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-%%
-%% Ein Beispiel der DANTE-Edition
-%%
-%% 1. Auflage
-%%
-%% Beispiel 09-07-60 auf Seite 468.
-%%
-%% Copyright (C) 2011 Herbert Voss
-%%
-%% It may be distributed and/or modified under the conditions
-%% of the LaTeX Project Public License, either version 1.3
-%% of this license or (at your option) any later version.
-%%
-%% See http://www.latex-project.org/lppl.txt for details.
-%%
-%%
-%% ====
-% Show page(s) 1
-%%
-%%
-\documentclass[]{exaarticle}
-\pagestyle{empty}
-\setlength\textwidth{352.81416pt}
-\usepackage[utf8]{inputenc}
-\setcounter{equation}{72}
-\renewcommand\theequation{9.\arabic{equation}}
-\AtBeginDocument{\setlength\parindent{0pt}}
-\StartShownPreambleCommands
-\usepackage{amsmath}
-\newcommand*\diff{\mathop{}\!\mathrm{d}}
-\StopShownPreambleCommands
-\begin{document}
-\begin{align}
-A_{1}
- &= \left|\int_0^1(f(x)-g(x))\diff x\right| +\left| \int _1^2(g(x)-h(x))\diff x
- \right|\nonumber\\
- &= \left|\int_0^1(x^2-3x)\diff x\right| +\left| \int _1^2(x^2-5x+6)\diff x
- \right|\nonumber
-\intertext{Jetzt werden die Stammfunktionen der beiden Integrale
- gebildet und anschließend die Werte berechnet:}
- &= \left|\frac{x^3}{3}-\frac{3}{2}x^2\right|_0^1+\left| \frac{x^3}{3}-
- \frac{5}{2}x^2+6x\right|_1^2\nonumber\\
- &= \left|\frac{1}{3}-\frac{3}{2}\right| +\left| \frac{8}{3}- \frac{20}{2}+12-
- \left(\frac{1}{3}-\frac{5}{2}+6\right) \right| \nonumber\\
- &= \left|-\frac{7}{6}\right| +\left| \frac{28}{6}-\frac{23}{6} \right| =\frac{7}{6}+
- \frac{5}{6}=2\,\text{FE}
-\end{align}
-\end{document}