What is the meaning of $$$ in LaTex?
The markup is completely wrong and should be removed, but to answer the question of what it means
\begin{equation}
$$$
\notag \displaystyle\int_{-1}^{2} x^3\:dx-\displaystyle \int_{-1}^{2} 2x\:dx \\ \\ \\
\notag \displaystyle \left. \frac{x^4}{4}\right|_{-1}^{2}-\left.\frac{2x^2}{2}\right|_{-1}^{2} \\ \\ \\
\notag \displaystyle \left. \frac{x^4}{4}\right|_{-1}^{2}-\left.x^2\right|_{-1}^{2}\\ \\ \\
\left(4-\frac{1}{4}\right)-(4-1)\\ \\ \\
\frac{15}{4}-3 \ = \ \textcolor{red}{\frac{3}{4}}
$$$
\end{equation}
\begin{equation}
starts a latex enviornment then does $$
to start display math.
Then you have $$
which ends display math with an empty display, but adding vertical space.
Then you have $
which starts inline math which is why you need \displaystyle
everywhere.
Then you have $
which ends the inline math,
Then you have $$
which starts another display math, adding vertical space.
Then you have \end{equation}
which closes the empty display math, adding more spurious vertical space and closing the latex environment.
The $$$
directives appear to be there to cope with and work around some rather questionable coding practices embedded in your code. The instances of $$$
occur in pairs. The first instance of each pair is interpreted by TeX as $$
(end of displaymath mode that was initiated by the preceding \begin{equation}
environment plus a line break) followed by $
(start of inline math mode). The second instance is interpreted by TeX as $
(end of inline math mode) followed by $$
(line break followed by start of displaymath mode, to be terminated later on, by the subsequent \end{equation}
instruction). This method "works", but just barely, and requires the use of lots and lots of \displaystyle
directives that clutter up the code.
Since you're using multi-line equations, you shouldn't be using equation
environments to begin with. Instead, you should use align*
environments along the lines shown in the following example. You should also get rid of the "\:
" and "\
" spacing directives. (Use \,
to offset the dummy variable of integration, not \:
.)
\documentclass[12pt]{article}
\usepackage[letterpaper]{geometry}
\geometry{top=1.5cm, bottom=2.5cm, hmargin=3cm}
\usepackage[fleqn]{amsmath}
\usepackage{xcolor}
\begin{document}
\section*{Solucionario}
\begin{enumerate}
\item $\displaystyle \int_{-1}^{2} x^3-2x\,dx $
\begin{align*}
&= \int_{-1}^{2} x^3\,dx-\int_{-1}^{2} 2x\,dx \\
&= \left. \frac{x^4}{4}\right|_{-1}^{2}-\left.\frac{2x^2}{2}\right|_{-1}^{2} \\
&= \left. \frac{x^4}{4}\right|_{-1}^{2}-\left.x^2\right|_{-1}^{2}\\
&= \left(4-\frac{1}{4}\right)-(4-1)\\
&= \frac{15}{4}-3 = \textcolor{red}{\frac{3}{4}}
\end{align*}
\item $\displaystyle \int_{0}^{1} x^{4/5}\,dx $
\begin{align*}
&= \left.\frac{x^{9/5}}{9/5}\right|_{0}^{1} \\
&= \left.\frac{5}{9}x^{9/5}\right|_{0}^{1} =
\textcolor{red}{\frac{5}{9}}
\end{align*}
\item $\displaystyle \int_{0}^{1} \frac{4}{t^2+1}\,dt $
\begin{align*}
&= 4\int_{0}^{1} \frac{1}{t^2+1}\,dt \\
&= \left.4 \tan^{-1} t\:\right|_{0}^{1}\\
&= \textcolor{red}{180}=\textcolor{red}{\pi}
\end{align*}
\end{enumerate}
\end{document}
Addendum: Here's another attempt to fix up your code; note the use of top-aligned aligned
environments to unify the multiline equations associated with each \item
. If you need to allow page breaks inside the aligned
environments, I suggest you add the instruction \allowdisplaybreaks
in the preamble.
\documentclass[12pt]{article}
\usepackage[letterpaper]{geometry}
\geometry{top=1.5cm, bottom=2.5cm, hmargin=3cm}
\usepackage{amsmath}
\usepackage{xcolor}
\begin{document}
\section*{Solucionario}
\begin{enumerate}
\item $\!\begin{aligned}[t]
&\int_{-1}^2 (x^3-2x)\,dx \\
&\quad= \int_{-1}^2 x^3\,dx -\int_{-1}^2 2x\,dx \\
&\quad= \frac{x^4}{4} \,\bigg\vert_{-1}^2
-\frac{2x^2}{2} \,\bigg\vert_{-1}^2 \\
&\quad= \frac{1}{4}\,x^4 \,\big\vert_{-1}^2
-x^2 \,\big\vert_{-1}^2\\
&\quad= \frac{1}{4}(16-1)-(4-1)\\[1ex]
&\quad= \frac{15}{4}-3 = \textcolor{red}{\frac{3}{4}}
\end{aligned}$
\item $\!\begin{aligned}[t]
&\int_0^1 x^{4/5}\,dx \\
&\quad= \frac{x^{9/5}}{9/5} \,\bigg\vert_0^1 \\
&\quad= \frac{5}{9}\,x^{9/5} \,\big\vert_0^1 =
\textcolor{red}{\frac{5}{9}}
\end{aligned}$
\item $\!\begin{aligned}[t]
&\int_0^1 \frac{4}{t^2+1}\,dt \\
&\quad= 4 \int_0^1 \frac{1}{t^2+1}\,dt \\
&\quad= 4 \tan^{-1} t \,\big\vert_0^1\\
&\quad= 4\,\frac{\pi}{4}=\textcolor{red}{\pi} = 180^\circ
\end{aligned}$
\end{enumerate}
\end{document}