Answer
$A \approx 1.7039$
Work Step by Step
$$\eqalign{
& y = \cos x,{\text{ }}y = x + 2{\sin ^4}x \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& {\text{Find the intersection points using the graph we obtain}} \cr
& x = - 1.92,{\text{ }}x = - 1.23,{\text{ }}x = 0.6 \cr
& {\text{We have the intervals }}\left[ { - 1.92, - 1.23} \right]{\text{ and }}\left[ { - 1.23,0.6} \right] \cr
& {\text{We can find the area integrating with respect to }}x \cr
& A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr
& {\text{From the graph }} \cr
& x + 2{\sin ^4}x \geqslant \cos x{\text{ on the interval }}\left[ { - 1.92, - 1.23} \right] \cr
& \cos x \geqslant x + 2{\sin ^4}x{\text{ on the interval }}\left[ { - 1.23,0.6} \right] \cr
& {\text{Therefore}} \cr
& A = \int_{ - 1.92}^{ - 1.23} {\left( {x + 2{{\sin }^4}x - \cos x} \right)} dx + \int_{ - 1.23}^{0.6} {\left( {\cos x - x - 2{{\sin }^4}x} \right)} dx \cr
& {\text{Integrating by a graphing calculator we obtain}} \cr
& A \approx 1.7039 \cr} $$
