Draw the rough sketch of the curve $y = 20 \cos 2x$; $\left( \text{where } \frac{\pi}{6} \leq x \leq \frac{\pi}{3} \right)$. Using integration, find the area of the region bounded by the curve $y = 20 \cos 2x$ from the ordinates $x = \frac{\pi}{6}$ to $x = \frac{\pi}{3}$ and the x-axis.

$20 \left( 1 - \frac{\sqrt{3}}{2} \right) \text{ sq. units.}$

The correct answer is Option (1) → $20 \left( 1 - \frac{\sqrt{3}}{2} \right) \text{ sq. units.}$

$y = 20 \cos 2x$; $\left\{ \frac{\pi}{6} \leq x \leq \frac{\pi}{3} \right\}$

$\text{Required area} = 20 \int\limits_{\pi/6}^{\pi/4} \cos 2x dx + \left| 20 \int\limits_{\pi/4}^{\pi/3} \cos 2x dx \right|$$

$= 20 \left[ \frac{\sin 2x}{2} \right]_{\pi/6}^{\pi/4} + 20 \left| \left[ \frac{\sin 2x}{2} \right]_{\pi/4}^{\pi/3} \right|$

$= 10 \left( 1 - \frac{\sqrt{3}}{2} \right) + 10 \left| \frac{\sqrt{3}}{2} - 1 \right|$

$= 20 \left( 1 - \frac{\sqrt{3}}{2} \right) \text{ sq. units.}$