3

I'm somewhat new to latex and I was trying to write an equation with an integral.

As seen in the image below the integral symbol generated by \int \dee x looks very small compared to the integrand.

enter image description here

The preview generated by vscode (IDE that I'm using to edit my file) shows the desired output.

enter image description here

Solutions that didn't help

  • Ihe closest I got after reading multiple similar questions is by using \displaystyle. The only drawback was that using \displaystyle inserted undesired line spacing.
  • I tried using the bigints package however I just couldn't get the latex file to compile.

Goal To produce an appropriately sized integral sign without any other changes to the layout (like increased linespacing) which is easily reproduced and implementable (yes, I can find some complicated way to change everything and eyeball what I want, but there just has got to be a better way that adds abstraction.)

Heres the output of pdflatex --version in case that's relevant.

zain@ubuntu:~/University$ pdflatex --version
pdfTeX 3.141592653-2.6-1.40.25 (TeX Live 2023/Debian)
kpathsea version 6.3.5
Copyright 2023 Han The Thanh (pdfTeX) et al.
There is NO warranty.  Redistribution of this software is covered by the terms of both the pdfTeX copyright and the Lesser GNU General Public License.
For more information about these matters, see the file named COPYING and the pdfTeX source.
Primary author of pdfTeX: Han The Thanh (pdfTeX) et al.
Compiled with libpng 1.6.40; using libpng 1.6.40
Compiled with zlib 1.2.13; using zlib 1.2.13
Compiled with xpdf version 4.04

edit

Thanks for the replies, from what I can see, my question was unclear so I'll elaborate here.

Here is the complete code of my document (without /displaystyle):

\usepackage[letterpaper, margin=0.05in]{geometry}
\usepackage{multicol}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{array}
\usepackage{xcolor,colortbl}
\usepackage{bigints}
\newcommand{\dee}{\mathop{\mathrm{d}\!}}

\begin{document}
\begin{multicols}{2}
    \section{Known Derivatives}
        \begin{enumerate}
            \item $\dfrac{\dee}{\dee x} k= 0$
            \item $\dfrac{\dee}{\dee x} x= 1$
            \item $\dfrac{\dee}{\dee x} x^2 = 2x$
            \item $\dfrac{\dee}{\dee x} x^3 = 3x^2$
            \item $\dfrac{\dee}{\dee x} x^n = nx^{n-1}$
            \item $\dfrac{\dee}{\dee x} e^x = e^x$
            \item $\dfrac{\dee}{\dee x} e^{kx} = ke^{kx}$
            \item $\dfrac{\dee}{\dee x} \ln(x) = \dfrac{\dee}{\dee x} \log_e(x) = \dfrac{1}{x}$
            \item $\dfrac{\dee}{\dee x} \sin x = \cos x$
            \item $\dfrac{\dee}{\dee x} \cos x = -\sin x$
            \item $\dfrac{\dee}{\dee x} \cos k x = -k \sin kx$
            \item $\dfrac{\dee}{\dee x} \tan x = \dfrac{\dee}{\dee x} \dfrac{\sin x}{\cos x} =\sec^2 x $
            \item $\dfrac{\dee}{\dee x} \tan kx = k\sec^2kx$
            \item $\dfrac{\dee}{\dee x} \csc x = \dfrac{\dee}{\dee x} \dfrac{1}{\sin x} = - \csc x \cot x$
            \item $\dfrac{\dee}{\dee x} \sec x = \dfrac{\dee}{\dee x} = \sec x \tan x$
            \item $\dfrac{\dee}{\dee x} \cot x = \dfrac{\cos x}{\sin x} = - \csc^2 x$
            \item $\dfrac{\dee}{\dee x} \arcsin x = \dfrac{1}{\sqrt{1-x^2}}$
            \item $\dfrac{\dee}{\dee x} \arccos x = -\dfrac{1}{\sqrt{1-x^2}}$
            \item $\dfrac{\dee}{\dee x} \arctan x = \dfrac{1}{1+x^2}$
        \end{enumerate}
    \columnbreak
    \section{Known Integrals}
    \begin{enumerate}
        \item $\int [af(x)+bg(x)] \dee x = a\int f(x)\ \dee{x}+b\int g(x)\ \dee{x}\ +\ C$
        \item $\int [f(x)+g(x)] \dee x = \int f(x)\ \dee{x}+\int g(x)\ \dee{x}\ +\ C$
        \item $\int [f(x)-g(x)] \dee x = \int f(x)\ \dee{x}-\int g(x)\ \dee{x}\ +\ C$
        \item $\int af(x) \dee x = a\int f(x)\ \dee{x}\ +\ C$
        \item $\int u(x)v'(x) \dee x = u(x)v(x)-\int u'(x)v(x)\ \dee{x}\ +\ C$
        \item $\int f\big(y(x)\big)y'(x) \dee x = F\big(y(x)\big)\\\hbox{ where }F(y)=\int f(y)\ \dee{y}$
        \item $\int a \dee x = ax\ +\ C$
        \item $\int x^a \dee x = \frac{x^{a+1}}{a+1}+C\hbox{ if }a\ne-1$
        \item $\int \dfrac{1}{x} \dee x = \ln|x|+C$
        \item $\int [g(x)^ag'(x)] \dee x= \frac{g(x)^{a+1}}{a+1}+C\hbox{ if }a\ne -1$
        \item $\int \dee x$
    \end{enumerate}
\end{multicols}
\end{document}

And here's the render

enter image description here

After adding displaystyle the render looks like: enter image description here

As you can see, the integral is more appropriately sized but there is a lot of unwanted space between the lines. I want line spacing similar to the first render and the integral signs similar to the second render.

  • Basically the implementation I'm (hopefully) looking for is something like \left[\right] in the context of how it adapts the height to match its contents.
6
  • 1
    Welcome, but \dee=\mathrm{d}x? Can you post a minimal working example? We can see your tex code?
    – Sebastiano
    Feb 6 at 20:28
  • 1
  • 2
    you should clarify your question, the integral is in two sizes \textstyle and \displaystyle the former is small enough to fit in a normal paragraph linespacing, the latter is not so will force the lines apart if you use it inline. so yor requirements to have a larger size and not change the line spacing appear contradictory Feb 6 at 20:31
  • 1
    Welcome to TeX.SE! Please show us a short compilable TeX code showing what you have tried. Then we do not have to guess what you are doing ...
    – Mensch
    Feb 6 at 20:36
  • 1
    that is the feature of \displaystyle it does not (directly) have any effect on vertical space, that is just larger as the lines don't fit because of the biger symbols. The textstyle only looks small as you are using \dfrac not \frac that is applying \displaystyle to get a larger fraction. the difference between your images does not look due to displaystyle, the first has unbalanced columns, the second has balanced columns forcing the second column to have stretched white space Feb 6 at 21:10

1 Answer 1

6

The spacing you show is not due to \displaystyle (or at least not directly) you have balanced columns so tex is stretching the white space to make them equal length. If you use multicols* for unbalanced columns, and use \newcolumn not \columnbreak, you get

enter image description here

\documentclass{article}
\usepackage[letterpaper, margin=0.05in]{geometry}
\usepackage{multicol}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{array}
\usepackage{xcolor,colortbl}
\usepackage{bigints}
\newcommand{\dee}{\mathop{\mathrm{d}\!}}

\begin{document}
\begin{multicols*}{2}
  \section{Known Derivatives}
        \begin{enumerate}
            \item $\displaystyle \dfrac{\dee}{\dee x} k= 0$
            \item $\displaystyle \dfrac{\dee}{\dee x} x= 1$
            \item $\displaystyle \dfrac{\dee}{\dee x} x^2 = 2x$
            \item $\displaystyle \dfrac{\dee}{\dee x} x^3 = 3x^2$
            \item $\displaystyle \dfrac{\dee}{\dee x} x^n = nx^{n-1}$
            \item $\displaystyle \dfrac{\dee}{\dee x} e^x = e^x$
            \item $\displaystyle \dfrac{\dee}{\dee x} e^{kx} = ke^{kx}$
            \item $\displaystyle \dfrac{\dee}{\dee x} \ln(x) = \dfrac{\dee}{\dee x} \log_e(x) = \dfrac{1}{x}$
            \item $\displaystyle \dfrac{\dee}{\dee x} \sin x = \cos x$
            \item $\displaystyle \dfrac{\dee}{\dee x} \cos x = -\sin x$
            \item $\displaystyle \dfrac{\dee}{\dee x} \cos k x = -k \sin kx$
            \item $\displaystyle \dfrac{\dee}{\dee x} \tan x = \dfrac{\dee}{\dee x} \dfrac{\sin x}{\cos x} =\sec^2 x $
            \item $\displaystyle \dfrac{\dee}{\dee x} \tan kx = k\sec^2kx$
            \item $\displaystyle \dfrac{\dee}{\dee x} \csc x = \dfrac{\dee}{\dee x} \dfrac{1}{\sin x} = - \csc x \cot x$
            \item $\displaystyle \dfrac{\dee}{\dee x} \sec x = \dfrac{\dee}{\dee x} = \sec x \tan x$
            \item $\displaystyle \dfrac{\dee}{\dee x} \cot x = \dfrac{\cos x}{\sin x} = - \csc^2 x$
            \item $\displaystyle \dfrac{\dee}{\dee x} \arcsin x = \dfrac{1}{\sqrt{1-x^2}}$
            \item $\displaystyle \dfrac{\dee}{\dee x} \arccos x = -\dfrac{1}{\sqrt{1-x^2}}$
            \item $\displaystyle \dfrac{\dee}{\dee x} \arctan x = \dfrac{1}{1+x^2}$
        \end{enumerate}
    \newcolumn
    \section{Known Integrals}
    \begin{enumerate}
        \item $\displaystyle \int [af(x)+bg(x)] \dee x = a\int f(x)\ \dee{x}+b\int g(x)\ \dee{x}\ +\ C$
        \item $\displaystyle \int [f(x)+g(x)] \dee x = \int f(x)\ \dee{x}+\int g(x)\ \dee{x}\ +\ C$
        \item $\displaystyle \int [f(x)-g(x)] \dee x = \int f(x)\ \dee{x}-\int g(x)\ \dee{x}\ +\ C$
        \item $\displaystyle \int af(x) \dee x = a\int f(x)\ \dee{x}\ +\ C$
        \item $\displaystyle \int u(x)v'(x) \dee x = u(x)v(x)-\int u'(x)v(x)\ \dee{x}\ +\ C$
        \item $\displaystyle \int f\big(y(x)\big)y'(x) \dee x = F\big(y(x)\big)\\\hbox{ where }F(y)=\int f(y)\ \dee{y}$
        \item $\displaystyle \int a \dee x = ax\ +\ C$
        \item $\displaystyle \int x^a \dee x = \frac{x^{a+1}}{a+1}+C\hbox{ if }a\ne-1$
        \item $\displaystyle \int \dfrac{1}{x} \dee x = \ln|x|+C$
        \item $\displaystyle \int [g(x)^ag'(x)] \dee x= \frac{g(x)^{a+1}}{a+1}+C\hbox{ if }a\ne -1$
        \item $\displaystyle \int \dee x$
    \end{enumerate}
\end{multicols*}
\end{document}
1
  • Thanks a lot! This was the exact fix I was looking for. Feb 6 at 21:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .