$f$在$(x_0,x_3)$上四阶可导，且在$[x_0,x_3]$上三阶导函数连续.
\begin{align*}
\int_{x_0}^{x_3}f(x)dx=\frac{3h}{8}[f(x_0)+3f(x_1)+3f(x_2)+f(x_3)]-\frac{3h^5}{80}f^{(4)}(\xi)
\end{align*}其中$h=x_1-x_0=x_2-x_1=x_3-x_2$.
下面先给出一种失败的方法.
证明:我们对$f(x)$进行牛顿插值.设立插值点
\begin{align*}
x_0,x_1,x_2,x_3
\end{align*}
则经过这几个插值点的牛顿插值多项式为\begin{align*}
f(x_0)+(x-x_0)f[x_0,x_1]+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+(x-x_0)(x-x_1)(x-x_2)f[x_0,x_1,x_2,x_3]
\end{align*}插值余项为
\begin{align*}
(x-x_0)(x-x_1)(x-x_2)(x-x_3)f[x,x_0,x_1,x_2,x_3]
\end{align*}
我们先来计算
\begin{align*}
\int_{x_0}^{x_3}\{f(x_0)+(x-x_0)f[x_0,x_1]+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+(x-x_0)(x-x_1)(x-x_2)f[x_0,x_1,x_2,x_3]\}dx
\end{align*}
易得
\begin{align*}
\int_{x_0}^{x_3}f(x_0)dx=3hf(x_0)
\end{align*}
\begin{align*}
\int_{x_0}^{x_3}(x-x_0)f[x_0,x_1]dx=f[x_0,x_1][\frac{1}{2}(x_3-x_0)^2]=\frac{9h}{2}(f(x_1)-f(x_0))
\end{align*}\begin{align*}
\int_{x_0}^{x_3}(x-x_0)(x-x_1)f[x_0,x_1,x_2]dx=(\frac{f(x_0)}{(-h)\times
(-2h)}+\frac{f(x_1)}{h\times (-h)}+\frac{f(x_2)}{2h\times
h})\int_{x_0}^{x_3}(x^2-(x_0+x_1)x+x_0x_1)dx
\end{align*}
我们知道，
\begin{align*}
\int_{x_0}^{x_3}[x^2-(x_0+x_1)x+x_0x_1]dx&=[\frac{1}{3}x_3^3-\frac{1}{2}(x_0+x_1)x_3^2+x_0x_1x_3]-[\frac{1}{3}x_0^3-\frac{1}{2}(x_0+x_1)x_0^2+x_0x_1x_0]\\&=\frac{1}{6}[2(x_3-x_0)(x_3^2+x_0^2+x_0x_3)-3(x_0x_3^2+x_1x_3^2-x_0^3-x_1x_0^2)+6x_0x_1(x_3-x_0)]\\&=\frac{1}{6}[2(x_3-x_0)(x_3^2+x_0^{2}+x_0x_3)-3[(x_0+x_1)(x_3^2-x_0^2)]+6x_0x_1(x_3-x_0)]\\&=\frac{1}{6}(x_3-x_0)(2x_3^2+2x_0^2+2x_0x_3-3(x_0+x_1)(x_3+x_0)+6x_0x_1)\\&=\frac{1}{6}(x_3-x_0)(2x_3^2-x_0^2-x_0x_3-3x_1x_3+3x_0x_1)\\&=\frac{1}{6}(x_3-x_0)^2(2x_3+x_0-3x_1)=\frac{9}{2}h^3
\end{align*}
因此，
\begin{align*}
\int_{x_0}^{x_3}(x-x_0)(x-x_1)f[x_0,x_1,x_2]dx&=(\frac{f(x_0)}{(-h)\times
(-2h)}+\frac{f(x_1)}{h\times (-h)}+\frac{f(x_2)}{2h\times
h})\int_{x_0}^{x_3}(x^2-(x_0+x_1)x+x_0x_1)dx\\&=\frac{9}{2}h^3(\frac{f(x_0)}{2h^2}-\frac{f(x_1)}{h^2}+\frac{f(x_2)}{2h^2})=\frac{9}{4}hf(x_0)-\frac{9}{2}hf(x_1)+\frac{9}{4}hf(x_2)
\end{align*}\begin{align*}
\int_{x_0}^{x_3}(x-x_0)(x-x_1)(x-x_2)f[x_0,x_1,x_2,x_3]dx
\end{align*}
我们来计算\begin{align*}
&\int_{x_0}^{x_3}(x-x_0)(x-x_1)(x-x_2)dx=\int_{x_0}^{x_1}(x-x_0)(x-x_1)(x-x_2)dx+\int_{x_1}^{x_2}(x-x_0)(x-x_1)(x-x_2)dx+\int_{x_2}^{x_3}(x-x_0)(x-x_1)(x-x_2)dx\\&=\int_{x_2}^{x_3}(x-x_0)(x-x_1)(x-x_2)dx\\&=(\frac{1}{4}x_3^4-\frac{1}{3}(x_0+x_1+x_2)x_{3}^3+\frac{1}{2}(x_0x_1+x_1x_2+x_2x_0)x_{3}^2-x_0x_1x_2x_{3})\\&-(\frac{1}{4}x_2^4-\frac{1}{3}(x_0+x_1+x_2)x_{2}^3+\frac{1}{2}(x_0x_1+x_1x_2+x_2x_0)x_{2}^2-x_0x_1x_2x_{2})\\&=\frac{1}{12}[3(x_3^2+x_2^2)(x_3+x_2)(x_3-x_2)\\&+4(x_0+x_1+x_2)(x_2-x_3)(x_2^2+x_3^2+x_2x_3)+6(x_0x_1+x_1x_2+x_2x_0)(x_3-x_2)(x_3+x_2)+12x_0x_1x_2(x_2-x_3)]\\&=\frac{1}{12}[(x_3-x_2)(3x_3^3+3x_3^2x_2+3x_2^2x_3+3x_2^3-4x_0x_2^2-4x_0x_3^2-4x_0x_2x_3-4x_1x_2^2-4x_1x_3^{2}\\&-4x_1x_2x_{3}-4x_2^3-4x_2x_3^2-4x_2^2x_3+6x_0x_1x_3+6x_0x_1x_2+6x_1x_2x_3+6x_1x_2^{2}+6x_0x_2x_3+6x_0x_2^2-12x_0x_1x_2)]\\&=\frac{1}{12}(x_3-x_2)(3x_3^3-x_2^3-x_3^2x_2-x_2^2x_3+2x_0x_2^2-4x_0x_3^2+2x_1x_{2}^2-4x_1x_3^2+2x_0x_2x_3+2x_1x_2x_3+6x_0x_1x_3-6x_0x_1x_2)\\&=\frac{1}{12}(x_3-x_2)[(x_3^3-x_2^3)+x_3^2(x_3-x_2)+x_3(x_3^2-x_2^2)-2x_0(x_3^2-x_2^2)-2x_0x_3(x_3-x_2)-2x_1(x_3^2-x_2^2)-2x_1x_3(x_3-x_2)+6x_0x_1(x_3-x_2)]\\&=\frac{1}{12}(x_3-x_2)^2[3x_3^2+x_2^2+2x_3x_2-4x_0x_3-2x_0x_2-4x_1x_3-2x_1x_2+6x_0x_1]\\&=\frac{1}{12}(x_3-x_2)^2[3(x_0+3h)^2+(x_0+2h)^2+2(x_0+3h)(x_0+2h)\\&-4x_0(x_0+3h)-2x_0(x_0+2h)-4(x_0+h)(x_0+3h)-2(x_0+h)(x_0+2h)+6x_0(x_0+h)]=\frac{9}{4}h^4
\end{align*}而
\begin{align*}
f[x_0,x_1,x_2,x_3]=\frac{f(x_0)}{(-h)(-2h)(-3h)}+\frac{f(x_1)}{h(-h)(-2h)}+\frac{f(x_2)}{2h\times
h(-h)}+\frac{f(x_3)}{3h\times 2h\times h}
\end{align*}可见，
\begin{align*}
\int_{x_0}^{x_3}(x-x_0)(x-x_1)(x-x_2)f[x_0,x_1,x_2,x_3]=\frac{9h}{4}[\frac{f(x_0)}{-6}+\frac{f(x_1)}{2}+\frac{f(x_2)}{-2}+\frac{f(x_3)}{6}]
\end{align*}综上所述，
\begin{align*}
\int_{x_0}^{x_3}\{f(x_0)+(x-x_0)f[x_0,x_1]+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+(x-x_0)(x-x_1)(x-x_2)f[x_0,x_1,x_2,x_3]\}dx=\frac{3}{8}hf(x_0)+\frac{9}{8}hf(x_1)+\frac{9}{8}hf(x_2)+\frac{3}{8}hf(x_3)
\end{align*}
下面我来证明
\begin{align*}
\int_{x_0}^{x_3} (x-x_0)(x-x_1)(x-x_2)(x-x_3)f[x,x_0,x_1,x_2,x_3]dx=\frac{-3}{80}h^5f^{(4)}(\xi)
\end{align*}由于
\begin{align*}
f[x,x_0,x_1,x_2,x_3]=\frac{f''''(\xi(x))}{4!}
\end{align*}下面我们来看
\begin{align*}
&\int_{x_0}^{x_3}(x-x_0)(x-x_1)(x-x_2)(x-x_3)dx\\&=\int_{x_0}^{x_3}[x^4-(x_0+x_1+x_2+x_3)x^3+(x_0x_1+x_1x_2+x_2x_3+x_3x_0+x_3x_1+x_0x_2)x^2-(x_1x_2x_3+x_0x_2x_3+x_0x_1x_3+x_0x_1x_2)x+x_0x_1x_2x_3]dx
\end{align*}这么麻烦的计算，只有傻瓜才会手工计算下去.反正我相信这样子继续算下去，再用一下加权积分中值定理,就能得出题目中的结论.这个结论和数值积分中的梯形法则在方法上没有任何区别.
下面给出另一种失败方案:
\begin{align*}
\int_{x_0}^{x_3} (x-x_0)(x-x_1)(x-x_2)(x-x_3)f[x,x_0,x_1,x_2,x_3]dx=\frac{-3}{80}h^5f^{(4)}(\xi)
\end{align*}我曾经想用数值积分中的辛普森方法及其误差估计 里的方法，即如下.
令
\begin{align*}
g(t)=\int_{x_0}^t(x-x_0)(x-x_1)(x-x_2)(x-x_3)f[x,x_0,x_1,x_2,x_3]dx
\end{align*}
则根据微积分第一基本定理，
\begin{align*}
g'(x_0)=g'(x_1)=g'(x_2)=g'(x_3)
\end{align*}
且$g(x_0)=0$.我们进行牛顿插值，设立插值点
\begin{align*}
x_0,x_0',x_1,x_1',x_2,x_2',x_3,x_3'
\end{align*}经过这几个插值点的牛顿插值多项式为
\begin{align*} p(x)= g(x_0)&+(x-x_0)g[x_0,x_0']\\&+(x-x_0)(x-x_0')g[x_0,x_0',x_1]\\&+(x-x_0)(x-x_0')(x-x_1)g[x_0,x_0',x_1,x_1']\\&+(x-x_0)(x-x_0')(x-x_1)(x-x_1')g[x_0,x_0',x_1,x_1',x_2]\\&+(x-x_0)(x-x_0')(x-x_1)(x-x_1')(x-x_2)g[x_0,x_0',x_1,x_1',x_2,x_2']\\&+(x-x_0)(x-x_0')(x-x_1)(x-x_1')(x-x_2)(x-x_2')g[x_0,x_0',x_1,x_1',x_2,x_2',x_3] \\&+(x-x_0)(x-x_0')(x-x_{1})(x-x_{1}')(x-x_2)(x-x_2')(x-x_3)f[x_0,x_0',x_1,x_1',x_2,x_2',x_3,x_3']\end{align*}令$x_0'\to x_0,x_1'\to x_1,x_2'\to x_2,x_3'\to x_3$,可得相应的Hermite
插值多项式为
\begin{align*}
q(x)=g(x_0)&+(x-x_0)g[x_0,x_0]\\&+(x-x_0)^2g[x_0,x_0,x_1]\\&+(x-x_0)^2(x-x_1)g[x_0,x_0,x_1,x_1]\\&+(x-x_0)^2(x-x_1)^2g[x_0,x_0,x_1,x_1,x_2]\\&+(x-x_0)^2(x-x_1)^2(x-x_2)g[x_0,x_0,x_1,x_1,x_2,x_2]\\&+(x-x_0)^2(x-x_1)^2(x-x_2)^2g[x_0,x_0,x_1,x_1,x_2,x_2,x_3]\\&+(x-x_0)^2(x-x_1)^2(x-x_2)^2(x-x_3)g[x_0,x_0,x_1,x_1,x_2,x_2,x_3,x_3] \end{align*}
令
\begin{align*}
k(x)=g(x)-q(x)
\end{align*}
可得
\begin{align*}
k(x_0)=k(x_1)=k(x_2)=k(x_3)=k'(x_0)=k'(x_1)=k'(x_2)=k'(x_3)=0
\end{align*}于是使用Rolle定理数次可得
\begin{align*}
k^{(5)}(\xi)=0
\end{align*}
也就是
\begin{align*}
g^{(7)}(\xi)=q^{(7)}(\xi)=7!g[x_0,x_0,x_1,x_1,x_2,x_2,x_3,x_3]
\end{align*}
由于
\begin{align*}
g^{(7)}(\xi)=f^{(6)}(\xi)
\end{align*}
因此
\begin{align*}
f^{(6)}(\xi)=7!g[x_0,x_0,x_1,x_{1},x_2,x_2,x_3,x_3]
\end{align*}但是到这里就放弃了，因为按照题目的意思，【$f$在$(x_0,x_3)$上四阶可导，且在$[x_0,x_3]$上三阶导函数连续.】
正确做法请见