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%%
%% An UIT Edition example
%%
%% Example 06-14-3 on page 116.
%%
%% 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}
\usepackage[utf8]{inputenc}
\setcounter{equation}{74}
\renewcommand\theequation{6.\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{Now the antiderivative of the two integrals is determined and the
values calculated:}
&= \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}
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