Quadratisches Polynom/Einschluss/1/Aufgabe/Lösung

Die Nullstellen des Polynoms ${}x^{2}-x-3$ sind

${}x_{1},x_{2}={\frac {1\pm {\sqrt {1+12}}}{2}}={\frac {1\pm {\sqrt {13}}}{2}}\,.$

Das bestimmte Integral zwischen diesen Grenzen ist

${}{\begin{aligned}\int _{\frac {1-{\sqrt {13}}}{2}}^{\frac {1+{\sqrt {13}}}{2}}x^{2}-x-3dx&=\left({\frac {1}{3}}x^{3}-{\frac {1}{2}}x^{2}-3x\right)|_{\frac {1-{\sqrt {13}}}{2}}^{\frac {1+{\sqrt {13}}}{2}}\\&={\frac {1}{3}}\cdot {\left(\left({\frac {1+{\sqrt {13}}}{2}}\right)^{3}-\left({\frac {1-{\sqrt {13}}}{2}}\right)^{3}\right)}-{\frac {1}{2}}\cdot {\left(\left({\frac {1+{\sqrt {13}}}{2}}\right)^{2}-\left({\frac {1-{\sqrt {13}}}{2}}\right)^{2}\right)}-3\cdot {\left({\frac {1+{\sqrt {13}}}{2}}-{\frac {1-{\sqrt {13}}}{2}}\right)}\\&={\frac {1}{3}}\cdot {\frac {6{\sqrt {13}}+26{\sqrt {13}}}{8}}-{\frac {1}{2}}\cdot {\sqrt {13}}-3{\sqrt {13}}\\&={\frac {32{\sqrt {13}}-12{\sqrt {13}}-72{\sqrt {13}}}{24}}\\&=-{\frac {52{\sqrt {13}}}{24}}\\&=-{\frac {13{\sqrt {13}}}{6}}.\,\end{aligned}}$

Der Flächeninhalt ist also

${\frac {13{\sqrt {13}}}{6}}.$

Zur gelösten Aufgabe