## Calculus with Applications (10th Edition)

$$A = \frac{1}{2}$$
\eqalign{ & y = 2{x^3} + {x^2} + x + 5,\,\,\,\,y = {x^3} + {x^2} + 2x + 5 \cr & {\text{Find the points of intersection of the graphs}} \cr & 2{x^3} + {x^2} + x + 5 = {x^3} + {x^2} + 2x + 5 \cr & 2{x^3} + {x^2} + x + 5 - {x^3} - {x^2} - 2x - 5 = 0 \cr & {x^3} - x = 0 \cr & x\left( {{x^2} - 1} \right) = 0 \cr & x\left( {x + 1} \right)\left( {x - 1} \right) = 0 \cr & x = 0,x = - 1,x = 1 \cr & {\text{Then with the critical values we have the intervals }}\left( { - 1,0} \right){\text{ and }}\left( {0,1} \right) \cr & {\text{For the interval }}\left( { - 1,0} \right){\text{ we have that }}2{x^3} + {x^2} + x + 5 > {x^3} + {x^2} + 2x + 5 \cr & {\text{and }} \cr & {\text{For the interval }}\left( {0,1} \right){\text{ we have that }}{x^3} + {x^2} + 2x + 5 > 2{x^3} + {x^2} + x + 5 \cr & \cr & {\text{The area is given by}}: \cr & A = \int_{ - 1}^0 {\left( {2{x^3} + {x^2} + x + 5 - {x^3} - {x^2} - 2x - 5} \right)} dx \cr & + \int_0^1 {\left( {{x^3} + {x^2} + 2x + 5 - 2{x^3} - {x^2} - x - 5} \right)} dx \cr & A = \int_{ - 1}^0 {\left( {{x^3} - x} \right)} dx + \int_0^1 {\left( {x - {x^3}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {\frac{1}{4}{x^4} - \frac{1}{2}{x^2}} \right)_{ - 1}^0 + \left( {\frac{1}{2}{x^2} - \frac{1}{4}{x^4}} \right)_0^1 \cr & A = \left( {\frac{1}{4}{{\left( 0 \right)}^4} - \frac{1}{2}{{\left( 0 \right)}^2}} \right) - \left( {\frac{1}{4}{{\left( { - 1} \right)}^4} - \frac{1}{2}{{\left( { - 1} \right)}^2}} \right) + \left( {\frac{1}{2}{{\left( 1 \right)}^2} - \frac{1}{4}{{\left( 1 \right)}^4}} \right) - \left( {\frac{1}{2}{{\left( 0 \right)}^2} - \frac{1}{4}{{\left( 0 \right)}^4}} \right) \cr & A = \left( 0 \right) - \left( { - \frac{1}{4}} \right) + \left( {\frac{1}{4}} \right) - \left( 0 \right) \cr & A = \frac{1}{2} \cr}