Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 32

$A = 8\pi - \frac{{16}}{3}$

$$\eqalign{ & y = {x^2},{\text{ }}y = \frac{{32}}{{{x^2} + 4}} \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{Find the intersection points}} \cr & y = y \cr & {x^2} = \frac{{32}}{{{x^2} + 4}} \cr & {x^4} + 4{x^2} = 32 \cr & {x^4} + 4{x^2} - 32 = 0 \cr & \left( {{x^2} + 8} \right)\left( {{x^2} - 4} \right) = 0 \cr & {x^2} - 4 = 0 \cr & x = \pm 2 \cr & {\text{We have the intervals }}\left[ { - 2,0} \right]{\text{, and }}\left[ {0,2} \right] \cr & \cr & {\text{We can find the area integrating with respect to }}x \cr & A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr & {\text{From the graph }} \cr & \frac{{32}}{{{x^2} + 4}} \geqslant {x^2}{\text{ for the intervals }}\left[ { - 2,0} \right]{\text{, and }}\left[ {0,2} \right] \cr & {\text{Therefore by symmetry}} \cr & A = 2\int_0^2 {\left( {\frac{{32}}{{{x^2} + 4}} - {x^2}} \right)} dx \cr & {\text{Integrating}} \cr & A = 2\left[ {32\left( {\frac{1}{2}\arctan \left( {\frac{x}{2}} \right)} \right) - \frac{1}{3}{x^3}} \right]_0^2 \cr & A = 2\left[ {16\arctan \left( {\frac{x}{2}} \right) - \frac{1}{3}{x^3}} \right]_0^2 \cr & {\text{Evaluate the limits}} \cr & A = 2\left[ {16\arctan \left( {\frac{2}{2}} \right) - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - 2\left[ {16\arctan \left( {\frac{0}{2}} \right) - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr & {\text{Simplifying}} \cr & A = 2\left[ {16\left( {\frac{\pi }{4}} \right) - \frac{8}{3}} \right] - 2\left[ 0 \right] \cr & A = 8\pi - \frac{{16}}{3} \cr} $$