The diagram shows part of the curve $y = x^3 + 3x^2 - 2$. The straight line $L$ cuts the curve at $A(-3, -2)$, the x-axis at $B$, and intersects the curve again at $C$.

1. Differentiate the equation of the curve to find an expression for the gradient of the tangent.

$y = x^3 + 3x^2 - 2$
$\frac{dy}{dx} = 3x^2 + 6x$

2. Equate the gradient expression to $-3$ to find the x-coordinate of $B$.

$3x^2 + 6x = -3$
$3x^2 + 6x + 3 = 0$
$x^2 + 2x + 1 = 0$
$(x + 1)^2 = 0$
$x = -1$

3. Substitute $x = -1$ into the curve's equation to find the y-coordinate of $B$.

When $x = -1$,
$y = (-1)^3 + 3(-1)^2 - 2$
$y = -1 + 3 - 2 = 0$
Therefore, the coordinates of $B$ are $(-1, 0)$.

1. Find the equation of line $L$ using $A(-3,-2)$ and $B(-1,0)$

Gradient of $L = \dfrac{0-(-2)}{-1-(-3)} = \dfrac{2}{2} = 1$.

Using $A(-3,-2)$: $y-(-2)=1\,(x-(-3))$, so $y = x+1$.

(Note: $B(-1,0)$ also lies on the curve since $(-1)^3+3(-1)^2-2 = 0$, and $L$ meets the curve at $A$ and $B$.)

2. Set up the area integral between the curve and the line

Over $-3 \le x \le -1$ the curve lies above $L$ (e.g. at $x=-2$: curve $=2$, line $=-1$).

$\text{Area} = \int_{-3}^{-1}\Big[(x^3+3x^2-2)-(x+1)\Big]\,dx = \int_{-3}^{-1}(x^3+3x^2-x-3)\,dx$

3. Integrate the expression

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

4. Evaluate the definite integral

At $x=-1$: $\frac{1}{4}-1-\frac{1}{2}+3 = 1.75$.

At $x=-3$: $\frac{81}{4}-27-\frac{9}{2}+9 = -2.25$.

$\text{Area} = 1.75-(-2.25) = 4 \text{ square units}$