You can use
allowdisplaybreaks[1] to allow displaybreaks. The other thing is, your equations are just too long to fit the page.
Code: Select all
\documentclass[12pt,oneside,letterpaper]{report}
\usepackage{amssymb,amsthm}
\usepackage[spanish,english]{babel}
\usepackage{amssymb,amsfonts}
\usepackage{amsmath}
\usepackage{amscd}
\usepackage[letterpaper,left=35mm,right=23mm]{geometry}
\usepackage{fancyhdr}
\usepackage{fancyhdr,endnotes}
\usepackage{multicol}
\usepackage{cite}
%\usepackage[sort&compress]{natbib}
\usepackage{}
%\usepackage[dvips]{graphicx}
\pagestyle{fancy}
\setlength{\headheight}{14.5pt}
\usepackage{algorithm}
%\usepackage{algorithmic}
%\usepackage{epsf,graphicx}
\usepackage{graphics,graphicx,psfrag}
\usepackage[ansinew]{inputenc}
\usepackage{float}
%\usepackage{subfigure}% Deprecated for a decade
\floatplacement{figure}{ht}
\usepackage{color}
\usepackage{amssymb,amsthm}
\usepackage[spanish,english]{babel}
\usepackage{amssymb,amsfonts}
\usepackage{amsmath}
\usepackage{amscd}
\usepackage[letterpaper,left=35mm,right=23mm]{geometry}
\usepackage{fancyhdr}
\usepackage{fancyhdr,endnotes}
\usepackage{multicol}
\usepackage{cite}
%\usepackage[sort&compress]{natbib}
%\usepackage[dvips]{graphicx}
\pagestyle{fancy}
\usepackage{algorithm}
%\usepackage{algorithmic}
%\usepackage{epsf,graphicx}
\usepackage{graphics,graphicx,psfrag}
\usepackage[ansinew]{inputenc}
\usepackage{float}
\usepackage{subfigure}
\floatplacement{figure}{ht}
\usepackage{color}
\usepackage{amssymb,amsthm}
\usepackage[spanish,english]{babel}
\usepackage{amssymb,amsfonts}
\usepackage{amsmath}
\usepackage{amscd}
\usepackage[letterpaper,left=35mm,right=23mm]{geometry}
\usepackage{fancyhdr}
\usepackage{fancyhdr,endnotes}
\usepackage{multicol}
\usepackage{cite}
%\usepackage[sort&compress]{natbib}
\usepackage{}
%\usepackage[dvips]{graphicx}
\pagestyle{fancy}
\usepackage{algorithm}
%\usepackage{algorithmic}
%\usepackage{epsf,graphicx}
\usepackage{graphics,graphicx,psfrag}
\usepackage[ansinew]{inputenc}
\usepackage{float}
%\usepackage{subfigure}
\floatplacement{figure}{ht}
\usepackage{color}
\begin{document}
\allowdisplaybreaks[1]
\begin{align*}
\int_{a}^{b}|s_4(x)|\,dx &= -\frac{1}{24}\int_{a}^{\frac{a+b}{2}} (x-a)^3\left(x-\frac{a+2b}{3}\right)\,dx - \frac{1}{24}\int_{\frac{a+b}{2}}^{b} (x-b)^3\left(x-\frac{2a+b}{3}\right)\,dx\\
&=-\frac{1}{24}\left[\int_{a}^{\frac{a+b}{2}}(x^3-3ax^2+3a^2x-a^3)\left(x-\frac{a+2b}{3}\right)\,dx\right.\\
&\left.+\int_{\frac{a+b}{2}}^{b}(x^3-3bx^2+3b^2x-b^3)\left(x-\frac{2a+b}{3}\right)\,dx\right]\\
&=-\frac{1}{24}\left[\int_{a}^{\frac{a+b}{2}}\left[x^4-3ax^3+3a^2x^2-a^3x-x^3\left(\frac{a+2b}{3}\right)\right.\right.\\
&\left.+ax^2(a+2b)-a^2x(a+2b)+\frac{a^3}{3}(a+2b)\right]\,dx\\
&+\left.\int_{\frac{a+b}{2}}^{b}\left[x^4-3bx^3+3b^2x^2-b^3x - \frac{x^3}{3}(2a+b)+bx^2(2a+b)-b^2x(2a+b)+\frac{b^3}{3}(2a+b)\right]\,dx\right]\\
&=-\frac{1}{24}\left[\frac{x^5}{5}-\frac{3}{4}ax^4+a^2x^3-\frac{a^3x^2}{2}-\frac{x^4}{12}(a+2b)+\frac{ax^3}{3}(a+2b)-\frac{a^2x^2}{2}\right.\\
&\left.\left.+\frac{a^3}{3}(a+2b)x\right]\right|_a^{\frac{a+b}{2}}-\frac{1}{24}\left[\frac{x^5}{5}-\frac{3}{4}bx^4+b^2x^3-\frac{b^3x^2}{2}-\frac{x^4}{12}(2a+b)\right.\\
&\left.\left.+\frac{bx^3}{3}(2a+b)-\frac{b^2x^2}{2}+\frac{b^3}{3}(2a+b)x\right]\right|_{\frac{a+b}{2}}^{b}\\
&=-\frac{1}{24}\left[\frac{1}{5}\left(\frac{a+b}{2}\right)^5-\frac{3a}{4}\left(\frac{a+b}{2}\right)^4+a^2\left(\frac{a+b}{2}\right)^3-\frac{a^3}{2}\left(\frac{a+b}{2}\right)^2\right.\\
&-\frac{1}{12}\left(\frac{a+b}{2}\right)^4(a+2b)+\frac{a}{3}\left(\frac{a+b}{2}\right)^3(a+2b)-\frac{a^2}{2}\left(\frac{a+b}{2}\right)^2(a+2b)\\
&+\frac{a^3}{3}(a+2b)\left(\frac{a+b}{2}\right)-\frac{a^5}{5}+\frac{3}{4}a^5-a^5+\frac{a^5}{2}+\frac{a^4}{12}(a+2b)-\frac{a^4}{3}(a+2b)\\
&+\frac{a^4}{2}(a+2b)-\frac{a^4}{3}(a+2b)\\
&+\frac{b^5}{5}-\frac{3}{4}b^5+b^5-\frac{b^5}{2}^-\frac{b^4}{12}(2a+b)+\frac{b^4}{3}(2a+b)-\frac{b^4}{2}(2a+b)+\frac{b^4}{3}(2a+b)\\
&-\frac{1}{5}\left(\frac{a+b}{2}\right)^5+\frac{3b}{4}\left(\frac{a+b}{2}\right)^4-b^2\left(\frac{a+b}{2}\right)^3+\frac{b^3}{2}\left(\frac{a+b}{2}\right)^2\\
&\left.+\frac{1}{12}\left(\frac{a+b}{2}\right)^4(2a+b)-\frac{b}{3}\left(\frac{a+b}{2}\right)^3(2a+b)+\frac{b^2}{2}\left(\frac{a+b}{2}\right)^2(2a+b)-\frac{b^3}{3}(2a+b)\left(\frac{a+b}{2}\right)\right]\\
&=-\frac{1}{24}\left[\frac{b^5-a^5}{20}+\frac{1}{12}(b^5-a^5+2ab^4-2ba^4)+2\left(\frac{a+b}{2}\right)^4(b-a)\right.\\
&\left.+\frac{4}{3}(a^2-b^2)\left(\frac{a+b}{2}\right)^3+\left(\frac{a+b}{2}\right)^2[b^3-a^3+b^2a-a^2b]+\left(\frac{a+b}{2}\right)\left(\frac{a^3}{3}(a+2b)-\frac{b^3}{3}(2a+b)\right)\right]\\
\end{align*}
\end{document}
\documentclass[]{article}
\usepackage{mathtools}
\allowdisplaybreaks[1]
\begin{document}
% \item[3c] Si $p=\infty$, entonces $q=1$ y se tiene lo siguiente
\begin{align*}
\int_{a}^{b}|s_4(x)|\,dx &= -\frac{1}{24}\int_{a}^{\frac{a+b}{2}} (x-a)^3\left(x-\frac{a+2b}{3}\right)\,dx - \frac{1}{24}\int_{\frac{a+b}{2}}^{b} (x-b)^3\left(x-\frac{2a+b}{3}\right)\,dx\\
&=-\frac{1}{24}\left[\int_{a}^{\frac{a+b}{2}}(x^3-3ax^2+3a^2x-a^3)\left(x-\frac{a+2b}{3}\right)\,dx\right.\\
&\left.+\int_{\frac{a+b}{2}}^{b}(x^3-3bx^2+3b^2x-b^3)\left(x-\frac{2a+b}{3}\right)\,dx\right]\\
&=-\frac{1}{24}\left[\int_{a}^{\frac{a+b}{2}}\left[x^4-3ax^3+3a^2x^2-a^3x-x^3\left(\frac{a+2b}{3}\right)\right.\right.\\
&\left.+ax^2(a+2b)-a^2x(a+2b)+\frac{a^3}{3}(a+2b)\right]\,dx\\
&+\left.\int_{\frac{a+b}{2}}^{b}\left[x^4-3bx^3+3b^2x^2-b^3x - \frac{x^3}{3}(2a+b)+bx^2(2a+b)-b^2x(2a+b)+\frac{b^3}{3}(2a+b)\right]\,dx\right]\\
&=-\frac{1}{24}\left[\frac{x^5}{5}-\frac{3}{4}ax^4+a^2x^3-\frac{a^3x^2}{2}-\frac{x^4}{12}(a+2b)+\frac{ax^3}{3}(a+2b)-\frac{a^2x^2}{2}\right.\\
&\left.\left.+\frac{a^3}{3}(a+2b)x\right]\right|_a^{\frac{a+b}{2}}-\frac{1}{24}\left[\frac{x^5}{5}-\frac{3}{4}bx^4+b^2x^3-\frac{b^3x^2}{2}-\frac{x^4}{12}(2a+b)\right.\\
&\left.\left.+\frac{bx^3}{3}(2a+b)-\frac{b^2x^2}{2}+\frac{b^3}{3}(2a+b)x\right]\right|_{\frac{a+b}{2}}^{b}\\
&=-\frac{1}{24}\left[\frac{1}{5}\left(\frac{a+b}{2}\right)^5-\frac{3a}{4}\left(\frac{a+b}{2}\right)^4+a^2\left(\frac{a+b}{2}\right)^3-\frac{a^3}{2}\left(\frac{a+b}{2}\right)^2\right.\\
&-\frac{1}{12}\left(\frac{a+b}{2}\right)^4(a+2b)+\frac{a}{3}\left(\frac{a+b}{2}\right)^3(a+2b)-\frac{a^2}{2}\left(\frac{a+b}{2}\right)^2(a+2b)\\
&+\frac{a^3}{3}(a+2b)\left(\frac{a+b}{2}\right)-\frac{a^5}{5}+\frac{3}{4}a^5-a^5+\frac{a^5}{2}+\frac{a^4}{12}(a+2b)-\frac{a^4}{3}(a+2b)\\
&+\frac{a^4}{2}(a+2b)-\frac{a^4}{3}(a+2b)\\
&+\frac{b^5}{5}-\frac{3}{4}b^5+b^5-\frac{b^5}{2}^-\frac{b^4}{12}(2a+b)+\frac{b^4}{3}(2a+b)-\frac{b^4}{2}(2a+b)+\frac{b^4}{3}(2a+b)\\
&-\frac{1}{5}\left(\frac{a+b}{2}\right)^5+\frac{3b}{4}\left(\frac{a+b}{2}\right)^4-b^2\left(\frac{a+b}{2}\right)^3+\frac{b^3}{2}\left(\frac{a+b}{2}\right)^2\\
&\left.+\frac{1}{12}\left(\frac{a+b}{2}\right)^4(2a+b)-\frac{b}{3}\left(\frac{a+b}{2}\right)^3(2a+b)+\frac{b^2}{2}\left(\frac{a+b}{2}\right)^2(2a+b)-\frac{b^3}{3}(2a+b)\left(\frac{a+b}{2}\right)\right]\\
&=-\frac{1}{24}\left[\frac{b^5-a^5}{20}+\frac{1}{12}(b^5-a^5+2ab^4-2ba^4)+2\left(\frac{a+b}{2}\right)^4(b-a)\right.\\
&\left.+\frac{4}{3}(a^2-b^2)\left(\frac{a+b}{2}\right)^3+\left(\frac{a+b}{2}\right)^2[b^3-a^3+b^2a-a^2b]+\left(\frac{a+b}{2}\right)\left(\frac{a^3}{3}(a+2b)-\frac{b^3}{3}(2a+b)\right)\right]\\
\end{align*}
\end{document}
The smart way: Calm down and take a deep breath, read posts and provided links attentively, try to understand and ask if necessary.
