Die Gleichung

${\displaystyle {}x^{2}-3x-2=0\,}$

ist äquivalent zu

${\displaystyle {}{\left(x-{\frac {3}{2}}\right)}^{2}=2+{\left({\frac {3}{2}}\right)}^{2}={\frac {17}{4}}\,,}$

somit sind die Nullstellen dieses Polynoms gleich

${\displaystyle {}x_{1,2}={\frac {\pm {\sqrt {17}}+3}{2}}\,.}$

Im Intervall ${\displaystyle {}[{\frac {-{\sqrt {17}}+3}{2}},{\frac {{\sqrt {17}}+3}{2}}]}$ ist ${\displaystyle {}f}$ negativ, sonst überall positiv. Der gesuchte Flächeninhalt ist deshalb der Betrag des bestimmten Integrals

${\displaystyle {}{\begin{aligned}\int _{\frac {-{\sqrt {17}}+3}{2}}^{\frac {{\sqrt {17}}+3}{2}}x^{2}-3x-2\,dx&=\left({\frac {1}{3}}x^{3}-{\frac {3}{2}}x^{2}-2x\right)|_{\frac {-{\sqrt {17}}+3}{2}}^{\frac {{\sqrt {17}}+3}{2}}\\&={\frac {1}{3}}{\left(\left({\frac {{\sqrt {17}}+3}{2}}\right)^{3}-\left({\frac {-{\sqrt {17}}+3}{2}}\right)^{3}\right)}-{\frac {3}{2}}{\left(\left({\frac {{\sqrt {17}}+3}{2}}\right)^{2}-\left({\frac {-{\sqrt {17}}+3}{2}}\right)^{2}\right)}-2{\left(\left({\frac {{\sqrt {17}}+3}{2}}\right)-\left({\frac {-{\sqrt {17}}+3}{2}}\right)\right)}\\&={\frac {1}{3}}{\left({\frac {17{\sqrt {17}}}{4}}+{\frac {27{\sqrt {17}}}{4}}\right)}-{\frac {3}{2}}\cdot 6{\sqrt {17}}-2{\sqrt {17}}\\&=-{\frac {22}{3}}{\sqrt {17}}.\end{aligned}}}$

${\displaystyle {}{\frac {22}{3}}{\sqrt {17}}}$