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+%%
+%% Ein Beispiel der DANTE-Edition
+%%
+%% 2. Auflage
+%%
+%% Beispiel 09-07-59 auf Seite 443.
+%%
+%% Copyright (C) 2016 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}