Answer
$$A = \frac{{31}}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the given graph we have}} \cr
& f\left( x \right) = 5 - {x^2}{\text{ and }}g\left( x \right) = {x^2} - 2x + 1 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}0 \leqslant x \leqslant 2 \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }}2 \leqslant x \leqslant 3 \cr
& {\text{The area is given by}} \cr
& A = \int_0^2 {\left[ {\left( {5 - {x^2}} \right) - \left( {{x^2} - 2x + 1} \right)} \right]} dx \cr
& {\text{ }} + \int_2^3 {\left[ {\left( {{x^2} - 2x + 1} \right) - \left( {5 - {x^2}} \right)} \right]} dx \cr
& A = \int_0^2 {\left( {5 - {x^2} - {x^2} + 2x - 1} \right)} dx + \int_2^3 {\left( {{x^2} - 2x + 1 - 5 + {x^2}} \right)} dx \cr
& A = \int_0^2 {\left( {4 - 2{x^2} + 2x} \right)} dx + \int_2^3 {\left( {2{x^2} - 2x - 4} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {4x - \frac{{2{x^3}}}{3} + {x^2}} \right]_0^2 + \left[ {\frac{{2{x^3}}}{3} - {x^2} - 4x} \right]_2^3 \cr
& A = \left[ {4\left( 2 \right) - \frac{{2{{\left( 2 \right)}^3}}}{3} + {{\left( 2 \right)}^2}} \right] + \left[ {\frac{{2{{\left( 3 \right)}^3}}}{3} - {{\left( 3 \right)}^2} - 4\left( 3 \right)} \right] \cr
& - \left[ {\frac{{2{{\left( 2 \right)}^3}}}{3} - {{\left( 2 \right)}^2} - 4\left( 2 \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{{20}}{3} - 3 + \frac{{20}}{3} \cr
& A = \frac{{31}}{3} \cr} $$
