قاعدة سمبسون

ولكي نحصل على صيغة سمبسون لابد بالتعويض في الصيغة الإستكمالية

${\displaystyle P_{2}(x)=c_{\circ }+c_{1}x+c_{2}x^{2}}$

${\displaystyle \int _{a}^{b}f(x)dx={\int _{x_{\circ }}^{x_{2}}}[{\frac {(x-x_{1})(x-x_{2})}{(x_{\circ }-x_{1})(x_{\circ }-x_{2})}}f(x_{\circ })+{\frac {(x-x_{\circ })(x-x_{2})}{(x_{1}-x_{\circ })(x_{1}-x_{2})}}f(x_{1})+{\frac {(x-x_{\circ })(x-x_{1})}{(x_{2}-x_{\circ })(x_{2}-x_{1})}}f(x_{2})]dx+{\int _{x_{\circ }}^{x_{2}}}{\frac {(x-x_{\circ })(x-x_{1})(x-x_{2})}{6}}f^{(3)}(\xi (x))dx}$

${\displaystyle f(x)=f(x_{1})+f'(x_{1})(x-x_{1})+{\frac {f''(x_{1})}{2}}(x-x_{1})^{2}+{\frac {f'''(x_{1})}{6}}(x-x_{1})^{3}+{\frac {f^{(4)}(\xi (x)}{24}}(x-x_{1})^{4}}$

and

${\displaystyle \int _{x_{\circ }}^{x_{2}}f(x)dx-\left[f(x_{1})(x-x_{1})+{\frac {f'(x_{1})}{2}}(x-x_{1})^{2}+{\frac {f''(x_{1})}{6}}(x-x_{1})^{3}+{\frac {f'''(\xi (x)}{24}}(x-x_{1})^{4}\right]_{x_{\circ }}^{x_{2}}+{\frac {1}{24}}\int _{x_{\circ }}^{x_{2}}f^{(4)}(\xi (x))(x-x_{1})^{4}dx}$

${\displaystyle {\frac {1}{24}}\int _{x_{\circ }}^{x_{2}}f^{(4)}(\xi (x))(x-x_{1})^{4}dx={\frac {f^{(4)}(\xi _{1})}{24}}\int _{x_{\circ }}^{x_{2}}(x-x_{1})^{4}dx=\left[{\frac {f^{(4)}(\xi _{1})}{120}}(x-x_{1})^{5}\right]_{x_{\circ }}^{x_{2}}}$

لبعض قيم ${\displaystyle \xi _{1}}$ في ${\displaystyle (x_{\circ },x_{1})}$

أيضا

${\displaystyle h=x_{2}-x_{1}=x_{1}-x_{\circ }}$

بالتالي :

${\displaystyle (x_{2}-x_{1})^{2}-(x_{\circ }-x_{1})^{2}=(x_{2}-x_{1})^{4}-(x_{\circ }-x_{1})^{4}=0}$

${\displaystyle (x_{2}-x_{1})^{3}-(x_{\circ }-x_{1})^{3}=2hand(x_{2}-x_{1})^{5}-(x_{\circ }-x_{1})^{5}=2h^{5}}$

${\displaystyle \int _{x_{\circ }}^{x_{2}}f(x)dx=2hf(x_{1})+{\frac {h^{3}}{3}}f''(x_{1})+{\frac {f^{(4)}(\xi _{1})}{60}}h^{5}}$

وبتعويض قيمة ${\displaystyle f''(x_{1})}$ يصبح لدينا

${\displaystyle \int _{x_{\circ }}^{x_{2}}f(x)dx=2hf(x_{1})+{\frac {h^{3}}{3}}\left\{{\frac {1}{h^{2}}}[f(x_{\circ })-2f(x_{1})+f(x_{2})]-{\frac {h^{2}}{12}}f^{(4)}(\xi _{2})\right\}+{\frac {f^{(4)}(\xi _{1})}{60}}h^{5}}$

إذاً

${\displaystyle \int _{x_{\circ }}^{x_{2}}f(x)dx={\frac {h}{3}}[f(x_{\circ })+4f(x_{1})+f(x_{2})]-{\frac {h^{5}}{12}}\left[{\frac {1}{3}}f^{(4)}(\xi _{2})-{\frac {1}{5}}f^{(4)}(\xi _{1})\right]}$ [1]

استخدام قاعدة سمبسون البسيطة لإيجاد قيمة التكامل ${\displaystyle \,\int _{0}^{2}x^{2}\,dx}$

الحل:

${\displaystyle \,\int _{x_{0}}^{x_{2}}f(x)\,dx}$

${\displaystyle ={\frac {h}{3}}[f(x_{0})+4f(x_{1})+f(x_{2})]}$

${\displaystyle x_{0}=0}$

${\displaystyle x_{1}=1}$

${\displaystyle x_{2}=2}$

${\displaystyle h=x_{1}-x_{0}=1-0=1}$

${\displaystyle \to \,\int _{0}^{2}x^{2}\,dx={\frac {1}{3}}[f(0)+4f(1)+f(2)]}$

${\displaystyle ={\frac {1}{3}}[(0)^{2}+4(1)^{2}+(2)^{2}]}$

${\displaystyle ={\frac {1}{3}}[8]={\frac {8}{3}}}$ [1]

${\displaystyle \int _{-0.5}^{0.5}xln(x+1)dx\,}$

${\displaystyle n=6}$

الحل بستخدام قاعدة سمبسون المركبة :

${\displaystyle x_{0}=-0.5}$

${\displaystyle x_{1}=-0.333}$

${\displaystyle x_{2}=-0.166}$

${\displaystyle x_{3}=0}$

${\displaystyle x_{4}=0.166}$

${\displaystyle x_{5}=0.333}$

${\displaystyle x_{6}=0.5}$

${\displaystyle h={\frac {0.5-(-0.5)}{6}}={\frac {1}{6}}}$

${\displaystyle \int _{-0.5}^{0.5}xln(x+1)dx\,={\frac {\frac {1}{6}}{3}}[f(-0.5)+4f(-0.333)+2f(-0.166)+4f(0)+2f(0.166)+4f(0.333)+f(0.5)]}$

${\displaystyle f(-0.5)=(-0.5)ln(-0.5+1)=0.346}$

${\displaystyle f(-0.333)=(-0.333)ln(-0.333+1)=0.134}$

${\displaystyle f(-0.166)=(-0.166)ln(-0.166+1)=0.030}$

${\displaystyle f(0)=(0)ln(0+1)=0}$

${\displaystyle f(0.166)=(0.166)ln(0.166+1)=0.025}$

${\displaystyle f(0.333)=(0.333)ln(0.333+1)=0.095}$

${\displaystyle f(0.5)=(0.5)ln(0.5+1)=0.202}$

${\displaystyle \int _{-0.5}^{0.5}xln(x+1)dx\,}$

${\displaystyle ={\frac {1}{18}}[0.346+0.536+0.06+0+0.05+0.38+0.202]}$

${\displaystyle ={\frac {1.574}{18}}=0.0874}$ [1]