数学 – 辛普森三点积分推导

结论

对于不超过三次的多项式函数

$$f(x) = k_0 x^3 + k_1 x^2 + k_2 x + k_3$$

其定积分

$$int_{ a }^{ b } { f(x) dx } = frac{(b - a)}{6} left[ f(a) + 4f(frac{a + b}{2}) + f(b) right]$$

推导

暴力推导：

$$F(x) = int {f(x) dx} = frac{k_0}{4} x^4 + frac{k_1}{3} x^3 + frac{k_2}{2} x^2 + k_3 x$$
$$% <![CDATA[ begin{aligned} int_{ a }^{ b } { f(x) dx } & = F(b) - F(a) \ & = left( frac{k_0}{4} b^4 + frac{k_1}{3} b^3 + frac{k_2}{2} b^2 + k_3 b right) - left( frac{k_0}{4} a^4 + frac{k_1}{3} a^3 + frac{k_2}{2} a^2 + k_3 a right) \ & = frac{k_0}{4} (b^4 - a^4) + frac{k_1}{3} (b^3 - a^3) + frac{k_2}{2} (b^2 - a^2) + k_3 (b - a) \ & = frac{1}{6} left( frac{3}{2} k_0 (b^4 - a^4) + 2 k_1 (b^3 - a^3) + 3 k_2 (b^2 - a^2) + 6 k_3 (b - a) right) \ & = frac{1}{6} left( frac{3}{2} k_0 (b - a)(b + a)(b^2+a^2) + 2 k_1 (b - a)(b^2 + ab + a^2) + 3 k_2 (b - a)(b + a) + 6 k_3 (b - a) right) \ & = frac{b - a}{6} left( frac{3}{2} k_0 (b + a)(b^2+a^2) + 2 k_1 (b^2 + ab + a^2) + 3 k_2 (b + a) + 6 k_3 right) \ & = frac{b - a}{6} left[ k_0 left( a^3 + b^3 + 4 left( frac{a+b}{2} right)^3 right) + k_1 left( a^2 + b^2 + 4 left( frac{a+b}{2} right)^2 right) + k_2 left( a + b + 4 left( frac{a+b}{2} right) right) + k_3 left( 1 + 1 + 4 right) right] \ & = frac{(b - a)}{6} left[ f(a) + 4f(frac{a + b}{2}) + f(b) right] end{aligned} %]]>$$

Q.E.D.

附加内容

辛普森公式的高阶形式（摘自维基百科）：

$$int _{a}^{b}f(x),dxapprox {frac {h}{3}}{bigg [}f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+2f(x_{4})+cdots +4f(x_)+f(x_{n}){bigg ]}$$

一个 YouTube 上的推导视频：

https://www.youtube.com/watch?v=7MoRzPObRf0