%% %% 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}