Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 31

$A = \frac{{13}}{5}$

$$\eqalign{ & y = {x^4},{\text{ }}y = 2 - \left| x \right| \cr & {\text{From the definition of absolute value}} \cr & y = 2 - \left| x \right| \to {\text{ }}2 - x{\text{ for }}x < 0{\text{ }} \cr & y = 2 - \left| x \right| \to 2 - x{\text{ for }}x > 0{\text{ }} \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{Find the intersection points using the graph}} \cr & \left( { - 1,1} \right){\text{ and }}\left( {1,1} \right) \cr & {\text{We have the intervals }}\left[ { - 1,0} \right]{\text{, and }}\left[ {0,1} \right] \cr & \cr & {\text{We can find the area integrating with respect to }}x \cr & A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr & {\text{From the graph }} \cr & 2 + x \geqslant {x^4}{\text{ for the intervals }}\left[ { - 1,0} \right] \cr & 2 - x \geqslant {x^4}{\text{ for the intervals }}\left[ {0,1} \right] \cr & {\text{Therefore}}{\text{, using the symmetry of the graph}} \cr & A = 2\int_0^1 {\left( {2 - x - {x^4}} \right)} dx \cr & {\text{Integrating}} \cr & A = 2\left[ {2x - \frac{1}{2}{x^2} - \frac{1}{5}{x^5}} \right]_0^1 \cr & {\text{Evaluate the limits}} \cr & A = 2\left[ {2\left( 1 \right) - \frac{1}{2}{{\left( 1 \right)}^2} - \frac{1}{5}{{\left( 1 \right)}^5}} \right] - 2\left[ 0 \right] \cr & {\text{Simplifying}} \cr & A = 2\left[ {2 - \frac{1}{2} - \frac{1}{5}} \right] - 0 \cr & A = \frac{{13}}{5} \cr} $$