图中显示了曲线 $y = \\sqrt{2x + 4}$ 的一部分，曲线与 $x$ 轴交于点 $A$，与 $y$ 轴交于点 $B$。曲线在点 $B$ 处的法线与 $x$ 轴交于点 $C$。

1. Find the y-coordinate of point B by setting x = 0

At point $B$, the curve crosses the $y$-axis, so $x = 0$.
$y = \sqrt{2(0) + 4}$
$y = \sqrt{4} = 2$

1. Find the derivative of the curve equation

$y = (2x + 4)^{\frac{1}{2}}$
$\frac{dy}{dx} = \frac{1}{2}(2x + 4)^{-\frac{1}{2}} \cdot 2 = \frac{1}{\sqrt{2x + 4}}$

2. Find the gradient of the tangent at point $B$

At $B$, $x = 0$.
Gradient of tangent, $m_t = \frac{1}{\sqrt{2(0) + 4}} = \frac{1}{2}$

3. Find the gradient and equation of the normal at $B$

Gradient of normal, $m_n = -\frac{1}{m_t} = -2$
Since $B$ is $(0, 2)$, the equation of the normal is:
$y - 2 = -2(x - 0)$
$y = -2x + 2$

4. Find the coordinates of $C$ by setting $y = 0$

At $C$, the normal meets the $x$-axis, so $y = 0$.
$0 = -2x + 2$
$2x = 2 \implies x = 1$
Coordinates of $C$ are $(1, 0)$.

1. Find the coordinates of points A, B, and C from the earlier parts

For $A$, $y = 0 \implies \sqrt{2x+4} = 0 \implies x = -2$. So $A(-2, 0)$.
For $B$, $x = 0 \implies y = \sqrt{4} = 2$. So $B(0, 2)$.
$\frac{dy}{dx} = \frac{1}{2}(2x+4)^{-\frac{1}{2}}(2) = \frac{1}{\sqrt{2x+4}}$.
At $x = 0$, gradient of tangent $= \frac{1}{2}$.
Gradient of normal $= -2$.
Equation of normal at $B$: $y - 2 = -2(x - 0) \implies y = -2x + 2$.
For $C$, $y = 0 \implies -2x + 2 = 0 \implies x = 1$. So $C(1, 0)$.

2. Set up the expression for the total shaded area

The shaded area is the sum of the area under the curve from $x = -2$ to $x = 0$ and the area of the triangle formed by the normal from $x = 0$ to $x = 1$.
$\text{Area} = \int_{-2}^{0} \sqrt{2x+4} \, dx + \text{Area of } \triangle OBC$

3. Evaluate the definite integral for the area under the curve

$\int_{-2}^{0} (2x+4)^{\frac{1}{2}} \, dx = \left[ \frac{(2x+4)^{\frac{3}{2}}}{\frac{3}{2} \times 2} \right]_{-2}^{0}$
$= \left[ \frac{(2x+4)^{\frac{3}{2}}}{3} \right]_{-2}^{0}$
$= \frac{4^{\frac{3}{2}}}{3} - 0 = \frac{8}{3}$

4. Calculate the area of the triangle and find the total area

$\text{Area of } \triangle OBC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1$
$\text{Total Area} = \frac{8}{3} + 1 = \frac{11}{3} = 3\frac{2}{3}$