# 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

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983