Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 515: 60

$$A = 9\ln 9 - 8$$

$$\eqalign{ & y = \frac{{{x^3}}}{{{x^2} + 1}}{\text{ and }}y = \frac{{8x}}{{{x^2} + 1}}{\text{ }} \cr & {\text{Calculate the intersection points}} \cr & \frac{{{x^3}}}{{{x^2} + 1}}{\text{ }} = \frac{{8x}}{{{x^2} + 1}} \cr & {x^5} + {x^3} = 8{x^3} + 8x \cr & {x^5} - 7{x^3} - 8x = 0 \cr & x\left( {{x^4} - 7{x^2} - 8} \right) = 0 \cr & x\left( {{x^2} - 8} \right)\left( {{x^2} + 1} \right) = 0 \cr & x = 0,{\text{ }}x = \pm \sqrt 8 = \pm 2\sqrt 2 \cr & {\text{From the graph we can see that }}\frac{{{x^3}}}{{{x^2} + 1}} > \frac{{8x}}{{{x^2} + 1}}{\text{ on }}\left[ { - 2\sqrt 2 ,0} \right] \cr & {\text{From the graph we can see that }}\frac{{8x}}{{{x^2} + 1}} > \frac{{{x^3}}}{{{x^2} + 1}}{\text{ on }}\left[ {0,2\sqrt 2 } \right] \cr & {\text{By symmetry the area is given by:}} \cr & A = 2\int_0^{2\sqrt 2 } {\left( {\frac{{8x}}{{{x^2} + 1}} - \frac{{{x^3}}}{{{x^2} + 1}}} \right)} dx \cr & A = 2\int_0^{2\sqrt 2 } {\left( {\frac{{8x - {x^3}}}{{{x^2} + 1}}} \right)} dx \cr & {\text{Integrate by substitution}} \cr & {\text{Let }}u = {x^2} + 1,{\text{ }}du = 2xdx,{\text{ }}dx = \frac{1}{{2x}}du \cr & x = 0 \Rightarrow u = 1 \cr & x = 2\sqrt 2 \Rightarrow u = 9 \cr & A = 2\int_1^9 {\left( {\frac{{8x - {x^3}}}{u}} \right)} \left( {\frac{1}{{2x}}} \right)du \cr & A = \int_1^9 {\left( {\frac{{x\left( {8 - {x^2}} \right)}}{u}} \right)} \left( {\frac{1}{x}} \right)du \cr & A = \int_1^9 {\left( {\frac{{8 - {x^2}}}{u}} \right)} du \cr & A = \int_1^9 {\left( {\frac{{8 - \left( {u - 1} \right)}}{u}} \right)} du \cr & A = \int_1^9 {\left( {\frac{{9 - u}}{u}} \right)} du = \int_1^9 {\left( {\frac{9}{u} - 1} \right)} du \cr & {\text{Integrating}} \cr & A = \left[ {9\ln u - u} \right]_1^9 \cr & A = \left[ {9\ln 9 - 9} \right] - \left[ {\ln 1 - 1} \right] \cr & A = 9\ln 9 - 9 + 1 \cr & A = 9\ln 9 - 8 \cr} $$