Proof
(related to Lemma: Trapezoid Rule)
- By hypothesis, the function f is a twice continuously differentiable function.
- First of all we will show that there is a mean value \xi\in[0,1] such that \int_0^1f(x)dx=\frac{f(0)+f(1)}{2}-\frac{1}{12}f^{\prime\prime}(\xi).\label{eq:E20226}\tag{1}
- For let \phi(x):=\frac{1}{2}x(1-x).
- Since we have the derivatives \phi^{\prime}(x):=\frac{1}{2}-x and \phi^{\prime\prime}(x):=-1, we have with partial integration \int_0^1f(x)dx=-\int_0^1f(x)\phi^{\prime\prime}(x)dx=\underbrace{-f(x)\phi^{\prime}(x)\;\Rule{2px}{4ex}{2ex}^1_0}_{=:A}+\underbrace{\int_{0}^{1}\phi^{\prime}(x)f’(x)dx}_{=:B}.
- The term A evaluates to A=-f(1)(\frac 12-1)+f(0)(\frac 12-0)=\frac 12f(1)+\frac 12f(0)=\frac 12(f(1)+f(0)).\label{eq:E20226a}\tag{2}
- The term B evaluates with yet another partial integration to B=\underbrace{f'(x)\phi(x)\;\Rule{2px}{4ex}{2ex}^1_0}_{=0}-\int_0^1\phi(x)f^{\prime\prime}(x)dx=-\int_0^1\phi(x)f^{\prime\prime}(x)dx.
- Since \phi(x)\ge 0 for all x\in[0,1], we can apply the mean value theorem for Riemann integrals and conclude that there exists a mean value \xi\in[0,1] such that B=-f^{\prime\prime}(\xi)\cdot\int_0^1\phi(x)dx.
- Thus, there exists a mean value \xi\in[0,1] such that B=-f^{\prime\prime}(\xi)\cdot\int_0^1\frac{1}{2}(x(1-x))dx=-f^{\prime\prime}(\xi)\frac 12\left(\int_0^1 xdx-\int_0^1 x^2dx\right) =-f^{\prime\prime}(\xi)\frac 12\left(\frac 12-\frac 13\right)=-f^{\prime\prime}(\xi)\frac 1{12}.\label{eq:E20226b}\tag{3}
- (\ref{eq:E20226}) follows now immediately from (\ref{eq:E20226a}) and (\ref{eq:E20226b}).
- Now, let n\ge 1 be a natural number and let h:=\frac{b-a}n.
- We define the values x_\nu:=a+h\nu for \nu=0,1,\ldots,n.
- By integration by substitution in (\ref{eq:E20226}), we get the result that there is a mean value \xi_\nu\in[x_{\nu-1},x_{\nu}] such that we can find the Riemann integral for \nu=1,\ldots,n \int_{a+(\nu-1)h}^{a+\nu h} f(\eta)d\eta=h\cdot \frac{f(x_{\nu-1})+f(x_\nu)}{2}-\frac{h^3}{12}f^{\prime\prime}(\xi_\nu).\label{eq:E20226c}\tag{4}
- Summing (\ref{eq:E20226c}) over all values of \nu=1,\ldots,n gives us \int_{a}^{b} f(x)dx=\left(\frac{f(a)}{2}+\sum_{\nu=1}^{n-1}f(x_\nu)+\frac{f(b)}{2}\right)h+R \label{eq:E20226d}\tag{5} with an absolute error term |R|=\frac{h^3}{12}\sum_{\nu=1}^n f^{\prime\prime}(\xi_\nu).
- Replacing the summands f^{\prime\prime}(\xi_\nu) by the supremum K:=\sup\{|f^{\prime\prime}(x)|\mid\text{for all } x\in[a,b]\} we get the approximation of the error term:
|R|\le \frac{h^3}{12}\sum_{\nu=1}^n K=\frac{h^3}{12}nK=\frac{(b-a)^3}{n^3}\frac{1}{12}nK=\frac{(b-a)^3}{n^2}\frac{1}{12}K=\frac{(b-a)h^2}{12}K.
∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983