Hans Georg Schaathun
Oktober 2016
$$I = \int_{x_1}^{x_2} f(x) dx$$
$$T = (x_2-x_1)\cdot\frac{y_1+y_2}2$$
$$\epsilon = I-T = \int_{x_1}^{x_2}g(x)dx$$
$$ \begin{align} \begin{split} \epsilon &=\int_{x_1}^{x_2} g(x)dx = -\frac12\int_{x_1}^{x_2} (x-x_1)(x_2-x)f''(x)dx \end{split} \end{align} $$
$$ \begin{align} \begin{split} & \int_{x_1}^{x_2} (x-x_1)(x_2-x)f''(x)dx \\=& \int_{x_1}^{x_2} (x-x_1)(x_2-x)g''(x)dx \\=& -2\int_{x_1}^{x_2} g(x)dx \end{split} \end{align} $$
$$|\epsilon| = \bigg|\frac12\int_{x_1}^{x_2} (x-x_1)(x_2-x)f''(x)dx\bigg|$$
$$\le \frac12\int_{x_1}^{x_2} (x-x_1)(x_2-x)|f''(x)|dx$$
$$\le \frac K2\int_{x_1}^{x_2} (x-x_1)(x_2-x)dx$$
$$\le \frac K2\int_{x_1}^{x_2} [-x^2 + (x_1+x_2)x - x_1x_2]dx$$
$$|\epsilon|\le \frac K2 \cdot\frac{(x_2-x_1)^3}6$$
$$\le \frac K{12} \cdot\bigg(\frac{(b-a)}{n}\bigg)^3$$
$$T = T_1 + T_2 + \ldots + T_n$$
$$\epsilon = \epsilon_1 + \epsilon_2 + \ldots + \epsilon_n$$
$$|\epsilon| \le n \frac K{12} \cdot\bigg(\frac{(b-a)}{n}\bigg)^3$$
$$|\epsilon| \le \frac K{12} \cdot\frac{(b-a)^3}{n^2}$$