Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 27

$A = \frac{1}{2}$

$$\eqalign{ & {\text{Let the functions }}y = \cos x,{\text{ }}y = \sin 2x,{\text{ on }}0 \leqslant x \leqslant \frac{\pi }{2} \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{Find the intersection points}} \cr & y = y \cr & \cos x = \sin 2x \cr & \cos x = 2\sin x\cos x \cr & 1 = 2\sin x \cr & \sin x = \frac{1}{2} \cr & {\text{For the interval }}0 \leqslant x \leqslant \frac{\pi }{2}{\text{ }}\sin x = \frac{1}{2}{\text{ at }}x = \frac{\pi }{6} \cr & {\text{We have the intervals }}\left[ {0,\frac{\pi }{6}} \right]{\text{ and }}\left[ {\frac{\pi }{6},\frac{\pi }{2}} \right] \cr & \cr & {\text{We can find the area integrating with respect to }}x \cr & A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr & {\text{From the graph }} \cr & \cos x \geqslant \sin 2x{\text{ on}}\left[ {0,\frac{\pi }{6}} \right]{\text{ and }}\sin 2x \geqslant \cos x{\text{ on }}\left[ {\frac{\pi }{6},\frac{\pi }{2}} \right] \cr & {\text{Therefore}} \cr & A = \int_0^{\pi /6} {\left( {\cos x - \sin 2x} \right)} dx + \int_{\pi /6}^{\pi /2} {\left( {\sin 2x - \cos x} \right)} dx \cr & {\text{Integrating}} \cr & A = \left[ {\sin x + \frac{1}{2}\cos 2x} \right]_0^{\pi /6} + \left[ { - \frac{1}{2}\cos 2x - \sin x} \right]_{\pi /6}^{\pi /2} \cr & {\text{Evaluate the limits}} \cr & A = \left[ {\sin \left( {\frac{\pi }{6}} \right) + \frac{1}{2}\cos \left( {\frac{\pi }{3}} \right)} \right] - \left[ { - \sin \left( 0 \right) - \frac{1}{2}\cos \left( 0 \right)} \right] \cr & + \left[ {\frac{1}{2}\cos \pi + \sin \left( {\frac{\pi }{2}} \right)} \right] - \left[ { - \sin \left( {\frac{\pi }{6}} \right) - \frac{1}{2}\cos \left( {\frac{\pi }{3}} \right)} \right] \cr & {\text{Simplifying}} \cr & A = \left[ {\frac{1}{2} + \frac{1}{4}} \right] - \left[ {0 + \frac{1}{2}} \right] + \left[ {\frac{1}{2} - 1} \right] - \left[ { - \frac{1}{2} - \frac{1}{4}} \right] \cr & A = \frac{3}{4} - \frac{1}{2} - \frac{1}{2} + \frac{3}{4} \cr & A = \frac{1}{2} \cr} $$