{16}{3} \right] - \left [ -\frac{128}{3} \right] = \left[ -\frac{36}{3} + \frac{16}{3} + \frac{128}{3} \right]$

Thus:

$= \frac{108}{3} = 36.$

Therefore, the total area of the shaded region is 36 square units.