Answer
$$A = 76$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2 - 2{x^2}{\text{; }}\left[ {0,5} \right] \cr
& {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr
& 2 - 2{x^2} = 0 \cr
& {\text{solve for }}x \cr
& 2{x^2} = 2 \cr
& x = \pm 1 \cr
& {\text{the graph crosses the }}x{\text{ - axis at }}x = \pm 1{\text{ and }}x = 1{\text{ is in the given interval }}\left[ {0,5} \right],{\text{ }} \cr
& f\left( 0 \right) = 2 - 2{\left( 0 \right)^2} = 2,{\text{ }}f\left( x \right){\text{ is positive for }}\left( {0,1} \right) \cr
& {\text{then}} \cr
& {\text{the definite integral of the total area will be}}: \cr
& A = \int_0^1 {\left( {2 - 2{x^2}} \right)} dx + \left| {\int_1^{10} {\left( {2 - 2{x^2}} \right)} dx} \right| \cr
& A = \left( {2x - \frac{{2{x^3}}}{3}} \right)_0^1 + \left| {\left( {2x - \frac{{2{x^3}}}{3}} \right)_1^5} \right| \cr
& A = \left( {2\left( 1 \right) - \frac{{2{{\left( 1 \right)}^3}}}{3}} \right) - \left( {2\left( 0 \right) - \frac{{2{{\left( 0 \right)}^3}}}{3}} \right) + \left| {\left( {2\left( 5 \right) - \frac{{2{{\left( 5 \right)}^3}}}{3}} \right) - \left( {2\left( 1 \right) - \frac{{2{{\left( 1 \right)}^3}}}{3}} \right)} \right| \cr
& {\text{simplifying}} \cr
& A = \frac{4}{3} - 0 + \left| { - \frac{{220}}{3} - \frac{4}{3}} \right| \cr
& A = \frac{4}{3} + \frac{{224}}{3} \cr
& A = 76 \cr} $$
