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I have been using multicol in my lists of exercises. But I'm trying to change the layout of the column. Rather than

a) xxx d) xxx g) xxx

b) xxx e) xxx h) xxx

c) xxx f) xxx i) xxx

for instance, I would like

a) xxx b) xxx c) xxx

d) xxx e) xxx f) xxx

g) xxx h) xxx i) xxx

Is it possible to use with my particular commands?

In time, I was trying to use Enumerate Horizontally with multicols but I believe this question is different and I don't know how to change it for me.

Here is a minimal working example.

\documentclass[a4paper,brazil, 12pt]{report}%{\article}
\usepackage[margin=1.5cm]{geometry}
\usepackage{amsmath,amsfonts,amscd,bezier}
\usepackage{amssymb}
\usepackage{babel}
\usepackage[latin1]{inputenc}
\usepackage{graphicx}
%\usepackage[dvips]{graphicx}
\usepackage{color}
\usepackage[shortlabels]{enumitem}
\usepackage{multicol}
\usepackage{tasks}
\usepackage{hyperref}

\newcommand{\dis}{\displaystyle}

\newcounter{theeq} \setcounter{theeq}{0}
\newcommand{\eq}{ 
    \

    \noindent 
    \refstepcounter{theeq}\textbf{\arabic{theeq}}. }


\begin{document}

\eq Calcule
\begin{multicols}{3}
    \begin{enumerate}[$(a)$, leftmargin=3.2em]
        \item $\dis \int  (\sqrt{x^3}+ \sqrt[3]{x^2}) \;dx$
        \item $\dis \int\, \dfrac{x^9 + 4x^3 -6x^2 + 3}{x^5}\ dx $
        \item $\dis \int\, \dfrac{1+x}{1+x^2}\;dx $
        \item $\int x\sqrt{1-x^2}\;dx$
        \item $\int x^2 \;dx$
        \item $\int \cos(x)\;dx$
    \end{enumerate}
\end{multicols}

Is it possible like bellow?

\setcounter{theeq}{0}

\eq Calcule
\begin{multicols}{3}
    \begin{enumerate}[$(a)$, leftmargin=3.2em]
        \item $\dis \int  (\sqrt{x^3}+ \sqrt[3]{x^2}) \;dx$
        \item $\dis \int\, \dfrac{x^9 + 4x^3 -6x^2 + 3}{x^5}\ dx $
        \item $\dis \int\, \dfrac{1+x}{1+x^2}\;dx $
    \end{enumerate}
\end{multicols}
\begin{multicols}{3}
    \begin{enumerate}[, leftmargin=3.2em]
        \item[$(d)$] $\int x\sqrt{1-x^2}\;dx$
        \item[$(e)$] $\int x^2 \;dx$
        \item[$(f)$] $\int \cos(x)\;dx$
    \end{enumerate}
\end{multicols}

\end{document}
  • thanks for edited my question. – Moura May 29 '19 at 18:55
1

There are lots of ways to do this sort of thing. just to illustrate the point...

\documentclass[a4paper,brazil, 12pt]{report}%{\article}
\usepackage[margin=1.5cm, showframe]{geometry}% added showframe for demonstration
\usepackage{amsmath,amsfonts,amscd,bezier}
\usepackage{amssymb}
\usepackage{babel}
\usepackage[latin1]{inputenc}
\usepackage{graphicx}
%\usepackage[dvips]{graphicx}
\usepackage{color}
%\usepackage[shortlabels]{enumitem}
\usepackage{multicol}
\usepackage{tasks}
\usepackage{hyperref}

\newcommand{\dis}{\displaystyle}

\newcounter{theeq} \setcounter{theeq}{0}
\newcommand{\eq}{\par\refstepcounter{theeq}\textbf{\arabic{theeq}}.~}

\newcounter{block}[theeq]% auto-reset
\newcommand{\block}[1]{% #1 = equation to solve (math mode)
  \refstepcounter{block}% assuming you might want to use \label inside #1
  \makebox[\labelwidth][r]{(\alph{block})\hspace{\labelsep}}% \labelwidth and \labelsep are used by \item
  \makebox[\dimexpr \textwidth/3-\leftskip/3-\labelwidth][l]{$\displaystyle #1$}%
  \allowbreak\ignorespaces}

\newenvironment{myindent}{\parindent=0pt
  \par
  \raggedright% for roundoff errors
  \leftskip=3.2em}{\par}

\begin{document}

\eq Calcule
\begin{myindent}
  \block{\int  (\sqrt{x^3}+ \sqrt[3]{x^2}) \;dx}%
  \block{\int\, \dfrac{x^9 + 4x^3 -6x^2 + 3}{x^5}\ dx}%
  \block{\int\, \dfrac{1+x}{1+x^2}\;dx }%
  \block{\int x\sqrt{1-x^2}\;dx}
  \block{\int x^2 \;dx}
  \block{\int \cos(x)\;dx}
\end{myindent}

\end{document}
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I was able to find the solution to the problems I wanted. If someone wants to know how I did, follow the code.

Thanky you for your time (and my apologies for the inconvenience).

\documentclass[a4paper,brazil, 12pt]{report}%{\article}
\usepackage[margin=1.5cm]{geometry}
\usepackage{amsmath,amsfonts,amscd,bezier}
\usepackage{amssymb}
\usepackage{babel}
\usepackage[latin1]{inputenc}
\usepackage{graphicx}
%\usepackage[dvips]{graphicx}
\usepackage{color}
\usepackage[shortlabels]{enumitem}
\usepackage{multicol}
\usepackage{tasks}
\usepackage{hyperref}

\everymath{\displaystyle}

\newcounter{theeq} \setcounter{theeq}{0}
\newcommand{\eq}{ 
    \

    \noindent 
    \refstepcounter{theeq}\textbf{\arabic{theeq}}. }


\begin{document}

\eq Calcule
\begin{multicols}{3}
    \begin{enumerate}[$(a)$, leftmargin=3.2em]\everymath{\displaystyle}
        \item $ \int  (\sqrt{x^3}+ \sqrt[3]{x^2}) \;dx$
        \item $ \int\, \dfrac{x^9 + 4x^3 -6x^2 + 3}{x^5}\ dx $
        \item $ \int\, \dfrac{1+x}{1+x^2}\;dx $
        \item $ \int x^2\sqrt{1-x}\;dx$
        \item $\int x^2 \;dx$
        \item $\int \cos(x)\;dx$
    \end{enumerate}
\end{multicols}



\setcounter{theeq}{0}

\eq Calcule
\begin{tasks}[counter-format=$(tsk[a])$,item-indent=3.5em, label-offset=1.00em, ](3)%
    \task $ \int  (\sqrt{x^3}+ \sqrt[3]{x^2}) \;dx$
    \task $ \int\, \dfrac{x^9 + 4x^3 -6x^2 + 3}{x^5}\ dx $
    \task $ \int\, \dfrac{1+x}{1+x^2}\;dx $
    \task \label{eqd} $ \int x^2\sqrt{1-x}\;dx$
    \task $\int x^2 \;dx$
    \task $\int \cos(x)\;dx$
\end{tasks}

(Hint: In \ref{eqd} use that $u = 1-x$)

\end{document}

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