${\displaystyle {}1B_{0}=1}$

${\displaystyle {}x+}$

${\displaystyle {}c}$

${\displaystyle {}0=\int _{0}^{1}(x+c)dx=\left({\frac {1}{2}}x^{2}+cx\right)|_{0}^{1}={\frac {1}{2}}+c\,}$

führt auf

${\displaystyle {}c=-{\frac {1}{2}}\,}$

und daher

${\displaystyle {}B_{1}=x-{\frac {1}{2}}\,.}$

${\displaystyle {}2B_{1}=2x-1}$

${\displaystyle {}x^{2}-x+a}$

${\displaystyle {}a}$

${\displaystyle {}{\begin{aligned}0&=\int _{0}^{1}(x^{2}-x+a)dx\\&=\left({\frac {1}{3}}x^{3}-{\frac {1}{2}}x^{2}+ax\right)|_{0}^{1}\\&={\frac {1}{3}}-{\frac {1}{2}}+a\\&=-{\frac {1}{6}}+a\end{aligned}}}$

führt auf

${\displaystyle {}a={\frac {1}{6}}\,}$

und daher

${\displaystyle {}B_{2}=x^{2}-x+{\frac {1}{6}}\,.}$

${\displaystyle {}3B_{2}=3x^{2}-3x+{\frac {1}{2}}}$

${\displaystyle {}x^{3}-{\frac {3}{2}}x^{2}+{\frac {1}{2}}x+b}$

${\displaystyle {}b}$

${\displaystyle {}{\begin{aligned}0&=\int _{0}^{1}(x^{3}-{\frac {3}{2}}x^{2}+{\frac {1}{2}}x+b)dx\\&=\left({\frac {1}{4}}x^{4}-{\frac {1}{2}}x^{3}+{\frac {1}{4}}x^{2}+bx\right)|_{0}^{1}\\&={\frac {1}{4}}-{\frac {1}{2}}+{\frac {1}{4}}+b\\&=b\end{aligned}}}$

führt auf

${\displaystyle {}b=0\,}$

und daher

${\displaystyle {}B_{3}=x^{3}-{\frac {3}{2}}x^{2}+{\frac {1}{2}}x\,.}$