Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 30

$A = \frac{{128}}{{15}}$

$$\eqalign{ & {\text{Let the functions }}y = {x^4} - 3{x^2}{\text{ and }}y = {x^2} \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{Find the intersection points}} \cr & y = y \cr & {x^4} - 3{x^2} = {x^2} \cr & {x^4} - 4{x^2} = 0 \cr & {x^2}\left( {{x^2} - 4} \right) = 0 \cr & x = 0,{\text{ }}x = \pm 2 \cr & {\text{We have the intervals }}\left[ { - 2,0} \right]{\text{, and }}\left[ {0,2} \right] \cr & \cr & {\text{We can find the area integrating with respect to }}x \cr & A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr & {\text{From the graph }} \cr & {x^2} \geqslant {x^4} - 3{x^2}{\text{ for the intervals }}\left[ { - 2,0} \right]{\text{, and }}\left[ {0,2} \right] \cr & {\text{Therefore}} \cr & A = 2\int_0^2 {\left( {{x^2} - \left[ {{x^4} - 3{x^2}} \right]} \right)} dx \cr & A = 2\int_0^2 {\left( {{x^2} - {x^4} + 3{x^2}} \right)} dx \cr & A = 2\int_0^2 {\left( {4{x^2} - {x^4}} \right)} dx \cr & {\text{Integrating}} \cr & A = 2\left[ {\frac{4}{3}{x^3} - \frac{1}{5}{x^5}} \right]_0^2 \cr & {\text{Evaluate the limits}} \cr & A = 2\left[ {\frac{4}{3}{{\left( 2 \right)}^3} - \frac{1}{5}{{\left( 2 \right)}^5}} \right] - 2\left[ {\frac{4}{3}{{\left( 0 \right)}^3} - \frac{1}{5}{{\left( 0 \right)}^5}} \right] \cr & {\text{Simplifying}} \cr & A = 2\left[ {\frac{{64}}{{15}}} \right] - 0 \cr & A = \frac{{128}}{{15}} \cr} $$