%% %% 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} $$\begin{array}{rl} A_1 &= \left|\int_0^1(f(x)-g(x))\diff x\right|+\left|\int_1^2(g(x)-h(x)) \diff x\right|\\ &= \left|\int_0^1 (x^2 -3x)\diff x\right|+\left|\int_1^2(x^2-5x+6) \diff x\right|\\ &= \left|\frac{x^3}{3}-\frac{3}{2}x^2\right|_0^1+\left|\frac{x^3}{3} -\frac{5}{2}x^2+6x\right|_1^2\\ &= \left|\frac{1}{3}-\frac{3}{2}\right|+\left|\frac{8}{3}-\frac{20}{2}+12 -\left(\frac{1}{3}-\frac{5}{2}+6\right)\right|\\ &= \left|-\frac{7}{6}\right|+\left|\frac{28}{6}-\frac{23}{6}\right| =\frac{7}{6}+\frac{5}{6}=2\,\textrm{AU} \end{array}$$ \end{document}