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# proof of Simpson’s rule

We want to derive Simpson’s rule for

$\int_{a}^{b}f(x)\,dx.$ |

We will use Newton and Cotes formulas for $n=2$. In this case, $x_{0}=a$, $x_{2}=b$ and $x_{1}=(a+b)/2$. We use Lagrange’s interpolation formula to get a polynomial $p(x)$ such that $p(x_{j})=f(x_{j})$ for $j=0,1,2$.

The corresponding interpolating polynomial is

$p(x)=f(x_{1})\frac{(x-x_{2})(x-x_{3})}{(x_{1}-x_{2})(x_{1}-x_{3})}+f(x_{2})% \frac{(x-x_{1})(x-x_{3})}{(x_{2}-x_{1})(x_{2}-x_{3})}+f(x_{3})\frac{(x-x_{1})(% x-x_{2})}{(x_{3}-x_{1})(x_{3}-x_{2})}.$ |

and thus

$\int_{a}^{b}f(x)\,dx\approx\int_{a}^{b}f(x_{1})\frac{(x-x_{2})(x-x_{3})}{(x_{1% }-x_{2})(x_{1}-x_{3})}+f(x_{2})\frac{(x-x_{1})(x-x_{3})}{(x_{2}-x_{1})(x_{2}-x% _{3})}+f(x_{3})\frac{(x-x_{1})(x-x_{2})}{(x_{3}-x_{1})(x_{3}-x_{2})}\,dx.$ |

Since integration is linear, we are concerned only with integrating each term in the sum. Now, taking $x_{j}=a+hj$ where $j=0,1,2$ and $h=|b-a|/2$, we can rewrite the quotients on the last integral as

$\int_{a}^{b}p(x)\,dx=hf(x_{0})\int_{0}^{2}\frac{(t-1)(t-2)}{(0-1)(0-2)}\,dt+hf% (x_{1})\int_{0}^{2}\frac{(t-0)(t-2)}{(1-0)(1-2)}\,dt+hf(x_{2})\int_{0}^{2}% \frac{(t-0)(t-1)}{(2-0)(2-1)}\,dt.$ |

and if we calculate the integrals on the last expression we get

$\int_{a}^{b}p(x)\,dx=hf(x_{0})\frac{1}{3}+hf(x_{1})\frac{4}{3}+hf(x_{2})\frac{% 1}{3},$ |

which is Simpson’s rule:

$\int_{a}^{b}f(x)\,dx\approx\frac{h}{3}(f(x_{0})+4f(x_{1})+f(x_{2})).$ |

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## Mathematics Subject Classification

65D32*no label found*41A55

*no label found*26A06

*no label found*28-00

*no label found*

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