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author | Karl Berry <karl@freefriends.org> | 2012-02-06 00:47:46 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2012-02-06 00:47:46 +0000 |
commit | f2fb432b14a3c467c6ddfee4a19874fea886307d (patch) | |
tree | 4fbba0fc79f546e17e07adbb3db78860e4d39ed8 /Master/texmf-dist/doc/latex/einfuehrung/09-07-60.ltx | |
parent | 27d1d1646fa5dc2daba1e09c2d2175c211d93def (diff) |
new examples from (voss) german book einfuehrung (2feb12)
git-svn-id: svn://tug.org/texlive/trunk@25301 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex/einfuehrung/09-07-60.ltx')
-rw-r--r-- | Master/texmf-dist/doc/latex/einfuehrung/09-07-60.ltx | 45 |
1 files changed, 45 insertions, 0 deletions
diff --git a/Master/texmf-dist/doc/latex/einfuehrung/09-07-60.ltx b/Master/texmf-dist/doc/latex/einfuehrung/09-07-60.ltx new file mode 100644 index 00000000000..ff8f1312e63 --- /dev/null +++ b/Master/texmf-dist/doc/latex/einfuehrung/09-07-60.ltx @@ -0,0 +1,45 @@ +%% +%% Ein Beispiel der DANTE-Edition +%% +%% Beispiel 09-07-60 auf Seite 468. +%% +%% Copyright (C) 2012 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{352.81416pt} +\usepackage[utf8]{inputenc} +\setcounter{equation}{72} +\renewcommand\theequation{9.\arabic{equation}} +\AtBeginDocument{\setlength\parindent{0pt}} + +\usepackage{amsmath} +\newcommand*\diff{\mathop{}\!\mathrm{d}} + +\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{Jetzt werden die Stammfunktionen der beiden Integrale + gebildet und anschließend die Werte berechnet:} + &= \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} |