Răspuns :

[tex]\displaystyle\bf\\aflam~punctele~de~intersectie~a~graficelor~:~\\-x=-x^2+x \Leftrightarrow -2x+x^2=0 \Leftrightarrow -x(2-x)=0 \implies x_1 = 0,~x_2=2.\\acum~ca~sa~aflam~aria~figurii~marginite~de~cele~doua~linii~procedam~astfel~:~\\\int\limits^2_0 {-x^2 +x - (-x)} \, dx =\int\limits^2_0 {-x^2+2x} \, dx = \int\limits^2_0 {2x-x^2} \, dx,~calculam~primitiva\\functiei,~\int\limits {2x-x^2} \, dx = \int{2x}dx-\int{x^2}dx=x^2-\frac{x^3}{3}.\\[/tex]

[tex]\displaystyle\bf\\\int\limits^2_0 {2x-x^2} \, dx = (x^2-\frac{x^3}{3}) \normalsize{|}_0^2=\frac{4}{3},~deci~\mathcal{A}=\frac{4}{3}.[/tex]