图中的图象显示了曲线 $y = \sqrt{2x+5}$ 的一部分。A 是曲线上的一个点，且 A 的 x 坐标为 2。曲线在点 A 处的法线与 x 轴交于点 B。

1. Find the $y$-coordinate of point A

When $x = 2$,
$y = \sqrt{2(2) + 5} = \sqrt{9} = 3$
So, point A is $(2, 3)$.

2. Differentiate the equation of the curve to find the gradient function

$y = (2x + 5)^{\frac{1}{2}}$
$\frac{dy}{dx} = \frac{1}{2}(2x + 5)^{-\frac{1}{2}} \cdot 2$
$\frac{dy}{dx} = \frac{1}{\sqrt{2x + 5}}$

3. Find the gradient of the tangent at A

At $x = 2$,
$\frac{dy}{dx} = \frac{1}{\sqrt{2(2) + 5}} = \frac{1}{3}$

4. Find the gradient of the normal and its equation

Gradient of normal $= -\frac{1}{\frac{1}{3}} = -3$
Equation of the normal at $(2, 3)$:
$y - 3 = -3(x - 2)$
$y - 3 = -3x + 6$
$y = -3x + 9$

1. Find the equation of the normal at A and the coordinates of B (from part a).

$\frac{dy}{dx} = \frac{1}{2}(2x+5)^{-\frac{1}{2}}(2) = \frac{1}{\sqrt{2x+5}}$
At $x = 2$, $\frac{dy}{dx} = \frac{1}{\sqrt{9}} = \frac{1}{3}$.
Gradient of normal $= -3$.
Equation of normal: $y - 3 = -3(x - 2) \implies y = -3x + 9$.
At $B$, $y = 0 \implies -3x + 9 = 0 \implies x = 3$. So $B$ is $(3, 0)$.

2. Set up the expression for the total shaded area.

The shaded area is the sum of the area under the curve from $x = 0$ to $x = 2$ and the area of the triangle formed by the normal from $x = 2$ to $x = 3$.
$\text{Total Area} = \int_{0}^{2} \sqrt{2x+5} \, dx + \text{Area of triangle}$

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

$\int_{0}^{2} (2x+5)^{\frac{1}{2}} \, dx = \left[ \frac{(2x+5)^{\frac{3}{2}}}{\frac{3}{2} \times 2} \right]_{0}^{2}$
$= \left[ \frac{1}{3}(2x+5)^{\frac{3}{2}} \right]_{0}^{2}$
$= \frac{1}{3}(9)^{\frac{3}{2}} - \frac{1}{3}(5)^{\frac{3}{2}}$
$= \frac{27}{3} - \frac{5\sqrt{5}}{3} = 9 - \frac{5\sqrt{5}}{3}$

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

$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (3 - 2) \times 3 = 1.5$
$\text{Total Area} = 9 - \frac{5\sqrt{5}}{3} + 1.5 = 10.5 - \frac{5\sqrt{5}}{3} \approx 6.77$