## Calculus with Applications (10th Edition)

$$A = \frac{3}{2}{\left( 2 \right)^{4/3}} - \frac{3}{7}{\left( 2 \right)^{7/3}}$$
\eqalign{ & y = {x^{4/3}},\,\,\,\,\,\,y = 2{x^{1/3}} \cr & {\text{Find the points of intersection of the graphs equating the functions}} \cr & {x^{4/3}} = 2{x^{1/3}} \cr & {x^{4/3}} - 2{x^{1/3}} = 0 \cr & {x^{1/3}}\left( {x - 2} \right) = 0 \cr & x = 0,\,\,x = 2 \cr & {\text{Then with the critical value we have the interval }}\left( {0,2} \right) \cr & {\text{For the interval }}\left( {0,2} \right){\text{ we have that }}2{x^{1/3}} > {x^{4/3}} \cr & \cr & {\text{The area between the curves is given by}}: \cr & A = \int_0^2 {\left( {{\text{ }}2{x^{1/3}} - {x^{4/3}}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {\frac{2}{{4/3}}{x^{4/3}} - \frac{{{x^{7/3}}}}{{7/3}}} \right)_0^2 \cr & A = \left( {\frac{3}{2}{x^{4/3}} - \frac{{3{x^{7/3}}}}{7}} \right)_0^2 \cr & A = \left( {\frac{3}{2}{{\left( 2 \right)}^{4/3}} - \frac{3}{7}{{\left( 2 \right)}^{7/3}}} \right) - \left( 0 \right) \cr & A = \frac{3}{2}{\left( 2 \right)^{4/3}} - \frac{3}{7}{\left( 2 \right)^{7/3}} \cr}