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+%% 
+%%  An UIT Edition example
+%% 
+%%  Example 06-03-15 on page 94.
+%% 
+%%  Copyright (C) 2010 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{375.57637pt}
+\AtBeginDocument{\setlength\parindent{0pt}}
+\setcounter{equation}{40}
+\renewcommand\theequation{6.\arabic{equation}}
+\StartShownPreambleCommands
+\newcommand*\diff{\mathop{}\!\mathrm{d}}
+\StopShownPreambleCommands
+\begin{document}
+\begin{equation}\begin{array}{rl}
+A_1 &= \left|\int_0^1(f(x)-g(x))\diff x\right|+\left|\int_1^2(g(x)-h(x))
+       \diff x\right|\\
+    &= \left|\int_0^1 (x^2 -3x)\diff x\right|+\left|\int_1^2(x^2-5x+6)
+       \diff x\right|\\
+    &= \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\\
+    &= \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|\\
+    &= \left|-\frac{7}{6}\right|+\left|\frac{28}{6}-\frac{23}{6}\right|
+       =\frac{7}{6}+\frac{5}{6}=2\,\textrm{AU}
+\end{array}\end{equation}
+\end{document}