我有
\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\begin{document}
1st period. Year 2011
Time: 3 hours
\newline
\textbf{Question 1.}
\textbf{a)} Give $z = ln(1-x^2y^2)$ . Calculate $A = x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y}$ .
\textbf{b} Find extreme values of function $A = e^y(y + x^3 - 3x^2)$
\textbf{Question 2.}
\textbf{a)} Change order of integral $\int_{0}^{2} dx \int_{-x}^{\sqrt{2x - x^2}} f(x,y) dy$
\end{document}
我想要这样的
(其他公式也一样)。如何做到这一点?
答案1
我会不是使用 displaystyle 数学模式,我当然不会将积分极限置于积分符号之下/之上。即使在内联数学模式下,积分和分数表达式也清晰易读。但是,我会微调积分表达式内的间距,并特别注意减少视觉格式化;而是使用 LaTeX 机制来生成一致的页面布局。哦,一定要写 ,而不仅仅是\ln
,ln
来表示自然对数运算符,并且不要在$
终止内联数学模式的符号和尾随标点符号(例如.
或,
)之间留空格。
\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
%%\usepackage{amsfonts} % is loaded automatically by 'amsymb' package
\usepackage{amssymb}
\usepackage{enumitem}
% create a custom-style enumerated list
\newlist{myenum}{enumerate}{1}
\setlist[myenum]{label=\bfseries\alph*), ref=\alph*),
wide=0pt, leftmargin=*}
\begin{document}
\noindent
1st period. Year 2011\\
Time: 3 hours
\subsubsection*{Question 1.}
\begin{myenum}
\item Give $z = \ln(1-x^2y^2)$. Calculate
$A = x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y}$.
\item Find extreme values of function $A = e^y(y + x^3 - 3x^2)$.
\end{myenum}
\subsubsection*{Question 2.}
\begin{myenum}
\item Change order of integral
$\int_{0}^{2}\! dx \int_{-x}^{\sqrt{2x - x^2}} \!\! f(x,y)\, dy$.
\end{myenum}
\end{document}
答案2
不要忘记,该amsmath
计划intlimits
还为那些想要“俄罗斯风格”定位整合极限的人提供了选择:
% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly
% declare the paper format.
\usepackage[T1]{fontenc} % Not always necessary, but recommended.
% End of standard header. What follows pertains to the problem at hand.
\usepackage[intlimits]{amsmath}
\begin{document}
In-line: \( \int_{-\infty}^{+\infty} e^{-t^{2}} dt = \sqrt{\pi} \).
In display:\[ \int_{-\infty}^{+\infty} e^{-t^{2}} dt = \sqrt{\pi} \]
\end{document}
请注意,在这种特殊情况下,最好省略\,
之前的 dt
。
结果是