The diagram shows the curve $y = 4 - 2\cos 2x$ for $0 \le x \le \frac{3}{2}\pi$ radians and a straight line passing through points $P$ and $R$. The coordinates of $P$ are $(0, -1)$ and $R$ is a minimum point of the curve. Find the area of the region bounded by the curve $y = 4 - 2\cos 2x$, the line segment $PR$ and the $y$-axis. Continuation of working space for question 10.

1. Find the coordinates of the minimum point $R$.

For $y = 4 - 2\cos 2x$ to be a minimum, $\cos 2x$ must be at its maximum value of $1$.
$y = 4 - 2(1) = 2$
$\cos 2x = 1 \implies 2x = 0, 2\pi, 4\pi, \dots$
From the diagram, $R$ is the first minimum after $x=0$, so $2x = 2\pi \implies x = \pi$.
The coordinates of $R$ are $(\pi, 2)$.

2. Find the equation of the line $PR$.

The line passes through $P(0, -1)$ and $R(\pi, 2)$.
Gradient $m = \frac{2 - (-1)}{\pi - 0} = \frac{3}{\pi}$
The $y$-intercept is $-1$.
Equation of line $PR$: $y = \frac{3}{\pi}x - 1$

3. Set up the definite integral for the area bounded by the curve and the line.

$\text{Area} = \int_{0}^{\pi} \left[ (4 - 2\cos 2x) - \left(\frac{3}{\pi}x - 1\right) \right] \, dx$
$\text{Area} = \int_{0}^{\pi} \left( 5 - 2\cos 2x - \frac{3}{\pi}x \right) \, dx$

4. Integrate the expression.

$\int \left( 5 - 2\cos 2x - \frac{3}{\pi}x \right) \, dx = 5x - \sin 2x - \frac{3}{2\pi}x^2$

5. Evaluate the definite integral to find the final area.

$\text{Area} = \left[ 5x - \sin 2x - \frac{3}{2\pi}x^2 \right]_{0}^{\pi}$
$= \left( 5\pi - \sin 2\pi - \frac{3}{2\pi}(\pi)^2 \right) - (0 - \sin 0 - 0)$
$= 5\pi - 0 - \frac{3}{2}\pi = \frac{7\pi}{2}$