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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}
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