Answer
$$A = \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& y = {x^3} - {x^2} + x + 1,\,\,\,\,y = 2{x^2} - x + 1 \cr
& {\text{Find the points of intersection of the graphs}} \cr
& {x^3} - {x^2} + x + 1 = 2{x^2} - x + 1 \cr
& {x^3} - {x^2} + x + 1 - 2{x^2} + x - 1 = 0 \cr
& {x^3} - 3{x^2} + 2x = 0 \cr
& x\left( {{x^2} - 3x + 2} \right) = 0 \cr
& x\left( {x - 1} \right)\left( {x - 2} \right) = 0 \cr
& x = 0,x = 1,x = 2 \cr
& {\text{Then with the critical values we have the intervals }}\left( {0,1} \right){\text{ and }}\left( {1,2} \right) \cr
& {\text{For the interval }}\left( {0,1} \right){\text{ we have that }}{x^3} - {x^2} + x + 1 > 2{x^2} - x + 1 \cr
& {\text{and }} \cr
& {\text{For the interval }}\left( {1,2} \right){\text{ we have that }}2{x^2} - x + 1 > {x^3} - {x^2} + x + 1 \cr
& \cr
& {\text{The area is given by}}: \cr
& A = \int_0^1 {\left( {{x^3} - {x^2} + x + 1 - 2{x^2} + x - 1} \right)} dx + \int_1^2 {\left( {2{x^2} - x + 1 - {x^3} + {x^2} - x - 1} \right)} dx \cr
& A = \int_0^1 {\left( {{x^3} - 3{x^2} + 2x} \right)} dx + \int_1^2 {\left( {3{x^2} - 2x - {x^3}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {\frac{1}{4}{x^4} - {x^3} + {x^2}} \right)_0^1 + \left( {{x^3} - {x^2} - \frac{1}{4}{x^4}} \right)_1^2 \cr
& A = \left( {\frac{1}{4}{{\left( 1 \right)}^4} - {{\left( 1 \right)}^3} + {{\left( 1 \right)}^2}} \right) - \left( {\frac{1}{4}{{\left( 0 \right)}^4} - {{\left( 0 \right)}^3} + {{\left( 0 \right)}^2}} \right) + \left( {{{\left( 2 \right)}^3} - {{\left( 2 \right)}^2} - \frac{1}{4}{{\left( 2 \right)}^4}} \right) \cr
& - \left( {{{\left( 1 \right)}^3} - {{\left( 1 \right)}^2} - \frac{1}{4}{{\left( 1 \right)}^4}} \right) \cr
& \cr
& A = \left( {\frac{1}{4}} \right) - \left( 0 \right) + \left( 0 \right) - \left( { - \frac{1}{4}} \right) \cr
& A = \frac{1}{2} \cr} $$
