The diagram shows part of the curve $y = 2\sin\frac{x}{2}$, where $A$ lies on the curve. The $x$-coordinate of $A$ is $\frac{4\pi}{3}$ and the normal to the curve at $A$ meets the $x$-axis at $B$ and the $y$-axis at $C$.

1. Find the coordinates of $A$ and the equation of the normal $AC$ (from part a).

At $A$, $x = \frac{4\pi}{3}$.
$y = 2\sin\left(\frac{4\pi}{6}\right) = 2\sin\left(\frac{2\pi}{3}\right) = \sqrt{3}$.
So $A = \left(\frac{4\pi}{3}, \sqrt{3}\right)$.
$\frac{dy}{dx} = \cos\frac{x}{2}$. At $x = \frac{4\pi}{3}$, $\frac{dy}{dx} = \cos\frac{2\pi}{3} = -\frac{1}{2}$.
Gradient of normal $= 2$.
Equation of normal $AC$: $y - \sqrt{3} = 2\left(x - \frac{4\pi}{3}\right) \implies y = 2x - \frac{8\pi}{3} + \sqrt{3}$.
At $C$, $x = 0$, so $y = \sqrt{3} - \frac{8\pi}{3}$.

2. Set up the expression for the area of the shaded region.

The shaded region is bounded above by the curve and below by the line $AC$ from $x = 0$ to $x = \frac{4\pi}{3}$.
$\text{Area} = \int_0^{\frac{4\pi}{3}} 2\sin\frac{x}{2} \, dx - \text{Area under line } AC$

3. Calculate the area under the curve.

$\int_0^{\frac{4\pi}{3}} 2\sin\frac{x}{2} \, dx = \left[ -4\cos\frac{x}{2} \right]_0^{\frac{4\pi}{3}}$
$= -4\cos\frac{2\pi}{3} - (-4\cos 0)$
$= -4\left(-\frac{1}{2}\right) + 4 = 2 + 4 = 6$

4. Calculate the area under the line $AC$ and find the total shaded area.

The integral of the line $AC$ from $x=0$ to $x=\frac{4\pi}{3}$ can be found using the area of a trapezium (signed area):
$\int_0^{\frac{4\pi}{3}} \left(2x - \frac{8\pi}{3} + \sqrt{3}\right) \, dx = \frac{1}{2} \times \frac{4\pi}{3} \times \left( y_C + y_A \right)$
$= \frac{2\pi}{3} \left( \sqrt{3} - \frac{8\pi}{3} + \sqrt{3} \right) = \frac{2\pi}{3} \left( 2\sqrt{3} - \frac{8\pi}{3} \right) = \frac{4\pi\sqrt{3}}{3} - \frac{16\pi^2}{9}$

$\text{Shaded Area} = 6 - \left( \frac{4\pi\sqrt{3}}{3} - \frac{16\pi^2}{9} \right) = 6 + \frac{16\pi^2}{9} - \frac{4\pi\sqrt{3}}{3}$