## Calculus with Applications (10th Edition)

$$A = \frac{1}{{12}}$$
\eqalign{ & y = {x^2},\,\,\,\,\,y = {x^3} \cr & {\text{Find the points of intersection of the graphs}} \cr & {x^2} = {x^3} \cr & {x^2} - {x^3} = 0 \cr & {\text{factor}} \cr & {x^2}\left( {1 - x} \right) = 0 \cr & x = 0{\text{ and }}x = 1 \cr & {\text{We have the interval }}\left( {0,1} \right) \cr & {\text{Evaluate both functions at }}x = 0.5 \cr & y\left( {0.5} \right) = {\left( {0.5} \right)^2} = 0.25 \cr & y\left( {0.5} \right) = {\left( {0.5} \right)^3} = 0.125 \cr & {\text{For the interval }}\left( {0,1} \right){\text{ we have that }}{x^2} \geqslant {x^3} \cr & \cr & {\text{The area is given by}}: \cr & A = \int_0^1 {\left( {{x^2} - {x^3}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {\frac{{{x^3}}}{3} - \frac{{{x^4}}}{4}} \right)_0^1 \cr & A = \left( {\frac{{{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 1 \right)}^4}}}{4}} \right) - \left( {\frac{{{{\left( 0 \right)}^3}}}{3} - \frac{{{{\left( 0 \right)}^4}}}{4}} \right) \cr & A = \left( {\frac{1}{{12}}} \right) - \left( 0 \right) \cr & A = \frac{1}{{12}} \cr}