Question Number 89735 by cindiaulia last updated on 19/Apr/20
Commented by jagoll last updated on 19/Apr/20
$$\mathrm{red}\:\mathrm{parabola}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\: \\ $$$$\mathrm{blue}\:\mathrm{parabola}\:\mathrm{y}\:=\:−\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4} \\ $$$$\mathrm{shaded}\:\mathrm{area}\::\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:=\:−\mathrm{x}^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{2x}\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\Delta\:=\:\mathrm{4}−\mathrm{4}.\mathrm{2}.\left(−\mathrm{4}\right)\:=\:\mathrm{36} \\ $$$$\mathrm{area}\:=\:\frac{\Delta\sqrt{\Delta}}{\mathrm{6a}^{\mathrm{2}} }\:=\:\frac{\mathrm{36}.\mathrm{6}}{\mathrm{6}.\mathrm{2}^{\mathrm{2}} }\:=\:\mathrm{9}\:\mathrm{sq}\:\mathrm{unit} \\ $$$$ \\ $$