The diagram shows the curve $y = \sin 2x + x$ for $0 \le x \le \pi$ radians. The point $M$ is the minimum point of the curve and $OM$ is a straight line. Show that the area of the shaded region is $\left(\frac{\sqrt{3}\pi}{6} + \frac{3}{4}\right)$ square units. Continuation of working space for question 10.

1. Find the derivative of the curve to locate stationary points.

$y = \sin 2x + x$
$\frac{dy}{dx} = 2\cos 2x + 1$

2. Set the derivative to zero and solve for $x$ to find the minimum point $M$.

For stationary points, $\frac{dy}{dx} = 0$:
$2\cos 2x + 1 = 0$
$\cos 2x = -\frac{1}{2}$
Since $0 \le x \le \pi$, $0 \le 2x \le 2\pi$.
$2x = \frac{2\pi}{3}$ or $2x = \frac{4\pi}{3}$
$x = \frac{\pi}{3}$ or $x = \frac{2\pi}{3}$

Using the second derivative test:
$\frac{d^2y}{dx^2} = -4\sin 2x$
At $x = \frac{\pi}{3}$, $\frac{d^2y}{dx^2} = -4\sin(\frac{2\pi}{3}) < 0$ (Maximum)
At $x = \frac{2\pi}{3}$, $\frac{d^2y}{dx^2} = -4\sin(\frac{4\pi}{3}) > 0$ (Minimum)
So, the $x$-coordinate of $M$ is $\frac{2\pi}{3}$.

3. Find the $y$-coordinate of $M$.

Substitute $x = \frac{2\pi}{3}$ into the equation of the curve:
$y = \sin\left(2\left(\frac{2\pi}{3}\right)\right) + \frac{2\pi}{3}$
$y = \sin\left(\frac{4\pi}{3}\right) + \frac{2\pi}{3}$
$y = -\frac{\sqrt{3}}{2} + \frac{2\pi}{3}$

4. Calculate the area under the curve from $x = 0$ to $x = \frac{2\pi}{3}$.

Area under curve $= \int_0^{\frac{2\pi}{3}} (\sin 2x + x) \, dx$
$= \left[ -\frac{1}{2}\cos 2x + \frac{x^2}{2} \right]_0^{\frac{2\pi}{3}}$
$= \left( -\frac{1}{2}\cos\left(\frac{4\pi}{3}\right) + \frac{1}{2}\left(\frac{2\pi}{3}\right)^2 \right) - \left( -\frac{1}{2}\cos(0) + 0 \right)$
$= \left( -\frac{1}{2}\left(-\frac{1}{2}\right) + \frac{4\pi^2}{18} \right) - \left( -\frac{1}{2}(1) \right)$
$= \frac{1}{4} + \frac{2\pi^2}{9} + \frac{1}{2}$
$= \frac{3}{4} + \frac{2\pi^2}{9}$

5. Calculate the area of the triangle under the line $OM$.

The region under the line $OM$ from $x=0$ to $x=\frac{2\pi}{3}$ is a right-angled triangle with base $\frac{2\pi}{3}$ and height $\left(-\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right)$.
Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \left(\frac{2\pi}{3}\right) \left(-\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right)$
$= \frac{\pi}{3} \left(-\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right)$
$= -\frac{\sqrt{3}\pi}{6} + \frac{2\pi^2}{9}$

6. Subtract the area of the triangle from the area under the curve to find the shaded area.

Shaded Area $= \text{Area under curve} - \text{Area of triangle}$
$= \left( \frac{3}{4} + \frac{2\pi^2}{9} \right) - \left( -\frac{\sqrt{3}\pi}{6} + \frac{2\pi^2}{9} \right)$
$= \frac{3}{4} + \frac{2\pi^2}{9} + \frac{\sqrt{3}\pi}{6} - \frac{2\pi^2}{9}$
$= \frac{\sqrt{3}\pi}{6} + \frac{3}{4}$ (Shown)