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diff --git a/fuvr2/n.lyx b/fuvr2/n.lyx
new file mode 100644
index 0000000..49459e8
--- /dev/null
+++ b/fuvr2/n.lyx
@@ -0,0 +1,190 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
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+\spacing single
+\use_hyperref false
+\papersize a5paper
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 0.2cm
+\topmargin 0.7cm
+\rightmargin 0.2cm
+\bottommargin 0.7cm
+\secnumdepth 3
+\tocdepth 3
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+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Funciones de una variable real II
+\end_layout
+
+\begin_layout Date
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+def
+\backslash
+cryear{2018}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "../license.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Bibliografía:
+\end_layout
+
+\begin_layout Itemize
+Análisis Matemático I, J.
+ M.
+ Mira & S.
+ Sánchez-Pedreño.
+\end_layout
+
+\begin_layout Itemize
+Funciones reales de una variable real: Notas de clase, B.
+ Cascales, L.
+ Oncina & S.
+ Sánchez-Pedreño (Curso 2017–18).
+\end_layout
+
+\begin_layout Chapter
+Cálculo diferencial
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n1.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Cálculo integral
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n2.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Series de potencias
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n3.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/fuvr2/n1.lyx b/fuvr2/n1.lyx
new file mode 100644
index 0000000..a8766da
--- /dev/null
+++ b/fuvr2/n1.lyx
@@ -0,0 +1,4028 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Una función 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+, siendo 
+\begin_inset Formula $I$
+\end_inset
+
+ un intervalo abierto, es 
+\series bold
+derivable
+\series default
+ en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ si existe
+\begin_inset Formula 
+\[
+f'(c):=\lim_{h\rightarrow0}\frac{f(c+h)-f(c)}{h}
+\]
+
+\end_inset
+
+y se dice derivable en 
+\begin_inset Formula $I$
+\end_inset
+
+ si es derivable en cada punto de 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Al valor 
+\begin_inset Formula $f'(c)$
+\end_inset
+
+ lo llamamos 
+\series bold
+derivada
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $c$
+\end_inset
+
+, y llamamos 
+\series bold
+cociente incremental
+\series default
+ a la expresión 
+\begin_inset Formula $\frac{f(c+h)-f(c)}{h}$
+\end_inset
+
+.
+ Otra definición de derivada es
+\begin_inset Formula 
+\[
+f'(c):=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}
+\]
+
+\end_inset
+
+ Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $I$
+\end_inset
+
+, llamamos 
+\series bold
+derivada de la función
+\series default
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ a la función 
+\begin_inset Formula $f':I\rightarrow\mathbb{R}$
+\end_inset
+
+ que a cada 
+\begin_inset Formula $x\in I$
+\end_inset
+
+ le hace corresponder 
+\begin_inset Formula $f'(x)$
+\end_inset
+
+.
+ Podemos definir la 
+\series bold
+derivada por la izquierda
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $c$
+\end_inset
+
+ como 
+\begin_inset Formula $f'(c^{-}):=f'_{-}(c):=\lim_{h\rightarrow0^{-}}\frac{f(c+h)-f(c)}{h}$
+\end_inset
+
+, y la 
+\series bold
+derivada por la derecha
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $c$
+\end_inset
+
+ como 
+\begin_inset Formula $f'(c^{+}):=f'_{+}(c):=\lim_{h\rightarrow0^{+}}\frac{f(c+h)-f(c)}{h}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c$
+\end_inset
+
+, llamamos 
+\series bold
+recta tangente
+\series default
+ a la curva 
+\begin_inset Formula $y=f(x)$
+\end_inset
+
+ en el punto 
+\begin_inset Formula $(c,f(c))$
+\end_inset
+
+ a la función dada por 
+\begin_inset Formula $g(x)=f(c)+f'(c)(x-c)$
+\end_inset
+
+.
+ Podemos formular que 
+\begin_inset Formula $f'(c)=m$
+\end_inset
+
+ diciendo que
+\begin_inset Formula 
+\[
+f(c+h)=f(c)+mh+h\phi(h)
+\]
+
+\end_inset
+
+donde 
+\begin_inset Formula $\phi:(-\delta,\delta)\backslash\{0\}\rightarrow\mathbb{R}$
+\end_inset
+
+ es una función tal que 
+\begin_inset Formula $\lim_{h\rightarrow0}\phi(h)=0$
+\end_inset
+
+.
+ Equivalentemente, podemos hacer uso de la 
+\series bold
+
+\begin_inset Quotes cld
+\end_inset
+
+o
+\begin_inset Quotes crd
+\end_inset
+
+ pequeña de Landau
+\series default
+, que representa una función cualquiera definida en un entorno reducido
+ o perforado del origen, 
+\begin_inset Formula $(-\delta,\delta)\backslash\{0\}$
+\end_inset
+
+, y cumple que 
+\begin_inset Formula $\lim_{h\rightarrow0}\frac{o(h)}{h}=0$
+\end_inset
+
+.
+ Así,
+\begin_inset Formula 
+\[
+f(c+h)=f(c)+mh+o(h)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ es 
+\series bold
+diferenciable
+\series default
+ en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ si existe una aplicación 
+\emph on
+lineal
+\emph default
+ 
+\begin_inset Formula $L:\mathbb{R}\rightarrow\mathbb{R}$
+\end_inset
+
+ llamada 
+\series bold
+diferencial
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $c$
+\end_inset
+
+, denotada 
+\begin_inset Formula $df(c)$
+\end_inset
+
+, tal que
+\begin_inset Formula 
+\[
+\lim_{h\rightarrow0}\frac{f(c+h)-f(c)-L(h)}{h}=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Se tiene que 
+\begin_inset Formula $f$
+\end_inset
+
+ es diferenciable en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ si y sólo si es derivable en 
+\begin_inset Formula $c$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $df(c)(x)=f'(c)x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $\alpha(h):=\frac{f(c+h)-f(c)-L(h)}{h}=\frac{f(c+h)-f(c)}{h}-L\left(\frac{h}{h}\right)=\frac{f(c+h)-f(c)}{h}-L(1)$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+f'(c)=\lim_{h\rightarrow0}\frac{f(c+h)-f(c)}{h}=\lim_{h\rightarrow0}\alpha(h)+L(1)=L(1)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\lim_{h\rightarrow0}\frac{f(c+h)-f(c)}{h}-f'(c)=\lim_{h\rightarrow0}\frac{f(c+h)-f(c)-f'(c)h}{h}=0
+\]
+
+\end_inset
+
+por lo que 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c$
+\end_inset
+
+ y 
+\begin_inset Formula $f'(c)=L(1)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:I\subseteq\mathbb{R}\rightarrow\mathbb{R}$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $c$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Se tiene que 
+\begin_inset Formula $f(c+h)-f(c)=(f'(c)+\phi(h))h$
+\end_inset
+
+, luego dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, existe 
+\begin_inset Formula $\delta'>0$
+\end_inset
+
+ tal que todo 
+\begin_inset Formula $|h|<\delta'$
+\end_inset
+
+ cumple que 
+\begin_inset Formula $|\phi(h)|<1$
+\end_inset
+
+, y tomando 
+\begin_inset Formula $\delta:=\min\{\delta',\frac{\varepsilon}{|f'(c)|+1}\}$
+\end_inset
+
+, si 
+\begin_inset Formula $|h|<\delta$
+\end_inset
+
+ entonces 
+\begin_inset Formula $|f(c+h)-f(c)|=|f'(c)+\phi(h)||h|\leq(|f'(c)+|\phi(h)|)|h|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Cálculo de derivadas
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f,g:I\rightarrow\mathbb{R}$
+\end_inset
+
+, siendo 
+\begin_inset Formula $I$
+\end_inset
+
+ un intervalo abierto, derivables en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(f+g)'(c)=f'(c)+g'(c)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\rightarrow0}\frac{(f+g)(c+h)-(f+g)(c)}{h}=\lim_{h\rightarrow0}\frac{f(c+h)-f(c)+g(c+h)-g(c)}{h}=f'(c)+g'(c)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(fg)'(c)=f'(c)g(c)+f(c)g'(c)$
+\end_inset
+
+.
+\begin_inset Formula 
+\begin{gather*}
+(fg)'(c)=\lim_{h\rightarrow0}\frac{f(c+h)g(c+h)-f(c)g(c)}{h}=\\
+=\lim_{h\rightarrow0}\frac{f(c+h)g(c+h)-f(c)g(c+h)+f(c)g(c+h)-f(c)g(c)}{h}=\\
+=\lim_{h\rightarrow0}g(c+h)\frac{f(c+h)-f(c)}{h}+\lim_{h\rightarrow0}f(c)\frac{g(c+h)-g(c)}{h}=g(c)f'(c)+f(c)g'(c)
+\end{gather*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $g(x)\neq0\forall x\in I\implies\left(\frac{f}{g}\right)'(c)=\frac{f'(c)g(c)-f(c)g'(c)}{g(c)^{2}}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{gathered}\lim_{h\rightarrow0}\frac{\frac{f(c+h)}{g(c+h)}-\frac{f(c)}{g(c)}}{h}=\lim_{h\rightarrow0}\frac{f(c+h)g(c)-f(c)g(c+h)}{hg(c)g(c+h)}=\\
+=\lim_{h\rightarrow0}\frac{f(c+h)g(c)-f(c)g(c)+f(c)g(c)-f(c)g(c+h)}{hg(c)g(c+h)}=\\
+=\lim_{h\rightarrow0}g(c)\frac{f(c+h)-f(c)}{hg(c)g(c+h)}+f(c)\frac{g(c)-g(c+h)}{hg(c)g(c+h)}=\frac{f'(c)g(c)}{g(c)^{2}}-\frac{f(c)g'(c)}{g(c)^{2}}
+\end{gathered}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(\alpha f)'(c)=\alpha f'(c)\forall\alpha\in\mathbb{R}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Sea 
+\begin_inset Formula $g(x)=\alpha$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in I$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+g'(c)=\lim_{h\rightarrow0}\frac{g(c+h)-g(c)}{h}=\lim_{h\rightarrow0}\frac{\alpha-\alpha}{h}=0
+\]
+
+\end_inset
+
+luego
+\begin_inset Formula 
+\[
+(\alpha f)'(c)=(fg)'(c)=f'(c)g(c)+f(c)g'(c)=f'(c)g(c)=\alpha f'(c)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Regla de la cadena:
+\series default
+ Sean 
+\begin_inset Formula $I,J$
+\end_inset
+
+ intervalos abiertos de 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $g:J\rightarrow\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $\text{Im}f\subseteq J$
+\end_inset
+
+, si 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ lo es en 
+\begin_inset Formula $f(c)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c$
+\end_inset
+
+ y
+\begin_inset Formula 
+\[
+(g\circ f)'(c)=g'(f(c))f'(c)
+\]
+
+\end_inset
+
+Para demostrarlo usamos que 
+\begin_inset Formula $f(c+h)=f(c)+hf'(c)+h\phi(h)$
+\end_inset
+
+ y 
+\begin_inset Formula $g(f(c)+k)=g(f(c))+kg'(f(c))+k\psi(k)$
+\end_inset
+
+.
+ Así,
+\begin_inset Formula 
+\begin{eqnarray*}
+g(f(c+h)) & = & g(f(c)+hf'(c)+h\phi(h))\\
+ & = & g(f(c))+(hf'(c)+h\phi(h))g'(f(c))+(hf'(c)+h\phi(h))\psi(hf'(c)+h\phi(h))\\
+ & = & g(f(c))+hf'(c)g'(f(c))+\\
+ &  & +h(\phi(h)g'(f(c))+(f'(c)+\phi(h))\psi(hf'(c)+h\phi(h)))
+\end{eqnarray*}
+
+\end_inset
+
+Si llamamos 
+\begin_inset Formula $\gamma(h)$
+\end_inset
+
+ al último sumando, vemos que 
+\begin_inset Formula $(g\circ f)(c+h)=(g\circ f)(c)+hf'(c)g'(f(c))+h\gamma(h)$
+\end_inset
+
+ con 
+\begin_inset Formula $\lim_{h\rightarrow0}\gamma(h)=0$
+\end_inset
+
+, lo que prueba el teorema.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f:I\rightarrow J$
+\end_inset
+
+ es una biyección derivable entre los intervalos 
+\begin_inset Formula $I$
+\end_inset
+
+ y 
+\begin_inset Formula $J$
+\end_inset
+
+ con 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ continua y 
+\begin_inset Formula $f'(x)\neq0\forall x\in I$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ es derivable y
+\begin_inset Formula 
+\[
+(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}
+\]
+
+\end_inset
+
+Sean 
+\begin_inset Formula $y=f(x),y_{0}=f(x_{0})$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\lim_{y\rightarrow y_{0}}\frac{f^{-1}(y)-f^{-1}(y_{0})}{y-y_{0}}=\lim_{y\rightarrow y_{0}}\frac{1}{\frac{y-y_{0}}{f^{-1}(y)-f^{-1}(y_{0})}}=\lim_{x\rightarrow x_{0}}\frac{1}{\frac{f(x)-f(x_{0})}{x-x_{0}}}=\frac{1}{f'(f^{-1}(y_{0}))}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Veamos algunas derivadas importantes.
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(x)=\sin x$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f'(x)=\cos x$
+\end_inset
+
+.
+ Si es 
+\begin_inset Formula $g(x)=\cos x$
+\end_inset
+
+, entonces 
+\begin_inset Formula $g'(x)=-\sin x$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Se tiene que
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\sin x=\sin\left(\frac{x+c}{2}+\frac{x-c}{2}\right)=\cos\frac{x+c}{2}\sin\frac{x-c}{2}+\sin\frac{x+c}{2}\cos\frac{x-c}{2}\\
+\sin c=\sin\left(\frac{x+c}{2}-\frac{x-c}{2}\right)=-\cos\frac{x+c}{2}\sin\frac{x-c}{2}+\sin\frac{x+c}{2}\cos\frac{x-c}{2}
+\end{array}
+\]
+
+\end_inset
+
+Por tanto,
+\begin_inset Formula 
+\[
+\lim_{x\rightarrow c}\frac{\sin x-\sin c}{x-c}=\lim_{x\rightarrow c}\frac{\cos\frac{x+c}{2}\sin\frac{x-c}{2}}{\frac{x-c}{2}}=\lim_{x\rightarrow c}\cos\frac{x+c}{2}\cdot1=\cos c
+\]
+
+\end_inset
+
+La derivada del coseno se obtiene de forma análoga.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f(x)=\tan x$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f'(x)=1+\tan^{2}x=\frac{1}{\cos^{2}x}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Como 
+\begin_inset Formula $f(x)=\frac{\sin x}{\cos x}$
+\end_inset
+
+, partiendo de la derivada del seno y del coseno,
+\begin_inset Formula 
+\[
+f'(x)=\frac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^{2}x}=\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}=1+\tan^{2}x=\frac{1}{\cos^{2}x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(x)=e^{x}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f'(x)=e^{x}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\rightarrow0}\frac{e^{x+h}-e^{x}}{h}=\lim_{h\rightarrow0}e^{x}\frac{e^{h}-1}{h}=e^{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $f:I\subseteq(0,+\infty)\rightarrow\mathbb{R}$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(x)=\log x$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f'(x)=\frac{1}{x}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+El logaritmo es la inversa de 
+\begin_inset Formula $g(x)=e^{x}$
+\end_inset
+
+, con 
+\begin_inset Formula $g'(x)=e^{x}$
+\end_inset
+
+, luego
+\begin_inset Formula 
+\[
+f'(x)=\frac{1}{e^{\log x}}=\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $f:I\subseteq(-1,1)\rightarrow(-\frac{\pi}{2},\frac{\pi}{2})$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(x)=\arcsin x$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f'(x)=\frac{1}{\sqrt{1-x^{2}}}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $g:I\subseteq(-1,1)\rightarrow(0,\pi)$
+\end_inset
+
+ dada por 
+\begin_inset Formula $g(x)=\arccos x$
+\end_inset
+
+, 
+\begin_inset Formula $g'(x)=\frac{-1}{\sqrt{1-x^{2}}}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Al ser 
+\begin_inset Formula $f$
+\end_inset
+
+ la inversa del seno y 
+\begin_inset Formula $\sin'(x)=\cos x$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+f'(x)=\frac{1}{\cos(\arcsin x)}=\frac{1}{\sqrt{1-\sin^{2}(\arcsin x)}}=\frac{1}{\sqrt{1-x^{2}}}
+\]
+
+\end_inset
+
+La derivada del arcocoseno se hace de forma análoga.
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $f:I\rightarrow(-\frac{\pi}{2},\frac{\pi}{2})$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(x)=\arctan x$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f'(x)=\frac{1}{1+x^{2}}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Esta función es la inversa de la tangente, y como 
+\begin_inset Formula $\tan'(x)=1+\tan^{2}x$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+f'(x)=\frac{1}{1+\tan^{2}(\arctan x)}=\frac{1}{1+x^{2}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Dado 
+\begin_inset Formula $\alpha\in\mathbb{R}$
+\end_inset
+
+, la derivada de 
+\begin_inset Formula $f(x)=x^{\alpha}$
+\end_inset
+
+ es 
+\begin_inset Formula $f'(x)=\alpha x^{\alpha-1}$
+\end_inset
+
+.
+ Para demostrarlo usamos 
+\series bold
+derivación logarítmica
+\series default
+: Tomamos logaritmos en la definición de 
+\begin_inset Formula $f$
+\end_inset
+
+ y derivamos la expresión resultante.
+\begin_inset Formula 
+\[
+\log(f(x))=\log(x^{\alpha})=\alpha\log x\implies\log(f(x))'=\frac{f'(x)}{f(x)}=\frac{\alpha}{x}\implies f'(x)=f(x)\frac{\alpha}{x}=\alpha x^{\alpha-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Derivabilidad en un intervalo
+\end_layout
+
+\begin_layout Standard
+Una función 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ definida en un intervalo 
+\begin_inset Formula $I$
+\end_inset
+
+ es 
+\series bold
+creciente
+\series default
+, 
+\series bold
+estrictamente creciente
+\series default
+, 
+\series bold
+decreciente
+\series default
+ o 
+\series bold
+estrictamente decreciente
+\series default
+ en 
+\begin_inset Formula $I$
+\end_inset
+
+ si para cualesquiera 
+\begin_inset Formula $x,y\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $x<y$
+\end_inset
+
+ se tiene, respectivamente, que 
+\begin_inset Formula $f(x)\leq f(y)$
+\end_inset
+
+, 
+\begin_inset Formula $f(x)<f(y)$
+\end_inset
+
+, 
+\begin_inset Formula $f(x)\geq f(y)$
+\end_inset
+
+ o 
+\begin_inset Formula $f(x)>f(y)$
+\end_inset
+
+.
+ Es creciente, estrictamente creciente, decreciente o estrictamente decreciente
+ en un punto 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ si existe un entorno perforado 
+\begin_inset Formula $V$
+\end_inset
+
+ de 
+\begin_inset Formula $c$
+\end_inset
+
+ tal que para 
+\begin_inset Formula $x\in I\cap V$
+\end_inset
+
+, si 
+\begin_inset Formula $m:=\frac{f(x)-f(c)}{x-c}$
+\end_inset
+
+ es, respectivamente, 
+\begin_inset Formula $m\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $m>0$
+\end_inset
+
+, 
+\begin_inset Formula $m\leq0$
+\end_inset
+
+ o 
+\begin_inset Formula $m<0$
+\end_inset
+
+.
+ Se tiene que 
+\begin_inset Formula $f$
+\end_inset
+
+ es creciente o decreciente en 
+\begin_inset Formula $I$
+\end_inset
+
+ si y sólo si lo es en cada punto de 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Trivial.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $f$
+\end_inset
+
+ creciente en cada 
+\begin_inset Formula $x\in I$
+\end_inset
+
+, es menester demostrar que, dados 
+\begin_inset Formula $x<y$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $f(x)\leq f(y)$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $A:=\{z\in(x,y]:f(x)\leq f(z)\}$
+\end_inset
+
+, como 
+\begin_inset Formula $A\neq\emptyset$
+\end_inset
+
+ porque 
+\begin_inset Formula $f$
+\end_inset
+
+ es creciente en 
+\begin_inset Formula $x$
+\end_inset
+
+ y 
+\begin_inset Formula $A$
+\end_inset
+
+ es acotado superiormente, podemos definir 
+\begin_inset Formula $\alpha:=\sup A$
+\end_inset
+
+, y basta probar que 
+\begin_inset Formula $\alpha=y$
+\end_inset
+
+ y 
+\begin_inset Formula $f(x)\leq f(\alpha)$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $f$
+\end_inset
+
+ es creciente en 
+\begin_inset Formula $\alpha$
+\end_inset
+
+, existe 
+\begin_inset Formula $\delta>0$
+\end_inset
+
+ con 
+\begin_inset Formula $f(z)\leq f(\alpha)$
+\end_inset
+
+ si 
+\begin_inset Formula $z\in(\alpha-\delta,\alpha)$
+\end_inset
+
+.
+ Pero por definición de 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ para alguno de esos valores es 
+\begin_inset Formula $f(x)\leq f(z)$
+\end_inset
+
+, luego 
+\begin_inset Formula $f(x)\leq f(\alpha)$
+\end_inset
+
+.
+ Si fuera 
+\begin_inset Formula $\alpha<y$
+\end_inset
+
+ existiría 
+\begin_inset Formula $z\in(\alpha,y]$
+\end_inset
+
+ con 
+\begin_inset Formula $f(\alpha)\leq f(z)$
+\end_inset
+
+ por el crecimiento de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $\alpha$
+\end_inset
+
+, pero entonces se tendría que 
+\begin_inset Formula $f(x)\leq f(\alpha)\leq f(z)$
+\end_inset
+
+, contradiciendo la definición de 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f$
+\end_inset
+
+ tiene un 
+\series bold
+máximo relativo
+\series default
+ o 
+\series bold
+local
+\series default
+ en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ si existe un entorno 
+\begin_inset Formula $V$
+\end_inset
+
+ de 
+\begin_inset Formula $c$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(x)\leq f(c)\forall x\in I\cap V$
+\end_inset
+
+, tiene un 
+\series bold
+mínimo relativo
+\series default
+ o 
+\series bold
+local
+\series default
+ en 
+\begin_inset Formula $c\in I$
+\end_inset
+
+ si existe un entorno 
+\begin_inset Formula $V$
+\end_inset
+
+ de 
+\begin_inset Formula $c$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(x)\geq f(c)\forall x\in I\cap V$
+\end_inset
+
+, y tiene un 
+\series bold
+extremo relativo
+\series default
+ o 
+\series bold
+local
+\series default
+ en 
+\begin_inset Formula $c$
+\end_inset
+
+ si tiene un máximo o mínimo relativo en 
+\begin_inset Formula $c$
+\end_inset
+
+.
+ Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f'(c)>0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es estrictamente creciente en 
+\begin_inset Formula $c$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+f'(c)=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}>0
+\]
+
+\end_inset
+
+por lo que existe un entorno reducido 
+\begin_inset Formula $V$
+\end_inset
+
+ de 
+\begin_inset Formula $c$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall x\in I\cap V,\frac{f(x)-f(c)}{x-c}>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f'(c)<0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es estrictamente decreciente en 
+\begin_inset Formula $c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $c$
+\end_inset
+
+ es un punto interior del intervalo 
+\begin_inset Formula $I$
+\end_inset
+
+ (no es un extremo) y 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable y tiene un extremo relativo en 
+\begin_inset Formula $c$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f'(c)=0$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Supongamos que el extremo es un máximo.
+ Existe un entorno 
+\begin_inset Formula $V$
+\end_inset
+
+ de 
+\begin_inset Formula $c$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall x\in I\cap V,f(x)\leq f(c)$
+\end_inset
+
+, luego para 
+\begin_inset Formula $x\in I\cap V$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\left\{ \begin{array}{ccccc}
+x<c & \implies & \frac{f(x)-f(c)}{x-c}\geq0 & \implies & f'(c^{-})=\lim_{x\rightarrow c^{-}}\frac{f(x)-f(c)}{x-c}\geq0\\
+x>c & \implies & \frac{f(x)-f(c)}{x-c}\leq0 & \implies & f(c^{+})=\lim_{x\rightarrow c^{+}}\frac{f(x)-f(c)}{x-c}\leq0
+\end{array}\right.
+\]
+
+\end_inset
+
+Pero como 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c$
+\end_inset
+
+, 
+\begin_inset Formula $0\leq f'(c^{-})=f'(c)=f'(c^{+})\leq0$
+\end_inset
+
+, luego 
+\begin_inset Formula $f'(c)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:(a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ derivable, 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+ es un 
+\series bold
+punto crítico
+\series default
+ o 
+\series bold
+estacionario
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ si 
+\begin_inset Formula $f'(c)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Teoremas del valor medio
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de Rolle:
+\series default
+ Sea 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continua en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ y derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(a)=f(b)$
+\end_inset
+
+ entonces existe 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+ con 
+\begin_inset Formula $f'(c)=0$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es constante, tomamos 
+\begin_inset Formula $c:=\frac{a+b}{2}$
+\end_inset
+
+.
+ Si no, supongamos por ejemplo que existe 
+\begin_inset Formula $x_{0}\in[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x_{0})>f(a)=f(b)$
+\end_inset
+
+.
+ Por el teorema de Weierstrass, 
+\begin_inset Formula $f$
+\end_inset
+
+ alcanza su máximo absoluto en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+, y por lo anterior debe alcanzarse en un punto interior 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+.
+ Pero por ser máximo absoluto es también máximo relativo y por tanto 
+\begin_inset Formula $f'(c)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema del valor medio de Cauchy:
+\series default
+ Sean 
+\begin_inset Formula $f,g:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continuas en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ y derivables en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+, entonces existe 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+ con 
+\begin_inset Formula $(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c)$
+\end_inset
+
+ (si 
+\begin_inset Formula $g(b)\neq g(a)$
+\end_inset
+
+ y 
+\begin_inset Formula $g'(c)\neq0$
+\end_inset
+
+ podemos expresar esto como 
+\begin_inset Formula $\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}$
+\end_inset
+
+).
+ 
+\series bold
+Demostración:
+\series default
+ Aplicamos el teorema de Rolle a 
+\begin_inset Formula $h(x):=f(x)(g(b)-g(a))-g(x)(f(b)-f(a))$
+\end_inset
+
+, pues 
+\begin_inset Formula $h(a)=h(b)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema del valor medio de Lagrange:
+\series default
+ Sea 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continua en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ y derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+, existe 
+\begin_inset Formula $\theta\in(a,b)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f'(\theta)(b-a)=f(b)-f(a)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Es un caso particular del teorema del valor medio de Cauchy tomando 
+\begin_inset Formula $g(x):=x$
+\end_inset
+
+.
+ El teorema de Rolle es un caso particular de este, por lo que estos tres
+ teoremas son equivalentes.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de los incrementos finitos:
+\series default
+ Sea 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continua en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ y derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+, si 
+\begin_inset Formula $|f'(x)|\leq M\forall x\in(a,b)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $|f(x)-f(y)|\leq M|x-y|$
+\end_inset
+
+ para cualesquiera 
+\begin_inset Formula $x,y\in[a,b]$
+\end_inset
+
+.
+ A efectos prácticos, esto significa que si 
+\begin_inset Formula $f'$
+\end_inset
+
+ es acotada entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es uniformemente continua.
+ 
+\series bold
+Demostración:
+\series default
+ Basta aplicar el teorema del valor medio de Lagrange a 
+\begin_inset Formula $f|_{[x,y]}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Para 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ continuas en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ y derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ se cumplen las siguientes propiedades:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall x\in(a,b),f'(x)=0\implies\exists k\in\mathbb{R}:\forall x\in(a,b),f(x)=k$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Aplicando el teorema de Lagrange en 
+\begin_inset Formula $[a,x]$
+\end_inset
+
+, existe 
+\begin_inset Formula $c\in(a,x)$
+\end_inset
+
+ con 
+\begin_inset Formula $\frac{f(x)-f(a)}{x-a}=f'(c)=0$
+\end_inset
+
+, luego 
+\begin_inset Formula $f(x)=f(a)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall x\in(a,b),f'(x)=g'(x)\implies\exists k\in\mathbb{R}:\forall x\in(a,b),f(x)=g(x)+k$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Sea 
+\begin_inset Formula $h(x):=f(x)-g(x)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $h'(x)=f'(x)-g'(x)=0$
+\end_inset
+
+ para 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+, luego 
+\begin_inset Formula $h(x)$
+\end_inset
+
+ es constante.
+\end_layout
+
+\begin_layout Enumerate
+Si para todo 
+\begin_inset Formula $x\in(a,b)$
+\end_inset
+
+ se tiene que 
+\begin_inset Formula $f'(x)\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $f'(x)>0$
+\end_inset
+
+, 
+\begin_inset Formula $f'(x)\leq0$
+\end_inset
+
+ o 
+\begin_inset Formula $f'(x)<0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es, respectivamente, creciente, estrictamente creciente, decreciente o
+ estrictamente decreciente.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:(a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ derivable y 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+ con 
+\begin_inset Formula $f'(c)=0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $\exists\delta>0:(\forall x\in(c-\delta,c)\subseteq(a,b),f'(x)\leq0\land\forall x\in(c,c+\delta)\subseteq(a,b),f'(x)\geq0)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ posee un mínimo relativo en 
+\begin_inset Formula $c$
+\end_inset
+
+.
+ Análogamente, si 
+\begin_inset Formula $\exists\delta>0:(\forall x\in(c-\delta,c)\subseteq(a,b),f'(x)\geq0\land\forall x\in(c,c+\delta)\subseteq(a,b),f'(x)\leq0)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ posee un máximo relativo en 
+\begin_inset Formula $c$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Para el primer caso, si 
+\begin_inset Formula $y\in(c-\delta,c)$
+\end_inset
+
+, existe 
+\begin_inset Formula $\eta\in(y,c)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(c)-f(y)=f'(\eta)(c-y)\leq0$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $f(c)\leq f(y)$
+\end_inset
+
+, mientras que si 
+\begin_inset Formula $y\in(c,c+\delta)$
+\end_inset
+
+, existe 
+\begin_inset Formula $\beta\in(c,y)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(y)-f(c)=f'(\beta)(y-c)\geq0$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $f(c)\leq f(y)$
+\end_inset
+
+; luego si 
+\begin_inset Formula $y\in(c-\delta,c+\delta)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f(y)\geq f(c)$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $f$
+\end_inset
+
+ tiene un mínimo relativo en 
+\begin_inset Formula $c$
+\end_inset
+
+.
+ El segundo caso se prueba de forma análoga.
+\end_layout
+
+\begin_layout Standard
+Con esto podemos probar la 
+\series bold
+desigualdad de Bernouilli
+\series default
+ de forma más general: dados 
+\begin_inset Formula $x>0$
+\end_inset
+
+ y 
+\begin_inset Formula $\alpha>1$
+\end_inset
+
+, 
+\begin_inset Formula $(1+x)^{\alpha}>1+\alpha x$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $f:(0,+\infty)\rightarrow\mathbb{R}$
+\end_inset
+
+ definida por 
+\begin_inset Formula $f(x)=(1+x)^{\alpha}-1-\alpha x$
+\end_inset
+
+ para un cierto 
+\begin_inset Formula $\alpha>1$
+\end_inset
+
+, como 
+\begin_inset Formula $f(0)=0$
+\end_inset
+
+, basta probar que 
+\begin_inset Formula $f$
+\end_inset
+
+ es estrictamente creciente si 
+\begin_inset Formula $\alpha>1$
+\end_inset
+
+, pero 
+\begin_inset Formula $f'(x)=\alpha((1+x)^{\alpha-1}-1)>0$
+\end_inset
+
+, probando la desigualdad.
+\end_layout
+
+\begin_layout Subsection
+Teorema de la función inversa
+\end_layout
+
+\begin_layout Standard
+La 
+\series bold
+propiedad de los valores intermedios
+\series default
+ afirma que, sea 
+\begin_inset Formula $f:(a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ derivable y 
+\begin_inset Formula $x,y\in(a,b)$
+\end_inset
+
+ con 
+\begin_inset Formula $x<y$
+\end_inset
+
+ y 
+\begin_inset Formula $f'(x)<\eta<f'(y)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\exists z\in(x,y):f'(z)=\eta$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $g:[x,y]\rightarrow\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $g(t)=f(t)-\eta t$
+\end_inset
+
+ continua y derivable, que por el teorema de Weierstrass (que usamos en
+ lugar del de Bolzano porque 
+\begin_inset Formula $g'$
+\end_inset
+
+ no tiene por qué ser continua), tiene un mínimo absoluto en un 
+\begin_inset Formula $z\in[x,y]$
+\end_inset
+
+.
+ Pero como 
+\begin_inset Formula $g'(x)<0$
+\end_inset
+
+ y 
+\begin_inset Formula $g'(y)>0$
+\end_inset
+
+, 
+\begin_inset Formula $z$
+\end_inset
+
+ no puede ser 
+\begin_inset Formula $x$
+\end_inset
+
+ ni 
+\begin_inset Formula $y$
+\end_inset
+
+, luego 
+\begin_inset Formula $z\in(x,y)$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $g'(z)=0$
+\end_inset
+
+, o dicho de otra forma, 
+\begin_inset Formula $f'(z)=\eta$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De aquí deducimos el 
+\series bold
+teorema de la función inversa:
+\series default
+ Sea 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ continua en el intervalo 
+\begin_inset Formula $I$
+\end_inset
+
+ y derivable en su interior con derivada no nula, entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es una biyección de 
+\begin_inset Formula $I$
+\end_inset
+
+ sobre un intervalo 
+\begin_inset Formula $J$
+\end_inset
+
+ y 
+\begin_inset Formula $f^{-1}:J\rightarrow\mathbb{R}$
+\end_inset
+
+ es continua y derivable en el interior de 
+\begin_inset Formula $J$
+\end_inset
+
+ con
+\begin_inset Formula 
+\[
+(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Por la propiedad anterior, bien 
+\begin_inset Formula $f'(x)>0$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x$
+\end_inset
+
+ o 
+\begin_inset Formula $f'(x)<0$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $f$
+\end_inset
+
+ es estrictamente monótona, de modo que es biyectiva de 
+\begin_inset Formula $I$
+\end_inset
+
+ sobre un intervalo 
+\begin_inset Formula $J$
+\end_inset
+
+ siendo 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ estrictamente monótona y continua.
+ Sean entonces 
+\begin_inset Formula $y,y_{0}\in J,x=f^{-1}(y),x_{0}=f^{-1}(y_{0})$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\lim_{y\rightarrow y_{0}}\frac{f^{-1}(y)-f^{-1}(y_{0})}{y-y_{0}}=\lim_{y\rightarrow y_{0}}\frac{1}{\frac{y-y_{0}}{f^{-1}(y)-f^{-1}(y_{0})}}=\lim_{x\rightarrow x_{0}}\frac{1}{\frac{f(x)-f(x_{0})}{x-x_{0}}}=\frac{1}{f'(f^{-1}(y_{0}))}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Regla de L'Hospital
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ derivables en 
+\begin_inset Formula $I=(a,b)$
+\end_inset
+
+ con 
+\begin_inset Formula $-\infty\leq a<b\leq+\infty$
+\end_inset
+
+, si 
+\begin_inset Formula $g$
+\end_inset
+
+ y 
+\begin_inset Formula $g'$
+\end_inset
+
+ no tienen ceros en 
+\begin_inset Formula $I$
+\end_inset
+
+ y se cumple que o bien 
+\begin_inset Formula $\lim_{x\rightarrow b^{-}}f(x)=\lim_{x\rightarrow b^{-}}g(x)=0$
+\end_inset
+
+ o 
+\begin_inset Formula $\lim_{x\rightarrow b^{-}}g(x)=\pm\infty$
+\end_inset
+
+, entonces, si existe 
+\begin_inset Formula $L:=\lim_{x\rightarrow b^{-}}\frac{f'(x)}{g'(x)}\in\overline{\mathbb{R}}$
+\end_inset
+
+, es también 
+\begin_inset Formula $L=\lim_{x\rightarrow b^{-}}\frac{f(x)}{g(x)}$
+\end_inset
+
+.
+ Por supuesto, esto también se cumple para límites por la derecha y por
+ tanto también para límites ordinarios.
+\end_layout
+
+\begin_layout Section
+Desarrollos de Taylor
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es derivable en el intervalo abierto 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ y 
+\begin_inset Formula $f'$
+\end_inset
+
+ también lo es, se dice que 
+\begin_inset Formula $f$
+\end_inset
+
+ es dos veces derivable en 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ y la derivada de 
+\begin_inset Formula $f'$
+\end_inset
+
+ se denota por 
+\begin_inset Formula $f^{(2)}:=f''$
+\end_inset
+
+, y por inducción, si 
+\begin_inset Formula $f$
+\end_inset
+
+ es 
+\begin_inset Formula $n-1$
+\end_inset
+
+ veces derivable y 
+\begin_inset Formula $f^{(n-1)}$
+\end_inset
+
+ es derivable, se dice que 
+\begin_inset Formula $f$
+\end_inset
+
+ es 
+\begin_inset Formula $n$
+\end_inset
+
+ veces derivable y llamamos 
+\begin_inset Formula $f^{(n)}:=(f^{(n-1)})'$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ es de clase 
+\begin_inset Formula ${\cal C}^{n}$
+\end_inset
+
+ en 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ si existe 
+\begin_inset Formula $f^{(n)}$
+\end_inset
+
+ y es continua, y es de clase 
+\begin_inset Formula ${\cal C}^{\infty}$
+\end_inset
+
+ en 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ si es de clase 
+\begin_inset Formula ${\cal C}^{n}$
+\end_inset
+
+ para todo 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Por ejemplo, los polinomios son funciones de clase 
+\begin_inset Formula ${\cal C}^{\infty}$
+\end_inset
+
+ en 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+, de modo que conociendo el valor de 
+\begin_inset Formula $P$
+\end_inset
+
+ y sus derivadas en un cierto punto es posible reconstruir el polinomio.
+ 
+\series bold
+Demostración:
+\series default
+ Dividiendo 
+\begin_inset Formula $P(x)=a_{n}x^{n}+\dots+a_{0}$
+\end_inset
+
+ por 
+\begin_inset Formula $(x-x_{0})^{n}$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $P(x)=b_{n}(x-x_{0})^{n}+Q_{n-1}(x)$
+\end_inset
+
+, donde 
+\begin_inset Formula $Q_{n-1}(x)$
+\end_inset
+
+ es de grado 
+\begin_inset Formula $n-1$
+\end_inset
+
+.
+ Por inducción se obtiene 
+\begin_inset Formula $P(x)=b_{n}(x-x_{0})^{n}+\dots+b_{0}$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $b_{0}=P(x_{0})$
+\end_inset
+
+, y derivando sucesivamente:
+\begin_inset Formula 
+\[
+\begin{array}{ccccc}
+P(x)=b_{n}(x-x_{0})^{n}+\dots+b_{0} &  & P(x_{0})=b_{0}\\
+P'(x)=nb_{n}(x-x_{0})^{n-1}+\dots+b_{1} &  & P'(x_{0})=b_{1} &  & b_{1}=\frac{P'(x_{0})}{1!}\\
+P''(x)=n(n-1)(x-x_{0})^{n-2}+\dots+2b_{2} &  & P''(x_{0})=2b_{2} &  & b_{2}=\frac{P''(x_{0})}{2!}\\
+P'''(x)=n(n-1)(n-2)(x-x_{0})^{n-3}+\dots+6b_{3} &  & P'''(x_{0})=6b_{3} &  & b_{3}=\frac{P'''(x_{0})}{3!}\\
+\vdots &  & \vdots &  & \vdots\\
+P^{(n)}(x)=n!b_{n} &  & P^{(n)}(x_{0})=n!b_{n} &  & b_{n}=\frac{P^{(n)}(x_{0})}{n!}
+\end{array}
+\]
+
+\end_inset
+
+Con lo que 
+\begin_inset Formula $P(x)=P(x_{0})+\frac{P'(x_{0})}{1!}(x-x_{0})+\dots+\frac{P^{(n)}(x_{0})}{n!}(x-x_{0})^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+polinomio de Taylor
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ de grado 
+\begin_inset Formula $n$
+\end_inset
+
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ a la siguiente expresión:
+\begin_inset Formula 
+\[
+P_{n}(f,x;x_{0})=f(x_{0})+\frac{f'(x_{0})}{1!}(x-x_{0})+\dots+\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n}
+\]
+
+\end_inset
+
+El 
+\series bold
+resto del polinomio
+\series default
+ es la diferencia entre la función y su polinomio de Taylor: 
+\begin_inset Formula $R_{n}(x;x_{0}):=f(x)-P_{n}(f,x;x_{0})$
+\end_inset
+
+.
+ Una función 
+\begin_inset Formula $g:(x_{0}-\delta,x_{0}+\delta)\backslash\{0\}\rightarrow\mathbb{R}$
+\end_inset
+
+ definida en un entorno reducido de 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ es una 
+\series bold
+
+\begin_inset Quotes cld
+\end_inset
+
+o
+\begin_inset Quotes crd
+\end_inset
+
+ pequeña
+\series default
+ de 
+\begin_inset Formula $|x-x_{0}|^{n}$
+\end_inset
+
+, escrito 
+\begin_inset Formula $g(x)=o(|x-x_{0}|^{n})$
+\end_inset
+
+ o informalmente 
+\begin_inset Formula $o(x-x_{0})^{n}$
+\end_inset
+
+, si 
+\begin_inset Formula $\lim_{x\rightarrow x_{0}}\frac{|g(x)|}{|x-x_{0}|^{n}}=0$
+\end_inset
+
+.
+ Así, si 
+\begin_inset Formula $g(x)=o(|x-x_{0}|^{n})$
+\end_inset
+
+ entonces 
+\begin_inset Formula $g(x)=o(|x-x_{0}|^{k})$
+\end_inset
+
+ para 
+\begin_inset Formula $1\leq k\leq n$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $\lim_{x\rightarrow x_{0}}\frac{|g(x)|}{|x-x_{0}|^{k}}=\lim_{x\rightarrow x_{0}}\frac{|g(x)|}{|x-x_{0}|^{n}}|x-x_{0}|^{n-k}=0\cdot0=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Resto de Landau y desarrollos limitados
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:(a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ es 
+\begin_inset Formula $n-1$
+\end_inset
+
+ veces derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ y existe la derivada 
+\begin_inset Formula $n$
+\end_inset
+
+-ésima en 
+\begin_inset Formula $x_{0}\in(a,b)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f(x)=P_{n}(f,x;x_{0})+o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Aplicando la regla de L'Hospital 
+\begin_inset Formula $n-1$
+\end_inset
+
+ veces y la definición de derivada 
+\begin_inset Formula $n$
+\end_inset
+
+-ésima de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\lim_{x\rightarrow x_{0}}\frac{f(x)-P_{n}(x)}{(x-x_{0})^{n}}=\dots=\lim_{x\rightarrow x_{0}}\frac{f^{(n-1)}(x)-P_{n}^{(n-1)}(x)}{n(n-1)\cdots2(x-x_{0})}
+\]
+
+\end_inset
+
+pero, al derivar 
+\begin_inset Formula $n-1$
+\end_inset
+
+ veces 
+\begin_inset Formula $P_{n}(x)$
+\end_inset
+
+, desaparecen todos los términos salvo los de grado 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $n-1$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $P_{n}^{(n-1)}(x)=(n-1)!\frac{f^{(n-1)}(x_{0})}{(n-1)!}+n!\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})=f^{(n-1)}(x_{0})+f^{(n)}(x_{0})(x-x_{0})$
+\end_inset
+
+.
+ Por tanto
+\begin_inset Formula 
+\begin{eqnarray*}
+\lim_{x\rightarrow x_{0}}\frac{f(x)-P_{n}(x)}{(x-x_{0})^{n}} & = & \lim_{x\rightarrow x_{0}}\frac{f^{(n-1)}(x)-f^{(n-1)}(x_{0})-f^{(n)}(x_{0})(x-x_{0})}{n!(x-x_{0})}\\
+ & = & \frac{1}{n!}\left(\lim_{x\rightarrow x_{0}}\frac{f^{(n-1)}(x)-f^{(n-1)}(x_{0})}{(x-x_{0})}-f^{(n)}(x_{0})\right)\\
+ & = & 0
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+A una expresión como la de arriba la llamamos 
+\series bold
+desarrollo limitado
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ de grado 
+\begin_inset Formula $n$
+\end_inset
+
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+, y cuando existe es única.
+ 
+\series bold
+Demostración:
+\series default
+ Supongamos que una expresión admite dos desarrollos limitados de orden
+ 
+\begin_inset Formula $n$
+\end_inset
+
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+: 
+\begin_inset Formula $f(x)=a_{0}+a_{1}(x-x_{0})+\dots+a_{n}(x-x_{0})^{n}+o(|x-x_{0}|^{n})=b_{0}+b_{1}(x-x_{0})+\dots+b_{n}(x-x_{0})^{n}+o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+ Igualando, 
+\begin_inset Formula $(b_{0}-a_{0})+(b_{1}-a_{1})(x-x_{0})+\dots+(b_{n}-a_{n})(x-x_{0})^{n}=o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+ Tomando límites cuando 
+\begin_inset Formula $x$
+\end_inset
+
+ tiende a 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+, 
+\begin_inset Formula $a_{0}=b_{0}$
+\end_inset
+
+.
+ Eliminando este sumando, dividiendo por 
+\begin_inset Formula $(x-x_{0})$
+\end_inset
+
+ y tomando límites de nuevo, queda 
+\begin_inset Formula $a_{1}=b_{1}$
+\end_inset
+
+, y así sucesivamente.
+\end_layout
+
+\begin_layout Standard
+Para calcular desarrollos limitados, muy útiles en el cálculo de límites
+ de cocientes sustituyendo a la regla de L'Hospital, sean 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ funciones de clase 
+\begin_inset Formula ${\cal C}^{n}$
+\end_inset
+
+ definidas en entornos de 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ e 
+\begin_inset Formula $y_{0}$
+\end_inset
+
+, respectivamente, y derivables 
+\begin_inset Formula $n$
+\end_inset
+
+ veces en dichos puntos:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $x_{0}=y_{0}$
+\end_inset
+
+, 
+\begin_inset Formula $(f+g)(x)=P_{n}(f,x;x_{0})+P_{n}(g,x;x_{0})+o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $x_{0}=y_{0}$
+\end_inset
+
+, 
+\begin_inset Formula $(fg)(x)=P_{n}(f,x;x_{0})P_{n}(g,x;x_{0})+o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+ Aquí hay que agrupar los términos convenientemente teniendo en cuenta que
+ los términos de grado mayor a 
+\begin_inset Formula $n$
+\end_inset
+
+ son 
+\begin_inset Formula $o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $x_{0}=y_{0}$
+\end_inset
+
+, 
+\begin_inset Formula $\left(\frac{f}{g}\right)(x)=\frac{P_{n}(f,x;x_{0})}{P_{n}(g,x;x_{0})}+o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+ Aquí hay que considerar la 
+\emph on
+fracción continua
+\emph default
+ de polinomios, que es igual que la división normal de polinomios pero tomando
+ los términos de menor grado del divisor y el dividendo en vez de los de
+ mayor grado, y terminando cuando el grado del término resultante del cociente
+ sea mayor que 
+\begin_inset Formula $n$
+\end_inset
+
+, pues a partir de ahí el resto de términos son 
+\begin_inset Formula $o(|x-x_{0}|^{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f(x)=P_{n}(f,x;x_{0})+o(|x-x_{0}|^{n})$
+\end_inset
+
+, el desarrollo limitado de orden 
+\begin_inset Formula $n-1$
+\end_inset
+
+ de 
+\begin_inset Formula $f'$
+\end_inset
+
+ es 
+\begin_inset Formula $f'(x)=(P_{n}(f,x;x_{0}))'+o(|x-x_{0}|^{n-1})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f(x_{0})=y_{0}$
+\end_inset
+
+ y la función 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ está definida en un entorno de 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ en el que admite un desarrollo limitado en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+, este se obtiene sustituyendo el desarrollo de 
+\begin_inset Formula $f$
+\end_inset
+
+ en el de 
+\begin_inset Formula $g$
+\end_inset
+
+ y agrupando los términos convenientemente tanto en la parte polinómica
+ de grado menor o igual a 
+\begin_inset Formula $n$
+\end_inset
+
+ como en la del resto de Landau.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:(a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ es 
+\begin_inset Formula $n$
+\end_inset
+
+ veces derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+, siendo 
+\begin_inset Formula $f'(x_{0})=\dots=f^{(n-1)}(x_{0})=0$
+\end_inset
+
+ y 
+\begin_inset Formula $f^{(n)}(x_{0})\neq0$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $n$
+\end_inset
+
+ es par, 
+\begin_inset Formula $f$
+\end_inset
+
+ presenta un máximo relativo en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ si 
+\begin_inset Formula $f^{(n)}(x_{0})<0$
+\end_inset
+
+ o un mínimo relativo si 
+\begin_inset Formula $f^{(n)}(x_{0})>0$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Como todas las derivadas en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ hasta 
+\begin_inset Formula $n-1$
+\end_inset
+
+ son 0,
+\begin_inset Formula 
+\[
+\begin{array}{c}
+f(x)=f(x_{0})+\frac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}+o((x-x_{0})^{n})\implies\\
+\implies\frac{f(x)-f(x_{0})}{(x-x_{0})^{n}}=\frac{1}{n!}f^{(n)}(x_{0})+\frac{o((x-x_{0})^{n})}{(x-x_{0})^{n}}
+\end{array}
+\]
+
+\end_inset
+
+Si 
+\begin_inset Formula $f^{(n)}(x_{0})<0$
+\end_inset
+
+, existe un entorno de 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ en el que el segundo miembro de la igualdad es estrictamente negativo y
+ por tanto también el primero, pero como 
+\begin_inset Formula $n$
+\end_inset
+
+ es par, esto significa que 
+\begin_inset Formula $f(x)-f(x_{0})<0$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $f(x)<f(x_{0})$
+\end_inset
+
+ y hay un máximo relativo.
+ El caso en que 
+\begin_inset Formula $f^{(n)}(x_{0})>0$
+\end_inset
+
+ es análogo.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $n$
+\end_inset
+
+ es impar, 
+\begin_inset Formula $f$
+\end_inset
+
+ no tiene extremo relativo en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Llegamos a que existe un entorno de 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ en el que el primer miembro de la igualdad es estrictamente positivo o
+ estrictamente negativo, pero cualquiera de las situaciones significa que
+ la función es estrictamente creciente a ambos lados o estrictamente decreciente
+ a ambos lados.
+\end_layout
+
+\begin_layout Subsection
+Fórmula de Taylor con resto de Lagrange
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:(a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ es 
+\begin_inset Formula $n$
+\end_inset
+
+ veces derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ y sean 
+\begin_inset Formula $x_{0},x\in(a,b)$
+\end_inset
+
+, entonces existe 
+\begin_inset Formula $c$
+\end_inset
+
+ estrictamente entre 
+\begin_inset Formula $x$
+\end_inset
+
+ y 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ tal que 
+\begin_inset Formula 
+\[
+R_{n-1}(x;x_{0})=\frac{f^{(n)}(c)}{n!}(x-x_{0})^{n}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Aplicando el teorema del valor medio de Cauchy a 
+\begin_inset Formula 
+\[
+F(t):=f(x)-\left(f(t)+\frac{1}{1!}f'(t)(x-t)+\dots+\frac{1}{(n-1)!}f^{(n-1)}(t)(x-t)^{n-1}\right)
+\]
+
+\end_inset
+
+ y 
+\begin_inset Formula $G(t):=(x-t)^{n}$
+\end_inset
+
+ entre 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ y 
+\begin_inset Formula $x$
+\end_inset
+
+, existe 
+\begin_inset Formula $c$
+\end_inset
+
+ estrictamente entre 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ y 
+\begin_inset Formula $x$
+\end_inset
+
+ tal que 
+\begin_inset Formula $(F(x_{0})-F(x))G'(c)=(G(x_{0})-G(x))F'(c)$
+\end_inset
+
+, pero 
+\begin_inset Formula $F(x)=0$
+\end_inset
+
+, 
+\begin_inset Formula $F(x_{0})=R_{n-1}(x;x_{0})$
+\end_inset
+
+, 
+\begin_inset Formula $G(x)=0$
+\end_inset
+
+ y 
+\begin_inset Formula $G(x_{0})=(x-x_{0})^{n}$
+\end_inset
+
+, luego 
+\begin_inset Formula $R_{n-1}(x;x_{0})G'(c)=(x-x_{0})^{n}F'(c)$
+\end_inset
+
+.
+ Ahora calculamos las derivadas de 
+\begin_inset Formula $G$
+\end_inset
+
+ y 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Se tiene que 
+\begin_inset Formula $G'(t)=-n(x-t)^{n-1}$
+\end_inset
+
+, 
+\begin_inset Formula $G'(c)=-n(x-c)^{n-1}$
+\end_inset
+
+ y
+\begin_inset Formula 
+\[
+\begin{array}{c}
+F'(t)=-\left(f'(t)+\frac{1}{1!}f''(t)(x-t)+\dots+\frac{1}{(n-1)!}f^{(n)}(t)(x-t)^{n-1}\right)+\\
++\left(\frac{1}{1!}f'(t)+\dots+\frac{n-1}{(n-1)!}f^{(n-1)}(t)(x-t)^{n-2}\right)=-\frac{1}{(n-1)!}f^{(n)}(t)(x-t)^{n-1}
+\end{array}
+\]
+
+\end_inset
+
+luego 
+\begin_inset Formula $F'(c)=-\frac{f^{(n)}(c)}{(n-1)!}(x-c)^{n-1}$
+\end_inset
+
+, y sustituyendo, 
+\begin_inset Formula 
+\[
+R_{n-1}(x;x_{0})=\frac{F'(c)}{G'(c)}(x-x_{0})^{n}=\frac{-\frac{f^{(n)}(c)}{(n-1)!}(x-c)^{n-1}}{-n(x-c)^{n-1}}(x-x_{0})^{n}=\frac{f^{(n)}(c)}{n!}(x-x_{0})^{n}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Esta forma de expresar el resto se llama 
+\series bold
+forma de Lagrange
+\series default
+, y a veces se escribe 
+\begin_inset Formula $c=x_{0}+\theta(x-x_{0})$
+\end_inset
+
+ para 
+\begin_inset Formula $0<\theta<1$
+\end_inset
+
+, de modo que si 
+\begin_inset Formula $x_{0}=0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $c=\theta x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Las funciones 
+\series bold
+analíticas
+\series default
+ son funciones de clase 
+\begin_inset Formula ${\cal C}^{\infty}$
+\end_inset
+
+ en las que 
+\begin_inset Formula $f$
+\end_inset
+
+ coincide con su polinomio de Taylor 
+\begin_inset Quotes cld
+\end_inset
+
+infinito
+\begin_inset Quotes crd
+\end_inset
+
+.
+ No todas las de clase 
+\begin_inset Formula ${\cal C}^{\infty}$
+\end_inset
+
+ cumplen esta propiedad, pues, por ejemplo, la función 
+\begin_inset Formula $g(x)=e^{-\frac{1}{x^{2}}}$
+\end_inset
+
+ si 
+\begin_inset Formula $x\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $g(0)=0$
+\end_inset
+
+ cumple que 
+\begin_inset Formula $g^{(n)}(0)=0$
+\end_inset
+
+ para todo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ y por tanto su 
+\series bold
+polinomio de Mac-Laurin
+\series default
+ (polinomio de Taylor en 
+\begin_inset Formula $x_{0}=0$
+\end_inset
+
+, 
+\begin_inset Formula $P_{n}(g,x;0)$
+\end_inset
+
+) es nulo.
+\end_layout
+
+\begin_layout Standard
+Desarrollos de Taylor importantes:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\dots+\frac{x^{n-1}}{(n-1)!}+\frac{e^{\theta x}}{n!}=\left(\sum_{k=0}^{n-1}\frac{x^{k}}{k!}\right)+\frac{e^{\theta x}}{n!}x^{n}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Para cualquier 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+, 
+\begin_inset Formula $f^{(n)}(x)=e^{x}$
+\end_inset
+
+, luego 
+\begin_inset Formula $f^{(n)}(0)=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\dots+\frac{\sin(\theta x+n\pi/2)}{n!}x^{n}=\left(\sum_{k=0}^{\lfloor(n-2)/2\rfloor}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}\right)+\frac{\sin(\theta x+n\pi/2)}{n!}x^{n}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{rccclrcl}
+f(x) & = & \sin x &  &  & f(0) & = & 0\\
+f'(x) & = & \cos x & = & \sin(x+\pi/2) & f'(0) & = & 1\\
+f''(x) & = & -\sin x & = & \sin(x+\pi) & f''(0) & = & 0\\
+f'''(x) & = & -\cos x & = & \sin(x+3\pi/2) & f'''(0) & = & -1\\
+f^{(4)}(x) & = & \sin x & = & \sin(x+2\pi) & f^{(4)}(0) & = & 0\\
+\vdots\\
+f^{(n)}(x) &  &  & = & \sin(x+n\pi/2) & f^{(n)}(0) & = & \sin(n\pi/2)
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\dots+\frac{\cos(\theta x+n\pi/2)}{n!}x^{n}=\left(\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac{(-1)^{k}x^{2k}}{(2k)!}\right)+\frac{\cos(\theta x+n\pi/2)}{n!}x^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\log(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\dots+\frac{(-1)^{n-1}}{n(1+\theta x)^{n}}x^{n}=\left(\sum_{k=1}^{n-1}\frac{(-1)^{k-1}x^{k}}{k}\right)+\frac{(-1)^{n-1}}{n(1+\theta x)^{n}}x^{n}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{rclrcl}
+f(x) & = & \log(1+x) & f(0) & = & 0\\
+f'(x) & = & (1+x)^{-1} & f'(0) & = & 1\\
+f''(x) & = & (-1)(1+x)^{-2} & f''(0) & = & -1=-1!\\
+f'''(x) & = & (-1)(-2)(1+x)^{-3} & f'''(0) & = & (-1)(-2)=2!\\
+\vdots\\
+f^{(n)}(x) & = & (-1)^{n-1}(n-1)!(1+x)^{-n} & f^{(n)} & = & (-1)^{n-1}(n-1)!
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $(1+x)^{\alpha}=1+\binom{\alpha}{1}x+\binom{\alpha}{2}x^{2}+\binom{\alpha}{3}x^{3}+\dots+\binom{\alpha}{n-1}x^{n-1}+\binom{\alpha}{n}\frac{(1+\theta x)^{\alpha}}{(1+\theta x)^{n}}x^{n}=1+\left(\sum_{k=1}^{n-1}\binom{\alpha}{k}x^{k}\right)+\binom{\alpha}{n}\frac{(1+\theta x)^{\alpha}}{(1+\theta x)^{n}}x^{n}$
+\end_inset
+
+, donde 
+\begin_inset Formula $\binom{\alpha}{k}:=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{rclrcl}
+f(x) & = & (1+x)^{\alpha} & f(0) & = & 1\\
+f'(x) & = & \alpha(1+x)^{\alpha-1} & f'(0) & = & \alpha\\
+f''(x) & = & \alpha(\alpha-1)(1+x)^{\alpha-2} & f''(0) & = & \alpha(\alpha-1)\\
+\vdots\\
+f^{(n)}(x) & = & \alpha\cdots(\alpha-n+1)(1+x)^{\alpha-n} & f^{(n)}(0) & = & \alpha\cdots(\alpha-n+1)
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Funciones convexas
+\end_layout
+
+\begin_layout Standard
+Una función 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ es 
+\series bold
+convexa
+\series default
+ en el intervalo 
+\begin_inset Formula $I$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall x,y\in I,t\in[0,1],f((1-t)x+ty)\leq(1-t)f(x)+tf(y)$
+\end_inset
+
+, y es 
+\series bold
+cóncava
+\series default
+ en 
+\begin_inset Formula $I$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall x,y\in I,t\in[0,1],f((1-t)x+ty)\geq(1-t)f(x)+tf(y)$
+\end_inset
+
+.
+ Geométricamente, 
+\begin_inset Formula $f$
+\end_inset
+
+ es convexa en 
+\begin_inset Formula $I$
+\end_inset
+
+ si para cualesquiera 
+\begin_inset Formula $x,y\in I$
+\end_inset
+
+, la secante que une los puntos 
+\begin_inset Formula $(x,f(x))$
+\end_inset
+
+ e 
+\begin_inset Formula $(y,f(y))$
+\end_inset
+
+ está por encima de la gráfica de la función en el intervalo 
+\begin_inset Formula $[x,y]$
+\end_inset
+
+, y cóncava si está por debajo.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\align center
+\begin_inset Graphics
+	filename pegado1.png
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Interpretación geométrica de la convexidad.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+La pendiente de la recta secante que pasa por 
+\begin_inset Formula $(x,f(x))$
+\end_inset
+
+ e 
+\begin_inset Formula $(y,f(y))$
+\end_inset
+
+ se denota 
+\begin_inset Formula $p_{x}(y):=\frac{f(y)-f(x)}{y-x}$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $f$
+\end_inset
+
+ es convexa en 
+\begin_inset Formula $I$
+\end_inset
+
+ si y sólo si para cualesquiera 
+\begin_inset Formula $a,x,b\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $a<x<b$
+\end_inset
+
+ se verifica 
+\begin_inset Formula $p_{a}(x)\leq p_{b}(x)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $x=a+t(b-a)=(1-t)a+tb$
+\end_inset
+
+ con 
+\begin_inset Formula $t\in(0,1)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $x-a=t(b-a)$
+\end_inset
+
+ y 
+\begin_inset Formula $x-b=(1-t)(a-b)$
+\end_inset
+
+, y se tiene que
+\begin_inset Formula 
+\begin{eqnarray*}
+p_{a}(x)\leq p_{b}(x) & \iff & \frac{f(x)-f(a)}{x-a}\leq\frac{f(x)-f(b)}{x-b}\\
+ & \iff & (f(x)-f(a))(x-b)\geq(f(x)-f(b))(x-a)\\
+ & \iff & f(x)(a-b)\geq f(a)(x-b)-f(b)(x-a)\\
+ & \iff & f(x)(a-b)\geq f(a)(1-t)(a-b)-f(b)t(b-a)\\
+ & \iff & f(x)\leq f(a)(1-t)+f(b)t
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ es convexa en un intervalo 
+\begin_inset Formula $I$
+\end_inset
+
+, entonces:
+\end_layout
+
+\begin_layout Enumerate
+Para cada 
+\begin_inset Formula $a\in I$
+\end_inset
+
+, 
+\begin_inset Formula $p_{a}:I\rightarrow\mathbb{R}$
+\end_inset
+
+ es creciente.
+\begin_inset Newline newline
+\end_inset
+
+Sean 
+\begin_inset Formula $a<x<y\in I$
+\end_inset
+
+, 
+\begin_inset Formula 
+\[
+\begin{array}{c}
+p_{a}(x)\leq p_{a}(y)\iff\frac{f(x)-f(a)}{x-a}\leq\frac{f(y)-f(a)}{y-a}\iff\\
+\iff f(x)-f(a)\leq\frac{f(y)-f(a)}{y-a}(x-a)\iff f(x)\leq f(a)+\frac{f(y)-f(a)}{y-a}(x-a)
+\end{array}
+\]
+
+\end_inset
+
+lo cual es cierto por ser 
+\begin_inset Formula $f$
+\end_inset
+
+ convexa.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Lema de las tres pendientes:
+\series default
+ 
+\begin_inset Formula $\forall a,x,b\in I,(a<x<b\implies p_{a}(x)\leq p_{a}(b)\leq p_{b}(x)$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Como 
+\begin_inset Formula $p_{a}$
+\end_inset
+
+ y 
+\begin_inset Formula $p_{b}$
+\end_inset
+
+ son crecientes, 
+\begin_inset Formula $p_{a}(x)\leq p_{a}(b)=p_{b}(a)\leq p_{b}(x)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en los puntos del interior del intervalo.
+\begin_inset Newline newline
+\end_inset
+
+Sea 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ un punto interior de 
+\begin_inset Formula $I$
+\end_inset
+
+ y 
+\begin_inset Formula $x',x\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $x'<x_{0}<x$
+\end_inset
+
+.
+ Por lo anterior, 
+\begin_inset Formula $p_{x_{0}}(x')=p_{x'}(x_{0})\leq p_{x}(x_{0})=p_{x_{0}}(x)$
+\end_inset
+
+, luego 
+\begin_inset Formula $p_{x_{0}}$
+\end_inset
+
+ es creciente y por tanto existe 
+\begin_inset Formula $\alpha:=\lim_{x\rightarrow x_{0}^{+}}p_{x_{0}}(x)$
+\end_inset
+
+.
+ Por otra parte, 
+\begin_inset Formula $f(x)=f(x_{0})+\frac{f(x)-f(x_{0})}{x-x_{0}}(x-x_{0})$
+\end_inset
+
+, y tomando límites, 
+\begin_inset Formula 
+\[
+\lim_{x\rightarrow x_{0}^{+}}f(x)=f(x_{0})+\lim_{x\rightarrow x_{0}^{+}}\frac{f(x)-f(x_{0})}{x-x_{0}}\lim_{x\rightarrow x_{0}^{+}}(x-x_{0})=f(x_{0})+\alpha\cdot0=f(x_{0})
+\]
+
+\end_inset
+
+lo que prueba la continuidad por la derecha de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+.
+ Podemos probar la continuidad por la izquierda de manera análoga.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, sea 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ derivable en el intervalo abierto 
+\begin_inset Formula $I$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+f\text{ es convexa en }I\iff f'\text{ es creciente en }I\iff\forall x_{0},x\in I,f(x)-f(x_{0})\geq f'(x_{0})(x-x_{0})
+\]
+
+\end_inset
+
+La última condición significa que para cada punto de 
+\begin_inset Formula $I$
+\end_inset
+
+, la gráfica de 
+\begin_inset Formula $f$
+\end_inset
+
+ está por encima de la tangente en dicho punto.
+\begin_inset Note Comment
+status open
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $a,b\in I$
+\end_inset
+
+ arbitrarios con 
+\begin_inset Formula $a<b$
+\end_inset
+
+, entonces 
+\begin_inset Formula 
+\begin{eqnarray*}
+f'(a) & = & \lim_{x\rightarrow a^{+}}\frac{f(x)-f(a)}{x-a}=\lim_{x\rightarrow a^{+}}p_{a}(x)\\
+f'(b) & = & \lim_{x'\rightarrow b^{-}}\frac{f(x')-f(b)}{x'-b}=\lim_{x'\rightarrow b^{-}}p_{b}(x')
+\end{eqnarray*}
+
+\end_inset
+
+ Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es convexa, 
+\begin_inset Formula $p_{a}(x)\leq p_{x'}(x)=p_{x}(x')\leq p_{b}(x')$
+\end_inset
+
+ para 
+\begin_inset Formula $a<x<x'<b$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $f'(a)\leq f'(b)$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $f'$
+\end_inset
+
+ es creciente.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $x_{0},x\in I$
+\end_inset
+
+, si 
+\begin_inset Formula $x_{0}<x$
+\end_inset
+
+, por el teorema del valor medio de Lagrange, 
+\begin_inset Formula $f(x)-f(x_{0})=f'(c)(x-x_{0})$
+\end_inset
+
+, pero como 
+\begin_inset Formula $f'$
+\end_inset
+
+ es creciente y 
+\begin_inset Formula $c\in(x_{0},x)$
+\end_inset
+
+, 
+\begin_inset Formula $f'(c)(x-x_{0})\geq f'(x_{0})(x-x_{0})$
+\end_inset
+
+.
+ El caso en que 
+\begin_inset Formula $x_{0}>x$
+\end_inset
+
+ se hace de forma análoga, y el caso en que 
+\begin_inset Formula $x_{0}=x$
+\end_inset
+
+ es trivial.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Si 
+\begin_inset Formula $f$
+\end_inset
+
+ no fuera convexa existirían 
+\begin_inset Formula $a,x_{0},b\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $a<x_{0}<b$
+\end_inset
+
+ tales que 
+\begin_inset Formula $f(x_{0})$
+\end_inset
+
+ estaría por encima de la secante entre 
+\begin_inset Formula $(a,f(a))$
+\end_inset
+
+ y 
+\begin_inset Formula $(b,f(b))$
+\end_inset
+
+, es decir, 
+\begin_inset Formula $p_{b}(x_{0})<p_{b}(a)=p_{a}(b)<p_{a}(x_{0})$
+\end_inset
+
+.
+ Ahora bien, la tangente de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ viene dada por 
+\begin_inset Formula $g(x)=f(x_{0})+f'(x_{0})(x-x_{0})$
+\end_inset
+
+, y si suponemos que 
+\begin_inset Formula $(b,f(b))$
+\end_inset
+
+ está por encima de la recta, entonces
+\begin_inset Formula 
+\[
+\begin{array}{c}
+f(b)>g(b)=f(x_{0})+f'(x_{0})(b-x_{0})\iff f(b)-f(x_{0})>f'(x_{0})(b-x_{0})\iff\\
+\iff f'(x_{0})<\frac{f(b)-f(x_{0})}{b-x_{0}}=p_{b}(x_{0})\overset{\text{hip.}}{<}p_{a}(x_{0})=\frac{f(x_{0})-f(a)}{x_{0}-a}\iff\\
+\iff f'(x_{0})(x_{0}-a)<f(x_{0})-f(a)\iff f(a)<f(x_{0})+f'(x_{0})(x_{0}-a)
+\end{array}
+\]
+
+\end_inset
+
+por lo que 
+\begin_inset Formula $(a,f(a))$
+\end_inset
+
+ queda por debajo de la tangente.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Convexidad local:
+\series default
+ 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ y derivable en 
+\begin_inset Formula $x_{0}\in I$
+\end_inset
+
+ es 
+\series bold
+convexa
+\series default
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ si 
+\begin_inset Formula $\exists\delta>0:\forall x\in B(x_{0},\delta)\cap I,f(x)\geq f(x_{0})+f'(x_{0})(x-x_{0})$
+\end_inset
+
+, y es 
+\series bold
+cóncava
+\series default
+ en 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ si 
+\begin_inset Formula $\exists\delta>0:\forall x\in B(x_{0},\delta)\cap I,f(x)\leq f(x_{0})+f'(x_{0})(x-x_{0})$
+\end_inset
+
+.
+ Decimos que 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ es un 
+\series bold
+punto de inflexión
+\series default
+ si existe 
+\begin_inset Formula $\delta>0$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $x\in B(x_{0},\delta)\cap I$
+\end_inset
+
+ entonces 
+\begin_inset Formula $x<x_{0}$
+\end_inset
+
+ implica 
+\begin_inset Formula $f(x)>f(x_{0})+f'(x_{0})(x-x_{0})$
+\end_inset
+
+ mientras que 
+\begin_inset Formula $x>x_{0}$
+\end_inset
+
+ implica 
+\begin_inset Formula $f(x)<f(x_{0})+f'(x_{0})(x-x_{0})$
+\end_inset
+
+ (o al revés).
+ Puede no darse ninguna de las tres situaciones como en el punto 
+\begin_inset Formula $x_{0}=0$
+\end_inset
+
+ en 
+\begin_inset Formula $f(x)=x^{2}\sin(1/x)$
+\end_inset
+
+.
+ Una función 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ derivable en el intervalo abierto 
+\begin_inset Formula $I$
+\end_inset
+
+ es convexa en 
+\begin_inset Formula $I$
+\end_inset
+
+ si y sólo si es convexa para cada 
+\begin_inset Formula $x\in I$
+\end_inset
+
+.
+\begin_inset Note Comment
+status open
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Ver teorema anterior.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Supongamos que existen 
+\begin_inset Formula $a,b,c\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $a<c<b$
+\end_inset
+
+ tales que 
+\begin_inset Formula $f(c)>f(a)+\frac{f(b)-f(a)}{b-a}(c-a)$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $g(x)=f(x)-\left(f(a)+\frac{f(b)-f(a)}{b-a}(x-a)\right)$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $0=g(a)=g(b)<g(c)$
+\end_inset
+
+, existe un máximo absoluto de 
+\begin_inset Formula $g$
+\end_inset
+
+ en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ y en este 
+\begin_inset Formula $g'(\xi)=0$
+\end_inset
+
+, así que 
+\begin_inset Formula $f'(\xi)=\frac{f(b)-f(a)}{b-a}$
+\end_inset
+
+ y 
+\begin_inset Formula $g(z)<g(\xi)\forall z\in[a,b]$
+\end_inset
+
+, luego
+\begin_inset Formula 
+\[
+g(b)=f(b)-\left(f(a)+\frac{f(b)-f(a)}{b-a}(b-a)\right)<f(\xi)-\left(f(a)+\frac{f(b)-f(a)}{b-a}(\xi-a)\right)
+\]
+
+\end_inset
+
+es decir,
+\begin_inset Formula 
+\[
+f(b)<f(\xi)+\frac{f(b)-f(a)}{b-a}(b-\xi)=f(\xi)+f'(\xi)(b-\xi)
+\]
+
+\end_inset
+
+lo que contradice la convexidad de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $\xi$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Como 
+\begin_inset Formula $f:I\rightarrow\mathbb{R}$
+\end_inset
+
+ es cóncava si y sólo si 
+\begin_inset Formula $-f$
+\end_inset
+
+ es convexa, todas las proposiciones sobre funciones convexas se pueden
+ aplicar a funciones cóncavas adaptándolas convenientemente.
+\end_layout
+
+\begin_layout Section
+Representación gráfica de funciones
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $y=f(x)$
+\end_inset
+
+.
+ La recta 
+\begin_inset Formula $x=a$
+\end_inset
+
+ es una 
+\series bold
+asíntota vertical
+\series default
+ de 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ si 
+\begin_inset Formula $\lim_{x\rightarrow a}f(x)=\pm\infty$
+\end_inset
+
+, sea el límite por la izquierda o por la derecha.
+ La recta 
+\begin_inset Formula $y=b$
+\end_inset
+
+ es una 
+\series bold
+asíntota horizontal
+\series default
+ de 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ si 
+\begin_inset Formula $\lim_{x\rightarrow\infty}f(x)=b$
+\end_inset
+
+, sea cuando 
+\begin_inset Formula $x$
+\end_inset
+
+ tiende a 
+\begin_inset Formula $-\infty$
+\end_inset
+
+ o a 
+\begin_inset Formula $+\infty$
+\end_inset
+
+.
+ Finalmente, la recta 
+\begin_inset Formula $y=mx+b$
+\end_inset
+
+ es una 
+\series bold
+asíntota oblicua
+\series default
+ de 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ si 
+\begin_inset Formula $\lim_{x\rightarrow\infty}(f(x)-(mx+b))=0$
+\end_inset
+
+, y entonces podemos calcular 
+\begin_inset Formula $m$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ como 
+\begin_inset Formula $m=\lim_{x\rightarrow\infty}\frac{f(x)}{x}$
+\end_inset
+
+ y 
+\begin_inset Formula $b=\lim_{x\rightarrow\infty}(f(x)-mx)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una función 
+\begin_inset Formula $f:D\subseteq\mathbb{R}\rightarrow\mathbb{R}$
+\end_inset
+
+ es 
+\series bold
+par
+\series default
+ o 
+\series bold
+simétrica respecto del eje de coordenadas 
+\series default
+si 
+\begin_inset Formula $f(-x)=f(x)\forall x\in D$
+\end_inset
+
+, y es 
+\series bold
+impar
+\series default
+ o 
+\series bold
+simétrica respecto del origen de coordenadas
+\series default
+ si 
+\begin_inset Formula $f(-x)=-f(x)\forall x\in D$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/fuvr2/n2.lyx b/fuvr2/n2.lyx
new file mode 100644
index 0000000..9d5d103
--- /dev/null
+++ b/fuvr2/n2.lyx
@@ -0,0 +1,3720 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Una 
+\series bold
+partición
+\series default
+ de 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ es una colección de puntos 
+\begin_inset Formula $a=t_{0}<t_{1}<\dots<t_{n}=b$
+\end_inset
+
+, y llamamos 
+\begin_inset Formula ${\cal P}([a,b])$
+\end_inset
+
+ al conjunto de todas las particiones de 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+.
+ Dada 
+\begin_inset Formula $\pi\equiv(t_{0}<\dots<t_{n})\in{\cal P}([a,b])$
+\end_inset
+
+, escribimos 
+\begin_inset Formula $M_{i}:=\sup\{f(t)\}_{t\in[t_{i-1},t_{i}]}$
+\end_inset
+
+ y 
+\begin_inset Formula $m_{i}:=\inf\{f(t)\}_{t\in[t_{i-1},t_{i}]}$
+\end_inset
+
+, y llamamos 
+\series bold
+suma superior
+\series default
+ y 
+\series bold
+suma inferior
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ correspondiente a 
+\begin_inset Formula $\pi$
+\end_inset
+
+, respectivamente, a
+\begin_inset Formula 
+\begin{eqnarray*}
+S(f,\pi):=\sum_{i=1}^{n}M_{i}(t_{i}-t_{i-1}) & \text{ y } & s(f,\pi):=\sum_{i=1}^{n}m_{i}(t_{i}-t_{i-1})
+\end{eqnarray*}
+
+\end_inset
+
+ 
+\end_layout
+
+\begin_layout Standard
+Obviamente 
+\begin_inset Formula $s(f,\pi)\leq S(f,\pi)$
+\end_inset
+
+ para cualquier 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+.
+ Dadas 
+\begin_inset Formula $\pi,\pi'\in{\cal P}([a,b])$
+\end_inset
+
+, decimos que 
+\begin_inset Formula $\pi'$
+\end_inset
+
+ es 
+\series bold
+más fina
+\series default
+ que 
+\begin_inset Formula $\pi$
+\end_inset
+
+ (
+\begin_inset Formula $\pi'\succ\pi$
+\end_inset
+
+) si 
+\begin_inset Formula $\pi'\supseteq\pi$
+\end_inset
+
+, y denotamos 
+\begin_inset Formula $\pi\lor\pi':=\pi\cup\pi'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $\pi\preceq\pi'$
+\end_inset
+
+ entonces 
+\begin_inset Formula $s(f,\pi)\leq s(f,\pi')$
+\end_inset
+
+ y 
+\begin_inset Formula $S(f,\pi)\geq S(f,\pi')$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Supongamos que 
+\begin_inset Formula $\pi'$
+\end_inset
+
+ tiene un punto más que 
+\begin_inset Formula $\pi$
+\end_inset
+
+, con 
+\begin_inset Formula $\pi\equiv t_{0}<\dots<t_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $\pi'\equiv t_{0}<\dots<t_{k-1}<p<t_{k}<\dots<t_{n}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $s(f,\pi)=\sum_{i\neq k}m_{i}(t_{i}-t_{i-1})+m_{k}(t_{k}-t_{k-1})=\sum_{i\neq k}m_{i}(t_{i}-t_{i-1})+m_{k}((t_{k}-p)+(p-t_{k-1}))\leq\sum_{i\neq k}m_{i}(t_{i}-t_{i-1})+\inf\{f(t)\}_{t\in[t_{k-1},p]}(p-t_{k-1})+\inf\{f(t)\}_{t\in[p,t_{k}]}(t_{k}-p)=s(f,\pi')$
+\end_inset
+
+.
+ La segunda afirmación se hace de forma análoga.
+\end_layout
+
+\begin_layout Standard
+Dadas 
+\begin_inset Formula $\pi,\pi'\in{\cal P}([a,b])$
+\end_inset
+
+, 
+\begin_inset Formula $s(f,\pi)\leq S(f,\pi')$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Como 
+\begin_inset Formula $\pi,\pi'\prec\pi\lor\pi'$
+\end_inset
+
+, entonces 
+\begin_inset Formula $s(f,\pi)\leq s(f,\pi\lor\pi')\leq S(f,\pi\lor\pi')\leq S(f,\pi')$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Llamamos pues 
+\series bold
+integral inferior
+\series default
+ e 
+\series bold
+integral superior
+\series default
+ (
+\series bold
+de Darboux
+\series default
+), respectivamente, a
+\begin_inset Formula 
+\begin{eqnarray*}
+\underline{\int_{a}^{b}}f:=\sup\{s(f,\pi)\}_{\pi\in{\cal P}([a,b])} & \text{ y } & \overline{\int_{a}^{b}}f:=\inf\{S(f,\pi)\}_{\pi\in{\cal P}([a,b])}
+\end{eqnarray*}
+
+\end_inset
+
+Decimos que 
+\begin_inset Formula $f$
+\end_inset
+
+ es 
+\series bold
+integrable Riemann
+\series default
+ en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+, escrito 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+, si las integrales superior e inferior coinciden y llamamos 
+\series bold
+integral Riemann
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+, escrito 
+\begin_inset Formula $\int_{a}^{b}f$
+\end_inset
+
+, a este valor.
+ Definimos, para 
+\begin_inset Formula $a<b$
+\end_inset
+
+, 
+\begin_inset Formula $\int_{b}^{a}f:=-\int_{a}^{b}f$
+\end_inset
+
+, e 
+\begin_inset Formula $\int_{a}^{a}f=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Caracterización
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, dada 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ acotada, 
+\begin_inset Formula $f\in{\cal R}[a,b]\iff\forall\varepsilon>0,\exists\pi\in{\cal P}([a,b]):S(f,\pi)-s(f,\pi)<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, como 
+\begin_inset Formula $\int_{a}^{b}f=\inf\{S(f,\pi)\}_{\pi\in{\cal P}([a,b])}$
+\end_inset
+
+, existe 
+\begin_inset Formula $\pi_{1}\in{\cal P}([a,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq S(f,\pi_{1})-\int_{a}^{b}f<\frac{\varepsilon}{2}$
+\end_inset
+
+, y análogamente existe 
+\begin_inset Formula $\pi_{2}\in{\cal P}([a,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq\int_{a}^{b}f-s(f,\pi_{2})<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\pi:=\pi_{1}\lor\pi_{2}$
+\end_inset
+
+ cumple ambas desigualdades, pues 
+\begin_inset Formula $S(f,\pi)\leq S(f,\pi_{1})$
+\end_inset
+
+ y 
+\begin_inset Formula $s(f,\pi)\geq s(f,\pi_{2})$
+\end_inset
+
+, y sumándolas obtenemos 
+\begin_inset Formula $S(f,\pi)-s(f,\pi)<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ y 
+\begin_inset Formula $\pi_{\varepsilon}\in{\cal P}([a,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $S(f,\pi_{\varepsilon})-s(f,\pi_{\varepsilon})<\varepsilon$
+\end_inset
+
+, por la definición de integral superior e inferior, 
+\begin_inset Formula $0\leq\overline{\int_{a}^{b}}f-\underline{\int_{a}^{b}}f\leq S(f,\pi_{\varepsilon})-s(f,\pi_{\varepsilon})\leq\varepsilon$
+\end_inset
+
+, lo que para 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ arbitrario implica que las integrales superior e inferior coinciden.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f\in{\cal R}[a,b]\iff\exists!\alpha\in\mathbb{R}:\forall\pi\in{\cal P}([a,b]),s(f,\pi)\leq\alpha\leq S(f,\pi)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $\alpha:=\int_{a}^{b}f$
+\end_inset
+
+, para toda 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+, 
+\begin_inset Formula $s(f,\pi)\leq\alpha\leq S(f,\pi)$
+\end_inset
+
+.
+ Si existiera 
+\begin_inset Formula $\beta\neq\alpha$
+\end_inset
+
+ que cumpliera la condición, como 
+\begin_inset Formula $\alpha=\sup\{s(f,\pi)\}_{\pi\in{\cal P}([a,b])}$
+\end_inset
+
+ se tendría 
+\begin_inset Formula $\beta>\alpha$
+\end_inset
+
+, pero análogamente que 
+\begin_inset Formula $\beta<\alpha$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Supongamos que existe un 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ que verifica la condición pero 
+\begin_inset Formula $f\notin{\cal R}[a,b]$
+\end_inset
+
+.
+ Entonces para cualquier 
+\begin_inset Formula $\pi\in{\cal R}[a,b]$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $s(f,\pi)\leq\underline{\int_{a}^{b}}f<\overline{\int_{a}^{b}}f\leq S(f,\pi)$
+\end_inset
+
+, por lo que existen infinitos números reales que verifican la condición
+ y por tanto 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ no es único.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Otro 
+\series bold
+teorema 
+\series default
+importante es que las funciones 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continuas son integrables en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+, y además, dados 
+\begin_inset Formula $z_{k,n}\in[a+\frac{b-a}{n}(k-1),a+\frac{b-a}{n}k]$
+\end_inset
+
+ cualesquiera,
+\begin_inset Formula 
+\[
+\lim_{n\rightarrow\infty}\frac{b-a}{n}\sum_{k=1}^{n}f(z_{k,n})=\int_{a}^{b}f
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Dado 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+, 
+\begin_inset Formula $S(f,\pi)-s(f,\pi)=\sum_{i=1}^{n}(M_{i}-m_{i})(t_{i}-t_{i-1})$
+\end_inset
+
+.
+ Ahora bien, dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, como 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ también es uniformemente continua, luego existe 
+\begin_inset Formula $\delta>0$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $|x-y|<\delta$
+\end_inset
+
+ entonces 
+\begin_inset Formula $|f(x)-f(y)|<\frac{\varepsilon}{2(b-a)}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $\frac{b-a}{n_{0}}<\delta$
+\end_inset
+
+.
+ Para todo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ definimos 
+\begin_inset Formula $\pi_{n}=(a<a+\frac{b-a}{n}<\dots<a+n\frac{b-a}{n}=b)\in{\cal P}([a,b])$
+\end_inset
+
+ y 
+\begin_inset Formula $t_{k,n}=a+k\frac{b-a}{n}$
+\end_inset
+
+, y tenemos que para 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+ es 
+\begin_inset Formula $t_{k,n}-t_{k-1,n}<\delta$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $M_{k,n}-m_{k,n}\leq\frac{\varepsilon}{2(b-a)}$
+\end_inset
+
+.
+ Entonces
+\begin_inset Formula 
+\[
+S(f,\pi_{n_{0}})-s(f,\pi_{n_{0}})\leq\sum_{i=1}^{n_{0}}\frac{\varepsilon}{2(b-a)}(t_{i,n_{0}}-t_{i-1,n_{0}})=\frac{\varepsilon}{2}<\varepsilon
+\]
+
+\end_inset
+
+De aquí que 
+\begin_inset Formula $f$
+\end_inset
+
+ es integrable.
+ Pero entonces existe un único 
+\begin_inset Formula $\alpha=\int_{a}^{b}f$
+\end_inset
+
+ tal que para 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+ es 
+\begin_inset Formula $s(f,\pi)\leq\alpha\leq S(f,\pi)$
+\end_inset
+
+, y en particular, 
+\begin_inset Formula $s(f,\pi_{n})\leq\alpha\leq S(f,\pi_{n})$
+\end_inset
+
+ para todo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $z_{k,n}\in[a+\frac{b-a}{n}(k-1),a+\frac{b-a}{n}k]$
+\end_inset
+
+ para 
+\begin_inset Formula $1\leq k\leq n$
+\end_inset
+
+ arbitrario y 
+\begin_inset Formula $a_{n}=\frac{b-a}{n}\sum_{k=1}^{n}f(z_{k,n})$
+\end_inset
+
+.
+ Por definición, 
+\begin_inset Formula $s(f,\pi_{n})\leq a_{n}\leq S(f,\pi_{n})$
+\end_inset
+
+, y dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, existe 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $S(f,\pi_{n})-s(f,\pi_{n})<\frac{\varepsilon}{2}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $S(f,\pi_{n})-\alpha\leq\frac{\varepsilon}{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $S(f,\pi_{n})-a_{n}<\frac{\varepsilon}{2}$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $|a_{n}-\alpha|\leq|a_{n}-S(f,\pi_{n})|+|S(f,\pi_{n})-\alpha|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dada 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ monótona y acotada entonces 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Dada 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+, 
+\begin_inset Formula $S(f,\pi)-s(f,\pi)=\sum_{i=1}^{n}(M_{i}-m_{i})(t_{i}-t_{i-1})$
+\end_inset
+
+, y dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, si por ejemplo 
+\begin_inset Formula $f$
+\end_inset
+
+ es monótona creciente y 
+\begin_inset Formula $f(a)<f(b)$
+\end_inset
+
+, dada 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $t_{i}-t_{i-1}<\frac{\varepsilon}{f(b)-f(a)}$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $M_{i}=f(t_{i})$
+\end_inset
+
+, 
+\begin_inset Formula $m_{i}=f(t_{i-1})$
+\end_inset
+
+ y 
+\begin_inset Formula $S(f,\pi)-s(f,\pi)=\sum_{i=1}^{n}(M_{i}-m_{i})(t_{i}-t_{i-1})\leq\sum_{i=1}^{n}(M_{i}-m_{i})\frac{\varepsilon}{f(b)-f(a)}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ es acotada y 
+\begin_inset Formula $f\in{\cal R}[c,b]\forall c>a$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $A>0$
+\end_inset
+
+ con 
+\begin_inset Formula $|f(x)|\leq A\forall x\in[a,b]$
+\end_inset
+
+, entonces 
+\begin_inset Formula $-A\leq\inf\{f(x)\}_{x\in[a,b]}\leq\sup\{f(x)\}_{x\in[a,b]}\leq A$
+\end_inset
+
+.
+ Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, sea 
+\begin_inset Formula $c\in(a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $c-a<\frac{\varepsilon}{4A}$
+\end_inset
+
+ y 
+\begin_inset Formula $\pi\in{\cal P}([c,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $S(f,\pi)-s(f,\pi)<\frac{\varepsilon}{2}$
+\end_inset
+
+, si tomamos 
+\begin_inset Formula $\pi'\in{\cal P}([a,b])$
+\end_inset
+
+ resultado de añadir a 
+\begin_inset Formula $\pi$
+\end_inset
+
+ el intervalo 
+\begin_inset Formula $[a,c]$
+\end_inset
+
+ con 
+\begin_inset Formula $M_{1}=\sup\{f(x)\}_{x\in[a,c]}$
+\end_inset
+
+ y 
+\begin_inset Formula $m_{1}=\inf\{f(x)\}_{x\in[a,c]}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $S(f,\pi')-s(f,\pi')=M_{1}(c-a)+S(f,\pi)-m_{1}(c-a)-s(f,\pi)\leq2A(c-a)+S(f,\pi)-s(f,\pi)\leq2A(c-a)+\frac{\varepsilon}{2}<2A\frac{\varepsilon}{4A}+\frac{\varepsilon}{2}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Sumas de Riemann
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $\pi\equiv(t_{0}<\dots<t_{n})\in{\cal P}([a,b])$
+\end_inset
+
+, llamamos 
+\series bold
+suma de Riemann
+\series default
+ asociada a la partición 
+\begin_inset Formula $\pi$
+\end_inset
+
+ y los puntos 
+\begin_inset Formula $z_{i}\in[t_{i-1},t_{i}]$
+\end_inset
+
+ a
+\begin_inset Formula 
+\[
+S(f,\pi,z_{i}):=\sum_{i=1}^{n}f(z_{i})(t_{i}-t_{i-1})
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f$
+\end_inset
+
+ es integrable Riemann en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ si y sólo si existe 
+\begin_inset Formula $A\in\mathbb{R}$
+\end_inset
+
+ tal que para 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ existe 
+\begin_inset Formula $\pi_{0}\in{\cal P}([a,b])$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $\pi_{0}\prec\pi$
+\end_inset
+
+, para cualesquiera 
+\begin_inset Formula $z_{i}\in[t_{i-1},t_{i}]$
+\end_inset
+
+ se cumple 
+\begin_inset Formula $|A-S(f,\pi,z_{i})|<\varepsilon$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $A=\int_{a}^{b}f$
+\end_inset
+
+.
+\begin_inset Note Comment
+status open
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $A=\int_{a}^{b}f$
+\end_inset
+
+, fijado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, sea 
+\begin_inset Formula $\pi_{0}\in{\cal P}([a,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $S(f,\pi_{0})-s(f,\pi_{0})<\varepsilon$
+\end_inset
+
+, si 
+\begin_inset Formula $\pi_{0}\prec\pi$
+\end_inset
+
+ entonces 
+\begin_inset Formula $S(f,\pi)-s(f,\pi)\leq S(f,\pi_{0})-s(f,\pi_{0})<\varepsilon$
+\end_inset
+
+, 
+\begin_inset Formula $s(f,\pi)\leq S(f,\pi,z_{i})\leq S(f,\pi)$
+\end_inset
+
+ y 
+\begin_inset Formula $s(f,\pi)\leq A\leq S(f,\pi)$
+\end_inset
+
+.
+ Pero esto implica que 
+\begin_inset Formula $0\leq A-s(f,\pi)\leq S(f,\pi)-s(f,\pi)\leq\varepsilon$
+\end_inset
+
+, 
+\begin_inset Formula $A-S(f,\pi,z_{i})\leq S(f,\pi)-s(f,\pi)\leq\varepsilon$
+\end_inset
+
+ y 
+\begin_inset Formula $S(f,\pi,z_{i})-A\geq s(f,\pi)-S(f,\pi)\geq-\varepsilon$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $|A-S(f,\pi,z_{i})|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, sea 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $|A-S(f,\pi,z_{i})|<\frac{\varepsilon}{2}$
+\end_inset
+
+ para puntos 
+\begin_inset Formula $z_{i}$
+\end_inset
+
+ con 
+\begin_inset Formula $M_{i}-f(z_{i})<\frac{\varepsilon}{2(b-a)}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $S(f,\pi)-S(f,\pi,z_{i})=\sum_{i=1}^{n}(M_{i}-f(z_{i}))(t_{i}-t_{i-1})\leq\sum_{i=1}^{n}\frac{\varepsilon}{2(b-a)}(t_{i}-t_{i-1})=\frac{\varepsilon}{2}$
+\end_inset
+
+, y como 
+\begin_inset Formula $|A-S(f,\pi,z_{i})|<\frac{\varepsilon}{2}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $|A-S(f,\pi)|<\varepsilon$
+\end_inset
+
+.
+ Análogamente se tiene que 
+\begin_inset Formula $|A-s(f,\pi)|<\varepsilon$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $|S(f,\pi)-s(f,\pi)|<2\varepsilon$
+\end_inset
+
+ y 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Queda ver que 
+\begin_inset Formula $A=\int_{a}^{b}f$
+\end_inset
+
+.
+ Supongamos que existe 
+\begin_inset Formula $\pi_{0}$
+\end_inset
+
+ con 
+\begin_inset Formula $s(f,\pi_{0})\leq S(f,\pi_{0})<A$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $\varepsilon=A-S(f,\pi_{0})$
+\end_inset
+
+, existe por hipótesis 
+\begin_inset Formula $\pi_{1}$
+\end_inset
+
+ tal que para 
+\begin_inset Formula $\pi\succ\pi_{1}$
+\end_inset
+
+ y elección de 
+\begin_inset Formula $z_{i}$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $|A-S(f,\pi,z_{i})|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula $\pi'=\pi_{0}\lor\pi_{1}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $S(f,\pi',z_{i})>A-\frac{\varepsilon}{2}=\frac{A+S(f,\pi_{0})}{2}>S(f,\pi_{0})$
+\end_inset
+
+, pero al mismo tiempo 
+\begin_inset Formula $S(f,\pi',z_{i})<S(f,\pi')\leq S(f,\pi_{0})$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Un conjunto 
+\begin_inset Formula $A\subseteq\mathbb{R}$
+\end_inset
+
+ tiene 
+\series bold
+medida cero
+\series default
+ si para cada 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ existe una sucesión 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+ de intervalos cerrados y acotados con 
+\begin_inset Formula $A\subseteq\bigcup_{n}I_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $\sum_{n=1}^{\infty}\text{long}(I_{n})\leq\varepsilon$
+\end_inset
+
+, donde 
+\begin_inset Formula $\text{long}([a,b]):=b-a$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $A$
+\end_inset
+
+ tiene medida cero y 
+\begin_inset Formula $B\subseteq A$
+\end_inset
+
+ entonces 
+\begin_inset Formula $B$
+\end_inset
+
+ tiene medida cero, y si 
+\begin_inset Formula $A$
+\end_inset
+
+ es numerable tiene medida cero tomando, para cada 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, la sucesión con 
+\begin_inset Formula $I_{n}=\{x_{n}-\frac{\varepsilon}{2^{n+1}},x_{n}+\frac{\varepsilon}{2^{n+1}}\}$
+\end_inset
+
+, pues 
+\begin_inset Formula $\sum_{n}\text{long}(I_{n})=\sum_{n}\frac{\varepsilon}{2^{n}}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de Lebesgue
+\series default
+ afirma que dada una función acotada 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+, si 
+\begin_inset Formula $D(f)\subseteq[a,b]$
+\end_inset
+
+ es el conjunto de puntos en los que 
+\begin_inset Formula $f$
+\end_inset
+
+ no es continua, entonces 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $D(f)$
+\end_inset
+
+ tiene medida cero.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $\pi=(t_{0}<\dots<t_{n})\in{\cal P}([a,b])$
+\end_inset
+
+, llamamos 
+\series bold
+norma
+\series default
+ de 
+\begin_inset Formula $\pi$
+\end_inset
+
+ a 
+\begin_inset Formula $\Vert\pi\Vert:=\max\{t_{i}-t_{i-1}\}_{1\leq i\leq n}$
+\end_inset
+
+.
+ Como 
+\series bold
+teorema
+\series default
+,
+\series bold
+ 
+\series default
+si 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ es acotada, son equivalentes:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+ y 
+\begin_inset Formula $A=\int_{a}^{b}f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists A\in\mathbb{R}:\forall\varepsilon>0,\exists\pi_{0}\in{\cal P}([a,b]):\forall\pi\succ\pi_{0},|A-S(f,\pi,z_{i})|<\varepsilon$
+\end_inset
+
+ para cualquier suma de Riemann correspondiente a 
+\begin_inset Formula $\pi$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists A\in\mathbb{R}:\forall\varepsilon>0,\exists\delta>0:\forall\pi:\Vert\pi\Vert<\delta,|A-S(f,\pi,z_{i})|<\varepsilon$
+\end_inset
+
+ para cualquier suma de Riemann correspondiente a 
+\begin_inset Formula $\pi$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Propiedades
+\end_layout
+
+\begin_layout Description
+Linealidad 
+\begin_inset Formula ${\cal R}[a,b]$
+\end_inset
+
+ es un 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+-espacio vectorial y el operador 
+\begin_inset Formula $\int_{a}^{b}$
+\end_inset
+
+ es lineal.
+\begin_inset Newline newline
+\end_inset
+
+Sean 
+\begin_inset Formula $f,g\in{\cal R}[a,b]$
+\end_inset
+
+, dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ existe 
+\begin_inset Formula $\pi_{0}\in{\cal P}([a,b])$
+\end_inset
+
+ tal que para 
+\begin_inset Formula $\pi_{0}\prec\pi$
+\end_inset
+
+ se tienen 
+\begin_inset Formula $\left|\int_{a}^{b}f-S(f,\pi,z_{i})\right|,\left|\int_{a}^{b}g-S(g,\pi,z_{i})\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula 
+\[
+\left|\int_{a}^{b}f+\int_{a}^{b}g-S(f+g,\pi,z_{i})\right|<\varepsilon
+\]
+
+\end_inset
+
+con lo que 
+\begin_inset Formula $\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $k\in\mathbb{R}$
+\end_inset
+
+, dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ y 
+\begin_inset Formula $\pi_{0}\in{\cal P}([a,b])$
+\end_inset
+
+ tal que para 
+\begin_inset Formula $\pi_{0}\prec\pi$
+\end_inset
+
+ se cumple 
+\begin_inset Formula $\left|\int_{a}^{b}f-S(f,\pi,z_{i})\right|<\frac{\varepsilon}{1+|k|}$
+\end_inset
+
+, entonces 
+\begin_inset Formula 
+\[
+\left|k\int_{a}^{b}f-S(kf,\pi,z_{i})\right|=|k|\left|\int_{a}^{b}f-S(f,\pi,z_{i})\right|<|k|\frac{\varepsilon}{1+|k|}<\varepsilon
+\]
+
+\end_inset
+
+luego 
+\begin_inset Formula $\int_{a}^{b}kf=k\int_{a}^{b}f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+Producto Si 
+\begin_inset Formula $f,g:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ son integrables Riemann, también lo es 
+\begin_inset Formula $fg$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Por el teorema de Lebesgue, si 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+, tendrá medida cero, pero 
+\begin_inset Formula $D(f^{2})\subseteq D(f)$
+\end_inset
+
+, pues si 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en un punto también lo es 
+\begin_inset Formula $f^{2}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $D(f^{2})$
+\end_inset
+
+ tiene medida cero, lo que nos da la integrabilidad de 
+\begin_inset Formula $f^{2}$
+\end_inset
+
+.
+ El caso general se sigue de que 
+\begin_inset Formula $fg=\frac{1}{2}\left((f+g)^{2}-f^{2}-g^{2}\right)$
+\end_inset
+
+ por la linealidad.
+\end_layout
+
+\begin_layout Description
+Monotonía Si 
+\begin_inset Formula $f(x)\leq g(x)$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\int_{a}^{b}f\leq\int_{a}^{b}g$
+\end_inset
+
+, y en particular si 
+\begin_inset Formula $m\leq f(x)\leq M$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+, entonces 
+\begin_inset Formula $m(b-a)\leq\int_{a}^{b}f\leq M(b-a)$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Para 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $s(f,\pi)\leq s(g,\pi)$
+\end_inset
+
+, y tomando supremos, 
+\begin_inset Formula $\int_{a}^{b}f\leq\int_{a}^{b}g$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+Valor
+\begin_inset space ~
+\end_inset
+
+medio Sea 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continua, existe 
+\begin_inset Formula $c\in[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $f(c)=\frac{1}{b-a}\int_{a}^{b}f$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Por el teorema de Weierstrass, existen 
+\begin_inset Formula $c_{1},c_{2}\in[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $f(c_{1})\leq f(x)\leq f(c_{2})$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+, y por la monotonía de la integral, 
+\begin_inset Formula $f(c_{1})\leq\frac{1}{b-a}\int_{a}^{b}f\leq f(c_{2})$
+\end_inset
+
+.
+ Entonces, aplicando la propiedad de los valores intermedios, existe 
+\begin_inset Formula $c\in[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $f(c)=\frac{1}{b-a}\int_{a}^{b}f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+Valor
+\begin_inset space ~
+\end_inset
+
+absoluto Si 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+ entonces 
+\begin_inset Formula $|f|\in{\cal R}[a,b]$
+\end_inset
+
+ y 
+\begin_inset Formula $\left|\int_{a}^{b}f\right|\leq\int_{a}^{b}|f|$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, sea 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+ con 
+\begin_inset Formula $S(f,\pi)-s(f,\pi)<\varepsilon$
+\end_inset
+
+, si 
+\begin_inset Formula $M'_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $m'_{i}$
+\end_inset
+
+ son el supremo y el ínfimo, respectivamente, de 
+\begin_inset Formula $|f|$
+\end_inset
+
+ en 
+\begin_inset Formula $[t_{i-1},t_{i}]$
+\end_inset
+
+, y 
+\begin_inset Formula $M_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+ son los de 
+\begin_inset Formula $f$
+\end_inset
+
+, entonces para 
+\begin_inset Formula $z,w\in[t_{i-1},t_{i}]$
+\end_inset
+
+ se tiene que 
+\begin_inset Formula $||f(z)|-|f(w)||\leq|f(z)-f(w)|\leq M_{i}-m_{i}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\sup\{|f(z)|-|f(w)|\}_{z,w\in[t_{i-1},t_{i}]}=M'_{i}-m'_{i}\leq M_{i}-m_{i}$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $S(|f|,\pi)-s(|f|,\pi)\leq S(f,\pi)-s(f,\pi)<\varepsilon$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $|f|\in{\cal R}[a,b]$
+\end_inset
+
+.
+ Ahora bien, 
+\begin_inset Formula $-|f(x)|\leq f(x)\leq|f(x)|$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\int_{a}^{b}-|f|=-\int_{a}^{b}|f|\leq\int_{a}^{b}f\leq\int_{a}^{b}|f|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+Aditividad
+\begin_inset space ~
+\end_inset
+
+respecto
+\begin_inset space ~
+\end_inset
+
+de
+\begin_inset space ~
+\end_inset
+
+intervalo Dada 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ acotada y 
+\begin_inset Formula $c\in[a,b]$
+\end_inset
+
+, 
+\begin_inset Formula $f\in{\cal R}[a,b]\iff f\in{\cal R}[a,c],{\cal R}[c,b]$
+\end_inset
+
+, y además 
+\begin_inset Formula $\int_{a}^{b}f=\int_{a}^{c}f+\int_{c}^{b}f$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Basta refinar una partición 
+\begin_inset Formula $\pi\in{\cal P}([a,b])$
+\end_inset
+
+ añadiéndole el punto 
+\begin_inset Formula $c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+Discontinuidades Si 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+ y 
+\begin_inset Formula $g:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ coincide con 
+\begin_inset Formula $f$
+\end_inset
+
+ salvo en un número finito de puntos, entonces 
+\begin_inset Formula $g\in{\cal R}[a,b]$
+\end_inset
+
+ y 
+\begin_inset Formula $\int_{a}^{b}f=\int_{a}^{b}g$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Supongamos que cambian en un punto 
+\begin_inset Formula $c\in[a,b]$
+\end_inset
+
+, y basta probar que 
+\begin_inset Formula $h:=g-f$
+\end_inset
+
+ es integrable.
+ Ahora bien, 
+\begin_inset Formula $h$
+\end_inset
+
+ es nula en todos los puntos salvo en 
+\begin_inset Formula $c$
+\end_inset
+
+, por lo que dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ podemos tomar 
+\begin_inset Formula $\pi\in{\cal P}[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $t_{i}-t_{i-1}\leq\frac{\varepsilon}{h(c)}$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $S(f,\pi,z_{i})\leq\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+El Teorema Fundamental del Cálculo
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+, llamamos 
+\series bold
+integral indefinida
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ a la función 
+\begin_inset Formula $F:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $F(x):=\int_{a}^{x}f$
+\end_inset
+
+.
+ El 
+\series bold
+TEOREMA FUNDAMENTAL DEL CÁLCULO
+\series default
+ afirma que, si 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+ y 
+\begin_inset Formula $F$
+\end_inset
+
+ es su integral indefinida, entonces 
+\begin_inset Formula $F$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ y si 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $F$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $c$
+\end_inset
+
+ y 
+\begin_inset Formula $F'(c)=f(c)$
+\end_inset
+
+, y esto también ocurre con los extremos del intervalo y las correspondientes
+ derivadas laterales.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $M:=\sup\{|f(x)|\}_{x\in[a,b]}$
+\end_inset
+
+, por las propiedades de la integral, 
+\begin_inset Formula $|F(x)-F(y)|=\left|\int_{x}^{y}f\right|\leq M|x-y|$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $F$
+\end_inset
+
+ es uniformemente continua en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+, pues dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ y 
+\begin_inset Formula $\delta=\frac{\varepsilon}{M}$
+\end_inset
+
+, si 
+\begin_inset Formula $|x-y|\leq\delta$
+\end_inset
+
+ entonces 
+\begin_inset Formula $|F(x)-F(y)|\leq\varepsilon$
+\end_inset
+
+.
+ Supongamos ahora que 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $h>0$
+\end_inset
+
+ con 
+\begin_inset Formula $c+h\in[a,b]$
+\end_inset
+
+, 
+\begin_inset Formula 
+\begin{multline*}
+\left|\frac{F(c+h)-F(c)}{h}-f(c)\right|=\left|\frac{\int_{a}^{c+h}f-\int_{a}^{c}f}{h}-\frac{1}{h}\int_{c}^{c+h}f(c)\right|=\left|\frac{1}{h}\int_{c}^{c+h}(f-f(c))\right|\leq\\
+\leq\frac{1}{h}\sup\{|f(t)-f(c)|\}_{t\in[c,c+h]}|h|=\sup\{|f(t)-f(c)|\}_{t\in[c,c+h]}
+\end{multline*}
+
+\end_inset
+
+y como 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $c$
+\end_inset
+
+, el último miembro de la desigualdad tiende a 0 cuando 
+\begin_inset Formula $h$
+\end_inset
+
+ tiende a 0, y lo mismo ocurre para 
+\begin_inset Formula $h<0$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $F'(c)=\lim_{h\rightarrow0}\frac{F(c+h)-F(c)}{h}=f(c)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dada 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+, decimos que 
+\begin_inset Formula $g:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ es una 
+\series bold
+primitiva
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ si 
+\begin_inset Formula $g$
+\end_inset
+
+ es derivable en 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ y para todo 
+\begin_inset Formula $x\in(a,b)$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $g'(x)=f(x)$
+\end_inset
+
+.
+ Por el teorema fundamental del cálculo, toda 
+\begin_inset Formula $f$
+\end_inset
+
+ continua en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ tiene primitivas en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+, donde la integral indefinida es una de ellas y el resto se obtienen sumando
+ a esta una constante.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $F$
+\end_inset
+
+ es la integral indefinida de 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ es otra primitiva de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+, entonces 
+\begin_inset Formula $(F-g)'(x)=F'(x)-g'(x)=f(x)-f(x)=0$
+\end_inset
+
+ para 
+\begin_inset Formula $x\in(a,b)$
+\end_inset
+
+, y por el teorema del valor medio, 
+\begin_inset Formula $F-g$
+\end_inset
+
+ es constante.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, la 
+\series bold
+fórmula de Barrow
+\series default
+ afirma que si 
+\begin_inset Formula $f\in{\cal R}[a,b]$
+\end_inset
+
+ admite una primitiva 
+\begin_inset Formula $g$
+\end_inset
+
+ en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\int_{a}^{b}f=g(b)-g(a)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración: 
+\series default
+Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ existe 
+\begin_inset Formula $\pi\equiv(t_{0}<\dots<t_{n})\in{\cal P}([a,b])$
+\end_inset
+
+ tal que para cualesquiera 
+\begin_inset Formula $z_{i}\in[t_{i-1},t_{i}]$
+\end_inset
+
+, 
+\begin_inset Formula $\left|\int_{a}^{b}f-S(f,\pi,z_{i})\right|<\varepsilon$
+\end_inset
+
+.
+ Por el teorema del valor medio aplicado a 
+\begin_inset Formula $g$
+\end_inset
+
+ en 
+\begin_inset Formula $[t_{i-1},t_{i}]$
+\end_inset
+
+, existe 
+\begin_inset Formula $z_{i}\in[t_{i-1},t_{i}]$
+\end_inset
+
+ con 
+\begin_inset Formula $g(t_{i})-g(t_{i-1})=g'(z_{i})(t_{i}-t_{i-1})=f(z_{i})(t_{i}-t_{i-1})$
+\end_inset
+
+, luego 
+\begin_inset Formula $g(b)-g(a)=\sum_{i=1}^{n}(g(t_{i})-g(t_{i-1}))=\sum_{i=1}^{n}g'(z_{i})(t_{i}-t_{i-1})=S(f,\pi,z_{i})$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $\left|\int_{a}^{b}f-(g(b)-g(a))\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Cálculo de primitivas
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int u^{n}u'\,dx=\frac{u^{n+1}}{n+1}+C\forall n\neq-1$
+\end_inset
+
+; 
+\begin_inset Formula $\int\frac{u'}{u}dx=\ln|u|+C\forall u\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int e^{u}u'\,dx=e^{u}+C$
+\end_inset
+
+; 
+\begin_inset Formula $\int a^{u}u'\,dx=\frac{a^{u}}{\ln a}+C\forall a>0,a\neq1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\cos u\,u'\,dx=\sin u+C$
+\end_inset
+
+; 
+\begin_inset Formula $\int\sin u\,u'\,dx=-\cos u+C$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\cosh u\,u'\,dx=\sinh u+C$
+\end_inset
+
+; 
+\begin_inset Formula $\int\sinh u\,u'\,dx=\cosh u+C$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\frac{u'}{\sin^{2}u}dx=\int\frac{u'}{\sinh^{2}u}dx=-\cot u+C$
+\end_inset
+
+; 
+\begin_inset Formula $\int\frac{u'}{\cos^{2}u}dx=\int\frac{u'}{\cosh^{2}u}dx=\tan u+C$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\frac{u'}{1+u^{2}}dx=\arctan u+C$
+\end_inset
+
+; 
+\begin_inset Formula $\int\frac{u'}{1-u^{2}}dx=\arg\tanh u+C$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\frac{u'}{\sqrt{1-u^{2}}}dx=\arcsin u+C=-\arccos u+C'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\frac{u'}{\sqrt{u^{2}+1}}dx=\arg\sinh u+C$
+\end_inset
+
+; 
+\begin_inset Formula $\int\frac{u'}{\sqrt{u^{2}-1}}dx=\arg\cosh u+C$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{eqnarray*}
+\cosh(x)=\frac{e^{x}+e^{-x}}{2} & \sinh(x)=\frac{e^{x}-e^{-x}}{2} & \cosh^{2}(x)-\sinh^{2}(x)=1\\
+\arg\cosh(x)=\ln(x+\sqrt{x^{2}-1}) & \arg\sinh(x)=\ln(x+\sqrt{x^{2}+1}) & \arg\tanh(x)=\frac{1}{2}\ln\frac{1+x}{1-x}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Integración por partes
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f,g\in{\cal R}[a,b]$
+\end_inset
+
+ con primitivas respectivas 
+\begin_inset Formula $F$
+\end_inset
+
+ y 
+\begin_inset Formula $G$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+\int_{a}^{b}Fg=F(b)G(b)-F(a)G(a)-\int_{a}^{b}fG
+\]
+
+\end_inset
+
+lo que suele escribirse como 
+\begin_inset Formula $\int u\,dv=uv-\int v\,du$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $(FG)'(x)=F'(x)G(x)+F(x)G'(x)=f(x)G(x)+F(x)g(x)$
+\end_inset
+
+, y por la fórmula de Barrow, 
+\begin_inset Formula $\int_{a}^{b}Fg+\int_{a}^{b}fG=\int_{a}^{b}(Fg+fG)=F(b)G(b)-F(a)G(a)$
+\end_inset
+
+, luego 
+\begin_inset Formula $\int_{a}^{b}Fg=F(b)G(b)-F(a)G(a)-\int_{a}^{b}fG$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Cambio de variable
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, sea 
+\begin_inset Formula $\varphi:[c,d]\rightarrow[a,b]\in{\cal C}^{1}[c,d]$
+\end_inset
+
+ con 
+\begin_inset Formula $\varphi(c)=a$
+\end_inset
+
+ y 
+\begin_inset Formula $\varphi(d)=b$
+\end_inset
+
+, sea 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continua, entonces
+\begin_inset Formula 
+\[
+\int_{a}^{b}f=\int_{c}^{d}(f\circ\varphi)\varphi'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $F$
+\end_inset
+
+ es una primitiva de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ entonces 
+\begin_inset Formula $F\circ\varphi$
+\end_inset
+
+ lo es de 
+\begin_inset Formula $(f\circ\varphi)\varphi'$
+\end_inset
+
+ en 
+\begin_inset Formula $[c,d]$
+\end_inset
+
+, luego 
+\begin_inset Formula $\int_{a}^{b}f=F(b)-F(a)=F(\varphi(d))-F(\varphi(c))=(F\circ\varphi)(d)-(F\circ\varphi)(c)=\int_{c}^{d}(f\circ\varphi)\varphi'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Esto da sentido a la notación de 
+\begin_inset Formula $\int_{a}^{b}f(x)dx:=\int_{a}^{b}f$
+\end_inset
+
+, porque entonces si 
+\begin_inset Formula $x=\varphi(t)$
+\end_inset
+
+ es fácil recordar 
+\begin_inset Formula $dx=\varphi'(t)dt$
+\end_inset
+
+ y entonces
+\begin_inset Formula 
+\[
+\int_{a}^{b}f(x)dx=\int_{c}^{d}f(\varphi(t))\varphi'(t)dt
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Funciones racionales
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $P(x)$
+\end_inset
+
+ y 
+\begin_inset Formula $Q(x)$
+\end_inset
+
+ polinomios y queremos resolver 
+\begin_inset Formula $\int_{a}^{b}\frac{P(x)}{Q(x)}dx$
+\end_inset
+
+.
+ Si el grado de 
+\begin_inset Formula $P(x)$
+\end_inset
+
+ es mayor o igual que el de 
+\begin_inset Formula $Q(x)$
+\end_inset
+
+ hacemos 
+\begin_inset Formula $\int_{a}^{b}\frac{P(x)}{Q(x)}dx=\int C(x)dx+\int\frac{R(x)}{Q(x)}dx$
+\end_inset
+
+ para que el grado del numerador sea menor que el del denominador.
+ Entonces descomponemos en fracciones simples.
+\end_layout
+
+\begin_layout Standard
+Descomponemos 
+\begin_inset Formula $Q(x)$
+\end_inset
+
+ como 
+\begin_inset Formula $Q(x)=\prod_{i=1}^{r}(x-a_{i})^{m_{i}}\prod_{i=1}^{s}(x^{2}+p_{i}x+q_{i})^{n_{i}}$
+\end_inset
+
+, donde 
+\begin_inset Formula $q_{i}>\frac{p_{i}^{2}}{4}$
+\end_inset
+
+ para que los factores sean irreducibles.
+ Entonces (si el grado de 
+\begin_inset Formula $P(x)$
+\end_inset
+
+ es menor que el de 
+\begin_inset Formula $Q(x)$
+\end_inset
+
+) podemos expresar la fracción como
+\begin_inset Formula 
+\[
+\frac{P(x)}{Q(x)}=\sum_{i=1}^{r}\sum_{j=1}^{m_{i}}\frac{A_{ij}}{(x-a_{i})^{j}}+\sum_{i=1}^{M}\sum_{j=1}^{n_{i}}\frac{M_{ij}x+N_{ij}}{(x^{2}+p_{i}x+q_{i})^{j}}
+\]
+
+\end_inset
+
+Resolvemos los 
+\begin_inset Formula $A_{k,i}$
+\end_inset
+
+, 
+\begin_inset Formula $M_{k,i}$
+\end_inset
+
+, 
+\begin_inset Formula $N_{k,i}$
+\end_inset
+
+ y nos queda hallar la integral de cada sumando como sigue:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\frac{A}{x-a}dx=A\ln|x-a|+C$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\frac{A}{(x-a)^{n}}dx=-\frac{A}{(n-1)(x-a)^{n-1}}+C$
+\end_inset
+
+, donde 
+\begin_inset Formula $n\in2,3,\dots$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\frac{Mx+N}{x^{2}+px+q}dx=\frac{M}{2}\ln\left(\left(x+\frac{p}{2}\right)^{2}+c^{2}\right)+\frac{N-\frac{Mp}{2}}{c}\arctan\left(\frac{x+\frac{p}{2}}{c}\right)+C$
+\end_inset
+
+, donde 
+\begin_inset Formula $c=\frac{\sqrt{4q-p^{2}}}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Funciones que contienen 
+\begin_inset Formula $\cos x$
+\end_inset
+
+ y 
+\begin_inset Formula $\sin x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+En general, haremos 
+\begin_inset Formula $t=\tan\frac{x}{2}$
+\end_inset
+
+ y entonces
+\begin_inset Formula 
+\begin{eqnarray*}
+\cos x=\frac{\cos(2\frac{x}{2})}{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}}=\frac{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}} & \overset{\text{div. }\cos^{2}\frac{x}{2}}{=} & \frac{1-\tan^{2}\frac{x}{2}}{\tan^{2}\frac{x}{2}+1}=\frac{1-t^{2}}{1+t^{2}}\\
+\sin x=\frac{\sin(2\frac{x}{2})}{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}} & \overset{\text{div. }\cos^{2}\frac{x}{2}}{=} & \frac{2\tan\frac{x}{2}}{\tan^{2}\frac{x}{2}+1}=\frac{2t}{1+t^{2}}\\
+x=2\arctan t & \text{ y } & dx=\frac{2}{1+t^{2}}dt
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si la función es de la forma 
+\begin_inset Formula $f(x)=g(\sin x)\cos x$
+\end_inset
+
+, siendo 
+\begin_inset Formula $g$
+\end_inset
+
+ una función racional, hacemos 
+\begin_inset Formula $t=\sin x$
+\end_inset
+
+, y si es 
+\begin_inset Formula $f(x)=g(\cos x)\sin x$
+\end_inset
+
+ hacemos 
+\begin_inset Formula $t=\cos x$
+\end_inset
+
+.
+ Si es 
+\begin_inset Formula $f(x)=g(\tan x)$
+\end_inset
+
+ hacemos 
+\begin_inset Formula $\tan x=t$
+\end_inset
+
+, y podemos llegar a esta situación cuando al sustituir 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ por 
+\begin_inset Formula $\cos x\tan x$
+\end_inset
+
+ quedan solo potencias pares de 
+\begin_inset Formula $\cos x$
+\end_inset
+
+, y hacemos 
+\begin_inset Formula $\cos^{2}x=\frac{1}{1+\tan^{2}x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+En el caso 
+\begin_inset Formula $f(x)=\cos^{n}x\sin^{m}x$
+\end_inset
+
+, si 
+\begin_inset Formula $n$
+\end_inset
+
+ es impar hacemos 
+\begin_inset Formula $t=\sin x$
+\end_inset
+
+, si 
+\begin_inset Formula $m$
+\end_inset
+
+ es impar, 
+\begin_inset Formula $t=\cos x$
+\end_inset
+
+, y si ambos son pares, usamos 
+\begin_inset Formula $\cos^{2}x=\frac{1+\cos(2x)}{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $\sin^{2}x=\frac{1-\cos(2x)}{2}$
+\end_inset
+
+ para 
+\begin_inset Quotes cld
+\end_inset
+
+reducir el grado
+\begin_inset Quotes crd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Funciones de la forma 
+\begin_inset Formula $f(e^{x})$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Hacemos el cambio 
+\begin_inset Formula $t=e^{x}$
+\end_inset
+
+ y 
+\begin_inset Formula $dt=e^{x}dx$
+\end_inset
+
+, y esto también sirve para el coseno y seno hiperbólicos (
+\begin_inset Formula $\cosh$
+\end_inset
+
+ y 
+\begin_inset Formula $\sinh$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Subsection
+Funciones que contienen 
+\begin_inset Formula $\sqrt{ax^{2}+2bx+c}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\begin_inset Formula $d:=\frac{ac-b^{2}}{a}$
+\end_inset
+
+ y se tiene 
+\begin_inset Formula $ax^{2}+2bx+c=a\left(x+\frac{b}{a}\right)^{2}+d$
+\end_inset
+
+.
+ Hacemos entonces el cambio de variable 
+\begin_inset Formula $t=x+\frac{b}{a}$
+\end_inset
+
+ y a continuación:
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $a>0$
+\end_inset
+
+ y 
+\begin_inset Formula $d>0$
+\end_inset
+
+ hacemos 
+\begin_inset Formula $at^{2}=d\tan^{2}u$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $\sqrt{at^{2}+d}=\sqrt{d\tan^{2}u+d}=\sqrt{d}\sqrt{1+\tan^{2}u}=\sqrt{d}\sqrt{\sec^{2}u}=\sqrt{d}\sec u$
+\end_inset
+
+ y 
+\begin_inset Formula $dt=\sqrt{\frac{d}{a}}\sec^{2}u\,du$
+\end_inset
+
+.
+ También podemos hacer 
+\begin_inset Formula $at^{2}=d\sinh^{2}u$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $\sqrt{at^{2}+d}=\sqrt{d\sinh^{2}u+d}=\sqrt{d}\sqrt{\sinh^{2}u+1}=\sqrt{d}\cosh u$
+\end_inset
+
+ y 
+\begin_inset Formula $dt=\sqrt{\frac{d}{a}}\cosh u\,du$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{sloppypar}
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $a>0$
+\end_inset
+
+ y 
+\begin_inset Formula $d<0$
+\end_inset
+
+ hacemos 
+\begin_inset Formula $at^{2}=-d\sec^{2}u$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $\sqrt{-d\sec^{2}u+d}=\sqrt{-d}\sqrt{\sec^{2}u+1}=\sqrt{-d}\tan u$
+\end_inset
+
+ y 
+\begin_inset Formula $dt=\sqrt{-\frac{d}{a}}\sec u\tan u\,du$
+\end_inset
+
+.
+ También podemos hacer 
+\begin_inset Formula $at^{2}=-d\cosh^{2}u$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $\sqrt{at^{2}+d}=\sqrt{-d\cosh^{2}u+d}=\sqrt{-d}\sqrt{\cosh^{2}u-1}=\sqrt{-d}\sinh u$
+\end_inset
+
+ y 
+\begin_inset Formula $dt=\sqrt{-\frac{d}{a}}\sinh u\,du$
+\end_inset
+
+.
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{sloppypar}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $a<0$
+\end_inset
+
+ y 
+\begin_inset Formula $d>0$
+\end_inset
+
+ hacemos 
+\begin_inset Formula $at^{2}=-d\sin^{2}u$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $\sqrt{at^{2}+d}=\sqrt{-d\sin^{2}u+d}=\sqrt{d}\sqrt{1-\sin^{2}u}=\sqrt{d}\cos u$
+\end_inset
+
+ y 
+\begin_inset Formula $dt=\sqrt{-\frac{d}{a}}\cos u\,du$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Aplicaciones
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f,g:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ continuas, si 
+\begin_inset Formula $f(a)=g(a)$
+\end_inset
+
+, 
+\begin_inset Formula $f(b)=g(b)$
+\end_inset
+
+ y 
+\begin_inset Formula $f(x)\geq g(x)$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+, se define el 
+\series bold
+área encerrada
+\series default
+ por las gráficas de 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ como 
+\begin_inset Formula $\int_{a}^{b}(f(x)-g(x))\,dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}\in{\cal C}^{1}[a,b]$
+\end_inset
+
+, la 
+\series bold
+longitud de la curva
+\series default
+ 
+\begin_inset Formula $C=\{(x,f(x))\}_{x\in[a,b]}$
+\end_inset
+
+ viene dada por 
+\begin_inset Formula $L=\int_{a}^{b}\sqrt{1+f'(x)^{2}}\,dx$
+\end_inset
+
+.
+ 
+\series bold
+Interpretación:
+\series default
+ Sea 
+\begin_inset Formula $\pi\equiv(a=x_{0}<\dots<x_{n}=b)\in{\cal P}([a,b])$
+\end_inset
+
+, sea 
+\begin_inset Formula $P_{i}=(x_{i},f(x_{i}))$
+\end_inset
+
+, una aproximación a la curva es
+\begin_inset Formula 
+\begin{multline*}
+\sum_{i=1}^{n}d(P_{i-1},P_{i})=\sum_{i=1}^{n}\sqrt{(f(x_{i})-f(x_{i-1}))^{2}+(x_{i}-x_{i-1})^{2}}=\\
+=\sum_{i=1}^{n}\sqrt{\left(\frac{f(x_{i})-f(x_{i-1})}{x_{i}-x_{i-1}}\right)^{2}+1}(x_{i}-x_{i-1})=\sum_{i=1}^{n}\sqrt{1+f'(\xi_{i})^{2}}(x_{i}-x_{i-1})
+\end{multline*}
+
+\end_inset
+
+ con 
+\begin_inset Formula $\xi_{i}\in(x_{i-1},x_{i})$
+\end_inset
+
+, que converge a 
+\begin_inset Formula $\int_{a}^{b}\sqrt{1+f'(x)^{2}}dx$
+\end_inset
+
+ cuando 
+\begin_inset Formula $\Vert\pi\Vert$
+\end_inset
+
+ tiende a 0.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+sólido de revolución
+\series default
+ al cuerpo obtenido al girar una función 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ alrededor del eje horizontal.
+ Su 
+\series bold
+volumen
+\series default
+ viene dado por 
+\begin_inset Formula $V=\pi\int_{a}^{b}f(x)^{2}\,dx$
+\end_inset
+
+, y su 
+\series bold
+área
+\series default
+ (lateral) por 
+\begin_inset Formula $A=2\pi\int_{a}^{b}f(x)\sqrt{1+f'(x)^{2}}\,dx$
+\end_inset
+
+.
+ 
+\series bold
+Interpretación:
+\series default
+ Sea 
+\begin_inset Formula $f$
+\end_inset
+
+ continua y positiva.
+ Para hallar el volumen tomamos 
+\begin_inset Formula $\pi\equiv(x_{0}<\dots<x_{n})\in{\cal P}([a,b])$
+\end_inset
+
+ y aproximamos el volumen por secciones cilíndricas con radio 
+\begin_inset Formula $f(x_{i})$
+\end_inset
+
+ y altura 
+\begin_inset Formula $x_{i}-x_{i-1}$
+\end_inset
+
+, con lo que su radio viene dado por 
+\begin_inset Formula $\pi f(x_{i})^{2}(x_{i}-x_{i-1})$
+\end_inset
+
+.
+ Sumando obtenemos 
+\begin_inset Formula $\sum_{i=1}^{n}\pi f(x_{i})^{2}(x_{i}-x_{i-1})$
+\end_inset
+
+, que converge a 
+\begin_inset Formula $\pi\int_{a}^{b}f(x)^{2}\,dx$
+\end_inset
+
+.
+ El área se obtiene con un razonamiento similar al usado para la longitud
+ de la curva.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+volumen
+\series default
+ del sólido resultante de girar alrededor del eje vertical la superficie
+ encerrada por las rectas 
+\begin_inset Formula $x=a$
+\end_inset
+
+, 
+\begin_inset Formula $x=b$
+\end_inset
+
+ e 
+\begin_inset Formula $y=f(x)$
+\end_inset
+
+ es 
+\begin_inset Formula $2\pi\int_{a}^{b}xf(x)\,dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Integrales impropias
+\end_layout
+
+\begin_layout Standard
+Una función 
+\begin_inset Formula $f:[a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ (
+\begin_inset Formula $b\leq+\infty$
+\end_inset
+
+) es 
+\series bold
+localmente integrable
+\series default
+ si 
+\begin_inset Formula $\forall u\in[a,b),f|_{[a,u]}\in{\cal R}[a,b]$
+\end_inset
+
+.
+ Si además existe 
+\begin_inset Formula $\lim_{u\rightarrow b^{-}}\int_{a}^{u}f(x)\,dx$
+\end_inset
+
+ diremos que la 
+\series bold
+integral impropia
+\series default
+ 
+\begin_inset Formula $\int_{a}^{b}f(x)\,dx$
+\end_inset
+
+ es convergente y su valor es este límite.
+ Análogamente, 
+\begin_inset Formula $f:(a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ (
+\begin_inset Formula $a\geq-\infty$
+\end_inset
+
+) es localmente integrable si 
+\begin_inset Formula $\forall u\in(a,b],f|_{[u,b]}\in{\cal R}[a,b]$
+\end_inset
+
+, y si además existe 
+\begin_inset Formula $\lim_{u\rightarrow a^{+}}\int_{u}^{b}f(x)\,dx$
+\end_inset
+
+ diremos que la integral impropia 
+\begin_inset Formula $\int_{a}^{b}f(x)\,dx$
+\end_inset
+
+ es convergente y su valor es este límite.
+ En ambos casos, si el límite es 
+\begin_inset Formula $+\infty$
+\end_inset
+
+ o 
+\begin_inset Formula $-\infty$
+\end_inset
+
+, diremos que la integral 
+\series bold
+diverge
+\series default
+, y si no existe el límite diremos que no existe la integral impropia.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, sea 
+\begin_inset Formula $f$
+\end_inset
+
+ localmente integrable en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+, 
+\begin_inset Formula $f$
+\end_inset
+
+ es integrable en sentido impropio en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ si y sólo si lo es en 
+\begin_inset Formula $[c,b)$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $\int_{a}^{b}f(x)\,dx=\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $a<c<t<b$
+\end_inset
+
+, 
+\begin_inset Formula $\int_{a}^{t}f(x)\,dx=\int_{a}^{c}f(x)\,dx+\int_{c}^{t}f(x)\,dx$
+\end_inset
+
+, por lo que existe 
+\begin_inset Formula $\lim_{t\rightarrow b^{-}}\int_{a}^{t}f(x)\,dx$
+\end_inset
+
+ si y sólo si existe 
+\begin_inset Formula $\lim_{t\rightarrow b^{-}}\int_{c}^{t}f(x)\,dx$
+\end_inset
+
+, lo que demuestra el teorema.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:(a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+ (
+\begin_inset Formula $a\geq-\infty,b\leq+\infty$
+\end_inset
+
+) es integrable Riemann en cada subintervalo cerrado de 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+, diremos que la integral impropia 
+\begin_inset Formula $\int_{a}^{b}f(x)\,dx$
+\end_inset
+
+ es convergente si para un 
+\begin_inset Formula $c\in(a,b)$
+\end_inset
+
+ son convergentes 
+\begin_inset Formula $\int_{a}^{c}f(x)\,dx$
+\end_inset
+
+ y 
+\begin_inset Formula $\int_{c}^{b}f(x)\,dx$
+\end_inset
+
+, y definimos
+\begin_inset Formula 
+\[
+\int_{a}^{b}f(x)\,dx:=\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El valor de esta integral no depende de 
+\begin_inset Formula $c$
+\end_inset
+
+.
+ La 
+\series bold
+condición de Cauchy
+\series default
+ afirma que, dada 
+\begin_inset Formula $f:[a,b)\rightarrow\mathbb{R}$
+\end_inset
+
+, existe 
+\begin_inset Formula $\lim_{x\rightarrow b^{-}}f(x)$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall\varepsilon>0,\exists b_{0}\in(a,b):\forall x_{1},x_{2}\in(b_{0},b):x_{1}<x_{2},|f(x_{1})-f(x_{2})|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+ y consecuencia de lo anterior, si 
+\begin_inset Formula $f$
+\end_inset
+
+ es localmente integrable en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+, la integral impropia 
+\begin_inset Formula $\int_{a}^{b}f(x)\,dx$
+\end_inset
+
+ es convergente si y sólo si 
+\begin_inset Formula $\forall\varepsilon>0,\exists b_{0}\in(a,b):\forall x_{1},x_{2}\in(b_{0},b):x_{1}<x_{2},\left|\int_{x_{1}}^{x_{2}}f(t)\,dt\right|<\varepsilon$
+\end_inset
+
+.
+ Más 
+\series bold
+teoremas
+\series default
+:
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ son integrables en sentido impropio en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+, dados 
+\begin_inset Formula $\lambda,\mu\in\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $\lambda f+\mu g$
+\end_inset
+
+ es integrable en sentido impropio con
+\begin_inset Formula 
+\[
+\int_{a}^{b}(\lambda f+\mu g)(t)\,dt=\lambda\int_{a}^{b}f(t)\,dt+\mu\int_{a}^{b}g(t)\,dt
+\]
+
+\end_inset
+
+Basta tomar límites cuando 
+\begin_inset Formula $x$
+\end_inset
+
+ tiende a 
+\begin_inset Formula $b$
+\end_inset
+
+ por la izquierda en la linealidad de integrales propias.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ son continuas en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ es derivable con derivada continua, sea 
+\begin_inset Formula $F$
+\end_inset
+
+ una primitiva de 
+\begin_inset Formula $f$
+\end_inset
+
+, la siguiente igualdad se cumple si existen dos de los tres límites e integrale
+s impropias en ella:
+\begin_inset Formula 
+\[
+\int_{a}^{b}f(t)g(t)\,dt=\lim_{x\rightarrow b^{-}}F(x)g(x)-F(a)g(a)-\int_{a}^{b}F(t)g'(t)\,dt
+\]
+
+\end_inset
+
+Basta tomar límites en la identidad dada por la regla de integración por
+ partes.
+\end_layout
+
+\begin_layout Subsection
+Integrales no negativas
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $f$
+\end_inset
+
+ es localmente integrable en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ y no negativa, 
+\begin_inset Formula $\int_{a}^{b}f(t)\,dt$
+\end_inset
+
+ converge si y sólo si 
+\begin_inset Formula $F(x)=\int_{a}^{x}f(t)\,dt$
+\end_inset
+
+ está acotada.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Como 
+\begin_inset Formula $f$
+\end_inset
+
+ es no negativa, 
+\begin_inset Formula $F$
+\end_inset
+
+ es creciente, y si no estuviese acotada sería 
+\begin_inset Formula $\lim_{x\rightarrow b^{-}}F(x)=+\infty$
+\end_inset
+
+ y la integral impropia divergería.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $F$
+\end_inset
+
+ está acotada existe 
+\begin_inset Formula $\lim_{x\rightarrow b^{-}}F(x)=\sup\{F(x)\}_{x\in[a,b)}$
+\end_inset
+
+, luego la integral impropia converge.
+\end_layout
+
+\begin_layout Standard
+Otro 
+\series bold
+teorema
+\series default
+ es que si 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ son localmente integrables en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ y no negativas y existe 
+\begin_inset Formula $K\in\mathbb{R}$
+\end_inset
+
+ y 
+\begin_inset Formula $V$
+\end_inset
+
+ entorno de 
+\begin_inset Formula $b$
+\end_inset
+
+ tal que 
+\begin_inset Formula $x\in V\implies f(x)\leq Kg(x)$
+\end_inset
+
+, entonces si 
+\begin_inset Formula $\int_{a}^{b}g(t)\,dt$
+\end_inset
+
+ converge, también lo hace 
+\begin_inset Formula $\int_{a}^{b}f(t)\,dt$
+\end_inset
+
+, por lo que si 
+\begin_inset Formula $\int_{a}^{b}f(t)\,dt$
+\end_inset
+
+ diverge también lo hace 
+\begin_inset Formula $\int_{a}^{b}g(t)\,dt$
+\end_inset
+
+ (y divergir también).
+ 
+\series bold
+Demostración:
+\series default
+ La convergencia depende sólo del comportamiento de las funciones en un
+ entorno, y en este 
+\begin_inset Formula $\int_{a}^{x}f(t)\,dt\leq K\int_{a}^{x}g(t)\,dt$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De aquí que si 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ son localmente integrables en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ y no negativas con 
+\begin_inset Formula $A:=\lim_{x\rightarrow b^{-}}\frac{f(t)}{g(t)}$
+\end_inset
+
+, entonces:
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $A\neq0,\infty$
+\end_inset
+
+, ambas integrales tienen el mismo carácter.
+\begin_inset Newline newline
+\end_inset
+
+Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ con 
+\begin_inset Formula $\varepsilon<A$
+\end_inset
+
+, existe 
+\begin_inset Formula $a_{\varepsilon}$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $a_{\varepsilon}\leq x\leq b$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $\left|\frac{f(x)}{g(x)}-A\right|\leq\varepsilon$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $A-\varepsilon\leq\frac{f(x)}{g(x)}\leq A+\varepsilon$
+\end_inset
+
+, luego para 
+\begin_inset Formula $x\in[a_{\varepsilon},b)$
+\end_inset
+
+ tenemos 
+\begin_inset Formula $(A-\varepsilon)g(x)\leq f(x)\leq(A+\varepsilon)g(x)$
+\end_inset
+
+, y no hay más que aplicar el teorema anterior.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $A=0$
+\end_inset
+
+, la convergencia de 
+\begin_inset Formula $\int_{a}^{b}g(t)\,dt$
+\end_inset
+
+ implica la de 
+\begin_inset Formula $\int_{a}^{b}f(t)\,dt$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Como antes, obtenemos 
+\begin_inset Formula $f(x)\leq\varepsilon g(x)$
+\end_inset
+
+ y aplicamos el teorema anterior.
+\end_layout
+
+\begin_layout Itemize
+Si 
+\begin_inset Formula $A=\infty$
+\end_inset
+
+, la convergencia de 
+\begin_inset Formula $\int_{a}^{b}f(t)\,dt$
+\end_inset
+
+ implica la de 
+\begin_inset Formula $\int_{a}^{b}g(t)\,dt$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Otro 
+\series bold
+teorema
+\series default
+ es que si 
+\begin_inset Formula $f$
+\end_inset
+
+ es no negativa y localmente integrable en 
+\begin_inset Formula $(0,1]$
+\end_inset
+
+ y existe 
+\begin_inset Formula $\alpha<1$
+\end_inset
+
+ con 
+\begin_inset Formula $\lim_{t\rightarrow0^{+}}f(t)t^{\alpha}$
+\end_inset
+
+ finito, 
+\begin_inset Formula $\int_{0}^{1}f(t)\,dt$
+\end_inset
+
+ es convergente, mientras que si existe 
+\begin_inset Formula $\alpha\geq1$
+\end_inset
+
+ con 
+\begin_inset Formula $\lim_{t\rightarrow0^{+}}f(t)t^{\alpha}$
+\end_inset
+
+ no nulo, la integral diverge.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $\lim_{t\rightarrow0^{+}}f(t)t^{\alpha}=\lim_{t\rightarrow0^{+}}\frac{f(t)}{\left(\frac{1}{t^{\alpha}}\right)}$
+\end_inset
+
+, y si 
+\begin_inset Formula $\alpha<1$
+\end_inset
+
+, la integral 
+\begin_inset Formula $\int_{0}^{1}\frac{dt}{t^{\alpha}}$
+\end_inset
+
+ es convergente y, por lo anterior, 
+\begin_inset Formula $\int_{0}^{1}f(t)\,dt$
+\end_inset
+
+ también.
+ De que 
+\begin_inset Formula $\int_{0}^{1}\frac{dt}{t^{\alpha}}$
+\end_inset
+
+ diverge si 
+\begin_inset Formula $t\geq1$
+\end_inset
+
+ se desprende la última afirmación.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $f$
+\end_inset
+
+ es no negativa y localmente integrable en 
+\begin_inset Formula $[a,+\infty)$
+\end_inset
+
+, si existe 
+\begin_inset Formula $\alpha>1$
+\end_inset
+
+ con 
+\begin_inset Formula $\lim_{t\rightarrow\infty}f(t)t^{\alpha}$
+\end_inset
+
+ finito, 
+\begin_inset Formula $\int_{a}^{\infty}f(t)\,dt$
+\end_inset
+
+ converge, mientras que si existe 
+\begin_inset Formula $\alpha\leq1$
+\end_inset
+
+ con 
+\begin_inset Formula $\lim_{t\rightarrow\infty}f(t)t^{\alpha}$
+\end_inset
+
+ no nulo, la integral diverge.
+\end_layout
+
+\begin_layout Subsection
+Convergencia absoluta
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f$
+\end_inset
+
+ localmente integrable en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+, decimos que la integral impropia de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ es 
+\series bold
+absolutamente convergente
+\series default
+ si 
+\begin_inset Formula $\int_{a}^{b}|f(t)|\,dt$
+\end_inset
+
+ es convergente.
+ La convergencia absoluta implica la convergencia.
+ 
+\series bold
+Demostración:
+\series default
+ Por el criterio de convergencia de Cauchy aplicado a 
+\begin_inset Formula $|f(t)|$
+\end_inset
+
+, dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ existe 
+\begin_inset Formula $b_{0}\in(a,b)$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $b_{0}<x_{1}<x_{2}<b$
+\end_inset
+
+ entonces 
+\begin_inset Formula $\int_{x_{1}}^{x_{2}}|f(t)|\,dt<\varepsilon$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\left|\int_{x_{1}}^{x_{2}}f(t)\,dt\right|<\varepsilon$
+\end_inset
+
+, lo que implica que 
+\begin_inset Formula $\int_{a}^{b}f(t)\,dt$
+\end_inset
+
+ es convergente.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ son funciones continuas en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ tiene derivada continua, si 
+\begin_inset Formula $F(x):=\int_{a}^{x}f(t)\,dt$
+\end_inset
+
+ está acotada superiormente por 
+\begin_inset Formula $K$
+\end_inset
+
+, 
+\begin_inset Formula $\int_{a}^{x}|g'(t)|\,dt$
+\end_inset
+
+ está acotada superiormente por 
+\begin_inset Formula $k$
+\end_inset
+
+ y 
+\begin_inset Formula $\lim_{t\rightarrow b^{-}}g(t)=0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\int_{a}^{b}f(t)g(t)\,dt$
+\end_inset
+
+ es convergente.
+ 
+\series bold
+Demostración:
+\series default
+ Basta probar la existencia de 
+\begin_inset Formula $\lim_{x\rightarrow b^{-}}F(x)g(x)$
+\end_inset
+
+ y de 
+\begin_inset Formula $\lim_{x\rightarrow b^{-}}\int_{a}^{x}F(t)g'(t)\,dt$
+\end_inset
+
+.
+ Las condiciones 
+\begin_inset Formula $F(x)\leq K$
+\end_inset
+
+ y 
+\begin_inset Formula $\lim_{t\rightarrow b^{-}}g(t)=0$
+\end_inset
+
+ aseguran que el primer límite es 0, y las dos primeras (
+\begin_inset Formula $F(x)\leq K$
+\end_inset
+
+ y 
+\begin_inset Formula $\int_{a}^{x}|g'(t)|dt\leq k$
+\end_inset
+
+) implican que 
+\begin_inset Formula $\int_{a}^{x}F(t)g'(t)\,dt$
+\end_inset
+
+ es absolutamente convergente, pues 
+\begin_inset Formula 
+\[
+\int_{a}^{x}|F(t)||g'(t)|\,dt\leq Kk
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+criterio de Dirichlet
+\series default
+ afirma que si 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ son continuas en 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ tiene derivada continua, si existe 
+\begin_inset Formula $K\in\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $\left|\int_{a}^{x}f(t)\,dt\right|\leq K$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ es monótona decreciente con 
+\begin_inset Formula $\lim_{t\rightarrow b^{-}}g(t)=0$
+\end_inset
+
+, la integral impropia 
+\begin_inset Formula $\int_{a}^{b}f(t)g(t)\,dt$
+\end_inset
+
+ es convergente.
+ 
+\series bold
+Demostración:
+\series default
+ Como 
+\begin_inset Formula $g$
+\end_inset
+
+ es decreciente, 
+\begin_inset Formula $g'(t)\leq0$
+\end_inset
+
+, luego 
+\begin_inset Formula $\int_{a}^{x}|g'(t)|\,dt=-\int_{a}^{x}g'(t)\,dt=g(a)-g(x)\overset{g(x)\geq0}{\leq}g(a)$
+\end_inset
+
+, y se tienen entonces todas las condiciones del teorema anterior.
+\end_layout
+
+\end_body
+\end_document
diff --git a/fuvr2/n3.lyx b/fuvr2/n3.lyx
new file mode 100644
index 0000000..5d9c1ab
--- /dev/null
+++ b/fuvr2/n3.lyx
@@ -0,0 +1,631 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+En este capítulo, 
+\begin_inset Formula $K$
+\end_inset
+
+ representa indistintamente a 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ o 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+.
+ Una 
+\series bold
+serie de potencias
+\series default
+ en torno a 
+\begin_inset Formula $z_{0}\in K$
+\end_inset
+
+ es una expresión de la forma
+\begin_inset Formula 
+\[
+\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}
+\]
+
+\end_inset
+
+donde 
+\begin_inset Formula $(a_{n})_{n=0}^{\infty}$
+\end_inset
+
+ es una sucesión de elementos de 
+\begin_inset Formula $K$
+\end_inset
+
+ y 
+\begin_inset Formula $z\in K$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+radio de convergencia
+\series default
+ de la serie al valor
+\begin_inset Formula 
+\[
+R:=\frac{1}{\limsup_{n}\sqrt[n]{|a_{n}|}}
+\]
+
+\end_inset
+
+donde 
+\begin_inset Formula $\limsup_{n}a_{n}$
+\end_inset
+
+ es el supremo de las subsucesiones convergentes de 
+\begin_inset Formula $(a_{n})$
+\end_inset
+
+.
+ Se entiende que si 
+\begin_inset Formula $\limsup_{n}\sqrt[n]{|a_{n}|}=0$
+\end_inset
+
+ se toma 
+\begin_inset Formula $R=\infty$
+\end_inset
+
+, y si 
+\begin_inset Formula $\limsup_{n}\sqrt[n]{|a_{n}|}=\infty$
+\end_inset
+
+ se toma 
+\begin_inset Formula $R=0$
+\end_inset
+
+.
+ Por el criterio de la raíz, o el del cociente, la serie converge sólo en
+ la bola abierta 
+\begin_inset Formula $B(z_{0};R)$
+\end_inset
+
+, llamada 
+\series bold
+disco de convergencia
+\series default
+\SpecialChar endofsentence
+
+\end_layout
+
+\begin_layout Standard
+La serie de funciones 
+\begin_inset Formula $\sum_{n=0}^{\infty}f_{n}$
+\end_inset
+
+ 
+\series bold
+converge uniformemente
+\series default
+ en un conjunto 
+\begin_inset Formula $A$
+\end_inset
+
+ a una función 
+\begin_inset Formula $f$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall\varepsilon>0,\exists n_{0}\in\mathbb{N}:\forall z\in A,m\geq n_{0};\left|f(z)-\sum_{n=0}^{m}f_{n}(z)\right|<\varepsilon$
+\end_inset
+
+.
+ El 
+\series bold
+criterio de Cauchy de convergencia uniforme
+\series default
+ afirma que una serie de funciones es uniformemente convergente en 
+\begin_inset Formula $A$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall\varepsilon>0,\exists n_{0}\in\mathbb{N}:\forall z\in A,n_{0}<p\leq q;\left|\sum_{n=p}^{q}f_{n}(z)\right|<\varepsilon$
+\end_inset
+
+, y el 
+\series bold
+criterio de Weierstrass
+\series default
+ afirma que si existe una serie de términos positivos 
+\begin_inset Formula $\sum_{n}b_{n}$
+\end_inset
+
+ convergente con 
+\begin_inset Formula $|f_{n}(z)|\leq b_{n}\forall z\in A,n\in\mathbb{N}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\sum_{n=0}^{\infty}f_{n}$
+\end_inset
+
+ converge uniformemente en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+La serie de potencias 
+\begin_inset Formula $\sum_{n}a_{n}(z-z_{0})^{n}$
+\end_inset
+
+ con radio de convergencia 
+\begin_inset Formula $R$
+\end_inset
+
+ converge absoluta y uniformemente en la bola cerrada 
+\begin_inset Formula $B[z_{0};r]$
+\end_inset
+
+ para cada 
+\begin_inset Formula $r<R$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $\sum_{n}f_{n}$
+\end_inset
+
+ converge uniformemente en 
+\begin_inset Formula $A$
+\end_inset
+
+ y las 
+\begin_inset Formula $f_{n}$
+\end_inset
+
+ son continuas en 
+\begin_inset Formula $A$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+criterio de Abel
+\series default
+ afirma que, dada una serie de potencias 
+\begin_inset Formula $\sum_{n}a_{n}z^{n}$
+\end_inset
+
+, si para 
+\begin_inset Formula $z=c$
+\end_inset
+
+ la serie converge, también converge uniformemente en 
+\begin_inset Formula $[0,c]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, si 
+\begin_inset Formula $f(z):=\sum_{n=0}^{\infty}a_{n}z^{n}$
+\end_inset
+
+ para 
+\begin_inset Formula $z\in B(0;R)$
+\end_inset
+
+, siendo 
+\begin_inset Formula $R$
+\end_inset
+
+ el radio de convergencia de la serie, entonces la serie 
+\begin_inset Formula $\sum_{n=1}^{\infty}na_{n}z^{n-1}$
+\end_inset
+
+, obtenida derivando formalmente la anterior, tiene radio de convergencia
+ 
+\begin_inset Formula $R$
+\end_inset
+
+, y de hecho esta serie converge a la derivada de 
+\begin_inset Formula $f$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es infinitamente derivable en el disco de convergencia y 
+\begin_inset Formula $a_{n}=\frac{f^{(n)}(0)}{n!}$
+\end_inset
+
+ para 
+\begin_inset Formula $n\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$
+\end_inset
+
+ con radio de convergencia 
+\begin_inset Formula $R$
+\end_inset
+
+, entonces la función 
+\begin_inset Formula $F$
+\end_inset
+
+ dada por 
+\begin_inset Formula $F(z):=\sum_{n=0}^{\infty}\frac{1}{n+1}a_{n}z^{n+1}$
+\end_inset
+
+ tiene radio de convergencia 
+\begin_inset Formula $R$
+\end_inset
+
+ y es primitiva de 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Funciones elementales
+\end_layout
+
+\begin_layout Standard
+La 
+\series bold
+exponencial compleja
+\series default
+ se define como
+\begin_inset Formula 
+\[
+e^{z}:=\sum_{n=0}^{\infty}\frac{1}{n!}z^{n}
+\]
+
+\end_inset
+
+Podemos ver que su radio de convergencia es infinito, 
+\begin_inset Formula $(e^{z})'=e^{z}$
+\end_inset
+
+ y 
+\begin_inset Formula $e^{z}e^{w}=e^{z+w}$
+\end_inset
+
+.
+ Además, 
+\begin_inset Formula $e^{x}>0$
+\end_inset
+
+ para todo 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+, y es estrictamente creciente con
+\begin_inset Formula 
+\begin{eqnarray*}
+\lim_{x\rightarrow\infty}e^{x}=+\infty & \text{ y } & \lim_{x\rightarrow-\infty}e^{x}=0
+\end{eqnarray*}
+
+\end_inset
+
+ Definimos el 
+\series bold
+seno
+\series default
+ y el 
+\series bold
+coseno
+\series default
+, respectivamente, como
+\begin_inset Formula 
+\begin{eqnarray*}
+\sin x:=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!} & \text{ y } & \cos x:=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Vemos que 
+\begin_inset Formula $e^{x+iy}=e^{x}e^{iy}=e^{x}(\cos y+i\sin y)$
+\end_inset
+
+, luego 
+\begin_inset Formula $\sin x=\text{Im}e^{ix}$
+\end_inset
+
+ y 
+\begin_inset Formula $\cos x=\text{Re}e^{ix}$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $|e^{iy}|^{2}=1$
+\end_inset
+
+, se tiene 
+\begin_inset Formula $\sin^{2}x+\cos^{2}x=1$
+\end_inset
+
+.
+ Además:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\sin'x=\cos x$
+\end_inset
+
+ y 
+\begin_inset Formula $\cos'x=-\sin x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\sin(-x)=-\sin x$
+\end_inset
+
+ y 
+\begin_inset Formula $\cos(-x)=\cos x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\sin(x+y)=\sin x\cos y+\cos x\sin y$
+\end_inset
+
+ y 
+\begin_inset Formula $\cos(x+y)=\cos x\cos y-\sin x\sin y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El conjunto 
+\begin_inset Formula $\{x>0:\cos x=0\}$
+\end_inset
+
+ es no vacío y de hecho tiene un primer elemento, que se denota 
+\begin_inset Formula $\frac{\pi}{2}$
+\end_inset
+
+.
+ Además, las funciones seno y coseno son 
+\begin_inset Formula $2\pi$
+\end_inset
+
+-periódicas, y 
+\begin_inset Formula $\psi:[0,2\pi)\rightarrow S$
+\end_inset
+
+ dada por 
+\begin_inset Formula $\psi(t)=e^{it}$
+\end_inset
+
+ es una biyección de 
+\begin_inset Formula $[0,2\pi)$
+\end_inset
+
+ sobre la circunferencia unidad 
+\begin_inset Formula $S\subseteq\mathbb{C}$
+\end_inset
+
+.
+ Tenemos 
+\begin_inset Formula $\sin0=0$
+\end_inset
+
+, 
+\begin_inset Formula $\sin\frac{\pi}{2}=1$
+\end_inset
+
+, 
+\begin_inset Formula $\sin\frac{\pi}{6}=\frac{1}{2}$
+\end_inset
+
+, 
+\begin_inset Formula $\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$
+\end_inset
+
+, 
+\begin_inset Formula $\sin\frac{\pi}{3}=\frac{1}{\sqrt{3}}$
+\end_inset
+
+, 
+\begin_inset Formula $\sin t=\cos\left(\frac{\pi}{2}-t\right)$
+\end_inset
+
+ y 
+\begin_inset Formula $\cos t=\sin\left(\frac{\pi}{2}-t\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Por la biyección 
+\begin_inset Formula $\psi$
+\end_inset
+
+, y como dado 
+\begin_inset Formula $z\in\mathbb{C}$
+\end_inset
+
+, 
+\begin_inset Formula $\frac{z}{|z|}\in S$
+\end_inset
+
+, existe un único 
+\begin_inset Formula $t\in[0,2\pi)$
+\end_inset
+
+, llamado 
+\series bold
+argumento principal
+\series default
+ de 
+\begin_inset Formula $z$
+\end_inset
+
+, tal que 
+\begin_inset Formula $z=|z|(\cos t+i\sin t)=|z|e^{it}$
+\end_inset
+
+.
+ Entonces:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $z_{1}z_{2}=|z_{1}|e^{it_{1}}|z_{2}|e^{it_{2}}=|z_{1}||z_{2}|e^{i(t_{1}+t_{2})}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\frac{1}{z}=z^{-1}=|z|^{-1}e^{-it}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $z^{n}=|z|^{n}e^{int}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Los 
+\begin_inset Formula $n$
+\end_inset
+
+ complejos de la forma 
+\begin_inset Formula $w=\sqrt[n]{|z|}e^{i\frac{2k\pi+t}{n}}$
+\end_inset
+
+ con 
+\begin_inset Formula $k=0,\dots,n-1$
+\end_inset
+
+ son los únicos con 
+\begin_inset Formula $w^{n}=z$
+\end_inset
+
+ para 
+\begin_inset Formula $z=|z|e^{it}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/fuvr2/pegado1.png b/fuvr2/pegado1.png
new file mode 100644
index 0000000..bc7c4e2
--- /dev/null
+++ b/fuvr2/pegado1.png