Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 23

$A = 4$

$$\eqalign{ & {\text{Let the functions }}y = \root 3 \of {2x} {\text{ and }}y = \frac{1}{2}x \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{Find the intersection points}} \cr & y = y \cr & \root 3 \of {2x} = \frac{1}{2}x \cr & {\text{Solve for }}x \cr & {\left( {\root 3 \of {2x} } \right)^3} = {\left( {\frac{1}{2}x} \right)^3} \cr & 2x = \frac{1}{8}{x^3} \cr & 16 = {x^2} \cr & x = \pm 4 \cr & \cr & {\text{We can find the area integrating with respect to }}x \cr & A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr & y = \root 3 \of {2x} \geqslant y = \frac{1}{2}x{\text{ into the interval }}\left[ {0,4} \right],{\text{ then}} \cr & {\text{By symmetry }} \cr & A = 2\int_0^4 {\left( {\root 3 \of {2x} - \frac{1}{2}x} \right)} dx \cr & {\text{Integrating}} \cr & A = 2\left[ {\root 3 \of 2 \left( {\frac{{{x^{4/3}}}}{{4/3}}} \right) - \frac{1}{2}\left( {\frac{{{x^2}}}{2}} \right)} \right]_0^4 \cr & A = 2\left[ {\frac{{3\root 3 \of 2 }}{4}{x^{4/3}} - \frac{1}{4}{x^2}} \right]_0^4 \cr & A = \frac{1}{2}\left[ {3\root 3 \of 2 {x^{4/3}} - {x^2}} \right]_0^4 \cr & {\text{Evaluate the limits}} \cr & A = \frac{1}{2}\left[ {3\root 3 \of 2 {{\left( 4 \right)}^{4/3}} - {{\left( 4 \right)}^2}} \right] - \frac{1}{2}\left[ {3\root 3 \of 2 {{\left( 0 \right)}^{4/3}} - {{\left( 0 \right)}^2}} \right] \cr & A = \frac{1}{2}\left[ {3\root 3 \of 2 {{\left( 4 \right)}^{4/3}} - 16} \right] - 0 \cr & A = \frac{1}{2}\left[ {3\left( {{2^{1/3}}} \right)\left( {{2^{8/3}}} \right) - 16} \right] \cr & A = 12 - 8 \cr & A = 4 \cr} $$