该图显示了曲线 $y = 2 \sin 2x + 1$ 的一部分。$P$ 是该曲线的最高点，而 $Q$ 是曲线上切线斜率为 $-4$ 的点。

1. Find the coordinates of the maximum point $P$ using the properties of the sine function.

For $y = 2 \sin 2x + 1$, the maximum value occurs when $\sin 2x = 1$.
$y_{\max} = 2(1) + 1 = 3$
$\sin 2x = 1 \implies 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4}$
Coordinates of $P$ are $\left(\frac{\pi}{4}, 3\right)$.

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

$\frac{dy}{dx} = 2(2 \cos 2x) = 4 \cos 2x$

3. Equate the gradient function to $-4$ to find the $x$-coordinate of $Q$.

At $Q$, $\frac{dy}{dx} = -4$
$4 \cos 2x = -4$
$\cos 2x = -1$
$2x = \pi \implies x = \frac{\pi}{2}$

4. Find the corresponding $y$-coordinate of $Q$.

When $x = \frac{\pi}{2}$,
$y = 2 \sin\left(2 \times \frac{\pi}{2}\right) + 1 = 2 \sin \pi + 1 = 2(0) + 1 = 1$
Coordinates of $Q$ are $\left(\frac{\pi}{2}, 1\right)$.

1. State the coordinates of $P$ and $Q$ from part (a)

From part (a),
$P = \left(\frac{\pi}{4}, 3\right)$
$Q = \left(\frac{\pi}{2}, 1\right)$

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 $x = \frac{\pi}{4}$ and the area of the trapezium under the line $PQ$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$.
$\text{Area} = \int_{0}^{\frac{\pi}{4}} (2 \sin 2x + 1) \, dx + \text{Area of trapezium}$

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

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

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

$\text{Area of trapezium} = \frac{1}{2} \times (3 + 1) \times \left(\frac{\pi}{2} - \frac{\pi}{4}\right)$
$= \frac{1}{2} \times 4 \times \frac{\pi}{4} = \frac{\pi}{2}$
$\text{Total Area} = 1 + \frac{\pi}{4} + \frac{\pi}{2} = 1 + \frac{3\pi}{4}$