Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.1 Area Between Two Curves - Exercises Set 5.1 - Page 353: 9

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

$$\eqalign{ & {\text{From the graph we can note that the area is given by}} \cr & A = \int_{\pi /4}^{\pi /2} {\left( {0 - \left( {\cos 2x} \right)} \right)} dx \cr & A = - \int_{\pi /4}^{\pi /2} {\cos 2x} dx \cr & {\text{Integrate}} \cr & A = - \frac{1}{2}\left[ {\sin 2x} \right]_{\pi /4}^{\pi /2} \cr & {\text{Evaluate}} \cr & A = - \frac{1}{2}\left[ {\sin 2\left( {\frac{\pi }{2}} \right) - \sin 2\left( {\frac{\pi }{4}} \right)} \right] \cr & {\text{Simplify}} \cr & A = - \frac{1}{2}\left[ {0 - 1} \right] \cr & A = \frac{1}{2} \cr} $$