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数组大小

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数组大小

我有以下代码：

\documentclass[12pt]{article}
\usepackage{mathtools}
\begin{document}
    $$  \begin{array}{cl}
        f(\hat{x}_n) & =\displaystyle  \int_{-1}^{0} \hat{x}_n d t  - \int_{0}^{1} \hat{x}_n d t  \\
         & = \displaystyle  \int_{-1}^{-\frac{1}{n+1}} d t   +\int_{-\frac{1}{n+1}}^{0} -(n+1)t d t   +\int_{0}^{\frac{1}{n+1}} (n+1)t d t  +\int_{\frac{1}{n+1}}^{1} d t \\
         & =\frac{n}{n+1} +\frac{n}{2n+21} +\frac{n}{2n+2}  +\frac{n}{n+1} \\
         & = \frac{2n+1}{n+1} 
    \end{array}
    $$
\end{document}

这给了我以下信息： 在此处输入图片描述

这非常丑陋，分数的大小太小，如果我改变它，它们会重叠，线条的相同空间非常小，积分看起来不太好。我该如何改进？

答案1

对于 align编号行和align*未编号行：

在此处输入图片描述

\documentclass[12pt]{article}
\usepackage{mathtools}
\begin{document}
\begin{align}
 f(\hat{x}_n) & = \int_{-1}^{0} \hat{x}_n d t  - \int_{0}^{1} \hat{x}_n d t  \\
              & = \int_{-1}^{-\frac{1}{n+1}} d t   +\int_{-\frac{1}{n+1}}^{0} -(n+1)t d t   
                  +\int_{0}^{\frac{1}{n+1}} (n+1)t d t  +\int_{\frac{1}{n+1}}^{1} d t \\
              & =\frac{n}{n+1} +\frac{n}{2n+21} +\frac{n}{2n+2}  +\frac{n}{n+1} \\
              & = \frac{2n+1}{n+1} 
\end{align}


\begin{align*}
 f(\hat{x}_n) & = \int_{-1}^{0} \hat{x}_n d t  - \int_{0}^{1} \hat{x}_n d t  \\
              & = \int_{-1}^{-\frac{1}{n+1}} d t   +\int_{-\frac{1}{n+1}}^{0} -(n+1)t d t   
                  +\int_{0}^{\frac{1}{n+1}} (n+1)t d t  +\int_{\frac{1}{n+1}}^{1} d t \\
              & =\frac{n}{n+1} +\frac{n}{2n+21} +\frac{n}{2n+2}  +\frac{n}{n+1} \\
              & = \frac{2n+1}{n+1} 
\end{align*}
\end{document}

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