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% (find-angg "LATEX/2015-2-C2-material.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2015-2-C2-material.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2015-2-C2-material.pdf"))
% (defun e () (interactive) (find-LATEX "2015-2-C2-material.tex"))
% (defun l () (interactive) (find-LATEX "2015-2-C2-material.lua"))
% (defun u () (interactive) (find-latex-upload-links "2015-2-C2-material"))
% (find-xpdfpage "~/LATEX/2015-2-C2-material.pdf")
% (find-lualatex-links "2015-2-C2-material")
% (find-LATEX "falta-misandria-a5.tex")
% (find-LATEXfile "2014-1-GA-P2-gab.tex")
% (find-LATEXfile "2015-1-GA-P2-gabarito.tex" "\\catcode")
% (find-LATEXfile "2015-1-GA-P2-gabarito.tex" "dednat6dir =")
% (find-angg "LATEX/2015-1-C2-lista-edrx-1.tex")
% file:///home/edrx/LATEX/2015-2-C2-material.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
% \usepackage{luacode}
%
\usepackage{edrx15} % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex % (find-angg "LATEX/edrxaccents.tex")
\input edrxheadfoot.tex % (find-dn4ex "edrxheadfoot.tex")
\input istanbuldefs.tex % (find-istfile "defs.tex")
\def\Diag#1{\directlua{tf:processuntil()}\diag{#1}}
\def\Ded #1{\directlua{tf:processuntil()}\ded{#1}}
\def\Exec#1{\directlua{tf:processuntil() #1}}
\def\Expr#1{\directlua{tf:processuntil() output(#1)}}
\def\Expr#1{\directlua{tf:processuntil() output(tostring(#1))}}
%
\begin{document}
% dednat6
%
\catcode`\^^J=10
\directlua{dednat6dir = "dednat6/"}
\directlua{dofile(dednat6dir.."dednat6.lua")}
\directlua{texfile(tex.jobname)}
\directlua{verbose()}
% \directlua{output(preamble1)}
\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
\def\pu{\directlua{pu()}}
% tikz
% mygrid
% (find-angg ".emacs.papers" "tikz")
\tikzset{mycurve/.style=very thick}
\tikzset{axis/.style=semithick}
\tikzset{tick/.style=semithick}
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\tikzset{cldot/.style={anydot, draw=black,fill=black}}
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\draw[step=1,grid] (#1-0.2, #2-0.2) grid (#3+0.2, #4+0.2);
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\def\tikzp#1{\mat{\begin{tikzpicture}#1\end{tikzpicture}}}
% (find-LATEX "edrxtikz.lua" "drawdots0")
\def\drawdots#1{\directlua{output(drawdots0("#1"))}}
{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2015.2
\par Material para exercícios - Eduardo Ochs
\par Versão: 7/dez/2015
\par Links importantes:
\par \url{http://angg.twu.net/2015.2-C2.html} (página do curso)
\par \url{http://angg.twu.net/2015.2-C2/2015.2-C2.pdf} (quadros)
\par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf}
(lista, atualizada)
\par {\tt eduardoochs@gmail.com} (meu e-mail)
}
\bsk
\bsk
% (find-LATEXgrep "grep --color -nH -e rigonom *.tex")
\def\ddx{\frac{d}{dx}}
\def\ddth{\frac{d}{d\theta}}
\def\arcsen{\operatorname{arcsen}}
\def\sen{\operatorname{sen}}
\def\sec{\operatorname{sec}}
\def\ln{\operatorname{ln}}
\def\subst#1{\left[\sm{#1}\right]}
\def\prims #1{∫#1\,ds}
\def\primth#1{∫#1\,dθ}
\def\ints #1#2#3{∫_{s=#1}^{s=#2}#3\,ds}
\def\intu #1#2#3{∫_{u=#1}^{u=#2}#3\,du}
\def\intx #1#2#3{∫_{x=#1}^{x=#2}#3\,dx}
\def\intth#1#2#3{∫_{θ=#1}^{θ=#2}#3\,dθ}
\def\difs #1#2#3{\left. #3 \right|_{s=#1}^{s=#2}}
\def\difu #1#2#3{\left. #3 \right|_{u=#1}^{u=#2}}
\def\difx #1#2#3{\left. #3 \right|_{x=#1}^{x=#2}}
\def\difth#1#2#3{\left. #3 \right|_{θ=#1}^{θ=#2}}
Substituição:
$\begin{array}{l}
\difx a b {g(h(x))} = \intx a b {g'(h(x))h'(x)} \\
\phantom{mmm}|\,| \\
\difu {h(a)} {h(b)} {g(u))} = \intu {h(a)} {h(b)} {g'(u)} \\
\end{array}
$
\msk
$\intx a b {f(g(x))g'(x)} = \intu {g(a)} {g(b)} {f(u)}$
% Substituição inversa:
%
% $
% \begin{array}{rcl}
% \intx {g¹(α)} {g¹(β)} {f(g(x))g'(x)} = \intu {α} {β} {f(u)} \\ \\
% \int {f(g(x))g'(x)} \,dx = \int {f(u)} \,du \quad \subst{u = g(x)} \\ \\
% \int {f(u)\frac{du}{dx}} \,dx = \int {f(u)} \,du \quad \subst{u = g(x)} \\
% \end{array}
% $
\bsk
Um exemplo de substituição:
$\begin{array}[t]{l}
\primth {s^3 c^3} \\
= \primth {s^3 c^2 c} \\
= \primth {s^3 (1-s^2) \frac{ds}{dθ}} \\
= \primth {(s^3 - s^5) \frac{ds}{dθ}} \\
= \prims {s^3 - s^5} \\
= \frac{s^4}{4} - \frac{s^6}{6} \\
\end{array}
%
\qquad
%
\def\intthab{\intthαβ◻}
\def\intsab {\ints{\senα}{\senβ}◻}
\def\difthab{\difthαβ◻}
\def\difsab {\difs{\senα}{\senβ}◻}
%
\def\intthab{(\intthαβ◻)}
\def\intsab {(\ints{\senα}{\senβ}◻)}
\def\difthab{(\difthαβ◻)}
\def\difsab {(\difs{\senα}{\senβ}◻)}
%
\begin{array}[t]{lll}
\primth {\sen^3θ \cos^3θ} & \intthab \\
= \primth {\sen^3θ \cos^2θ \cosθ} & \intthab \\
= \primth {\sen^3θ (1-\sen^2θ) \frac{d\senθ}{dθ}} & \intthab \\
= \primth {(\sen^3θ - \sen^5θ) \frac{d\senθ}{dθ}} & \intthab \\
= \prims {s^3 - s^5} & \intsab \\
= \frac{s^4}{4} - \frac{s^6}{6} & \difsab \\
= \frac{\sen^4θ}{4} - \frac{\sen^6θ}{6} & \difthab \\
\end{array}
$
\bsk
\bsk
\newpage
% Indefinite integrals
\def\ints #1{∫#1\,ds}
\def\intt #1{∫#1\,dt}
\def\intu #1{∫#1\,du}
\def\intx #1{∫#1\,dx}
\def\intz #1{∫#1\,dz}
\def\intth#1{∫#1\,dθ}
% Definite integrals
\def\Ints #1#2#3{∫_{s=#1}^{s=#2}#3\,ds}
\def\Intt #1#2#3{∫_{t=#1}^{t=#2}#3\,dt}
\def\Intu #1#2#3{∫_{u=#1}^{u=#2}#3\,du}
\def\Intx #1#2#3{∫_{x=#1}^{x=#2}#3\,dx}
\def\Intz #1#2#3{∫_{z=#1}^{z=#2}#3\,dz}
\def\Intth#1#2#3{∫_{θ=#1}^{θ=#2}#3\,dθ}
% Difference
\def\Difs #1#2#3{\left. #3 \right|_{s=#1}^{s=#2}}
\def\Difu #1#2#3{\left. #3 \right|_{u=#1}^{u=#2}}
\def\Difx #1#2#3{\left. #3 \right|_{x=#1}^{x=#2}}
\def\Difth#1#2#3{\left. #3 \right|_{θ=#1}^{θ=#2}}
Substituição:
\msk
$\begin{array}{l}
\Difx a b {g(h(x))} = \Intx a b {g'(h(x))\frac{d\,h(x)}{dx}} \\
\phantom{mmm}|\,| \\
\Difu {h(a)} {h(b)} {g(u))} = \Intu {h(a)} {h(b)} {g'(u)} \\
\end{array}
$
\msk
Fórmulas:
\msk
$\begin{array}{l}
\Intx a b {f(g(x))\frac{d\,g(x)}{dx}} \\
= \Intx a b {f(u)\frac{du}{dx}} \\
= \Intu {g(a)} {g(b)} {f(u)}
\end{array}
\qquad
\begin{array}{ll}
\intx {f(g(x))\frac{d\,g(x)}{dx}} \\
= \intx {f(u)\frac{du}{dx}} & \subst{u=g(x)} \\
= \intu {f(u)} & \subst{u=g(x)}
\end{array}
$
\bsk
Substituição inversa:
$\def\a{{h¹(α)}}
\def\b{{h¹(β)}}
\begin{array}{l}
\Difx \a \b {g(h(x))} = \Intx \a \b {g'(h(x))\frac{d\,h(x)}{dx}} \\
\phantom{mmm}|\,| \\
\Difu {h(\a)} {h(\b)} {g(u))} = \Intu {h(\a)} {h(\b)} {g'(u)} \\
\phantom{mmm}|\,| \\
\Difu α β {g(u))} = \Intu α β {g'(u)} \\
\end{array}
$
\msk
Fórmulas:
\msk
$\def\a{{g¹(α)}}
\def\b{{g¹(β)}}
\begin{array}{l}
\Intu α β {f(u)} \\
= \Intx \a \b {f(u)\frac{du}{dx}} \\
= \Intx \a \b {f(g(x))\frac{d\,g(x)}{dx}}
\end{array}
\qquad
\begin{array}{ll}
\intu {f(u)} \\
= \intx {f(u)\frac{du}{dx}} & \subst{u=g(x)\\x=g¹(u)} \\
= \intx {f(g(x))\frac{d\,g(x)}{dx}} & \subst{x=g¹(u)} \\
\end{array}
$
\bsk
Substituição trigonométrica:
\msk
$
\begin{array}{ll}
\Ints a b {F(s, \sqrt{1-s^2})} \\
= \Intth {\arcsen a} {\arcsen b} {F(\senθ, \sqrt{1-\sen^2θ}) \frac{d\senθ}{dθ}} \\
= \Intth {\arcsen a} {\arcsen b} {F(\senθ, \cosθ) \cosθ} \\
\end{array}
\qquad
\begin{array}{ll}
\ints {F(s, \sqrt{1-s^2})} \\
= \intth {F(s, \sqrt{1-s^2}) \frac{ds}{dθ}} & \subst{s=\senθ \\ θ=\arcsenθ} \\
= \intth {F(s, c) c} & \subst{s=\senθ \\ c=\cosθ \\ θ=\arcsenθ} \\
\end{array}
$
\msk
$
\begin{array}{ll}
\Intz a b {F(z, \sqrt{z^2-1})} \\
= \Intth {\arcsec a} {\arcsec b} {F(\secθ, \sqrt{\sec^2θ-1}) \frac{d\secθ}{dθ}} \\
= \Intth {\arcsec a} {\arcsec b} {F(\secθ, \tanθ) \secθ \tanθ} \\
\end{array}
\qquad
\begin{array}{ll}
\intz {F(z, \sqrt{z^2-1})} \\
= \intth {F(z, \sqrt{z^2-1}) \frac{dz}{dθ}} & \subst{z=\secθ \\ θ=\arcsec z} \\
= \intth {F(z, t) zt} & \subst{z=\secθ \\ θ=\arcsec z \\ t=\tanθ} \\
\end{array}
$
\msk
$
\begin{array}{ll}
\Intt a b {F(t, \sqrt{1+t^2})} \\
= \Intth {\arctan a} {\arctan b} {F(\tanθ, \sqrt{1+\tan^2θ}) \frac{d\tanθ}{dθ}} \\
= \Intth {\arctan a} {\arctan b} {F(\tanθ, \secθ) \sec^2θ} \\
\end{array}
\qquad
\begin{array}{ll}
\intt {F(t, \sqrt{1+t^2})} \\
= \intth {F(t, \sqrt{1+t^2}) \frac{dt}{dθ}} & \subst{t=\tanθ \\ θ=\arctan t} \\
= \intth {F(t, z) z^2} & \subst{t=\tanθ \\ θ=\arctan t \\ z=\secθ} \\
\end{array}
$
% & \subst{u = g(x)}
\newpage
% (find-LATEX "2010-1-C2-prova-1.tex")
Algumas fórmulas:
Integração por partes:
$\int_{x=a}^{x=b} f'(x)g(x)\,dx = f(x)g(x) \big|_{x=a}^{x=b} - \int_{x=a}^{x=b} f(x)g'(x)\,dx$
\msk
% Mudança de variável:
%
% $\int_{x=a}^{y=b} \frac{dg}{du} \frac{du}{dx} \,dx = \int_{u=u(a)}^{u=u(b)} \frac{dg}{du} \, du$
%
% $\int_{x=a}^{y=b} g'(u(x))u'(x)\,dx = \int_{u=u(a)}^{u=u(b)} g'(u) \, du$
%
% $\int_{x=a}^{y=b} f(u(x))u'(x)\,dx = \int_{u=u(a)}^{u=u(b)} f(u) \, du$
%
% \msk
Integrais de $(\sen θ)^m (\cos θ)^n$ com um expoente ímpar ($s = \sen θ$, $c= \cos θ$):
$\int s^n c^{2k+1} dθ = \int s^n c^{2k} · c \,dθ =
\subst{\sen θ = s \\
\cos^2 θ = 1 - s^2 \\
\cos θ\,dθ = ds \\
θ = \arcsen s}
\int s^n (1-s^2)^k \, ds$
$\int c^n s^{2k+1} dθ = \int c^n s^{2k} · s \,dθ =
\subst{\cos θ = c \\
\sen^2 θ = 1 - c^2 \\
- \sen θ\,dθ = dc \\
θ = \arccos s}
- \int c^n (1-c^2)^k \, dc$
\msk
Substituição trigonométrica:
$\int F(s, \sqrt{1 - s^2})\,ds =
\subst{s = \sen θ \\
\sqrt{1-s^2} = \cos θ \\
ds = \cos θ \, dθ \\
θ = \arcsen s}
\int F(\sen θ, \cos θ) \cos θ \, dθ$
$\int F(t, \sqrt{1 + t^2})\,dt =
\subst{t = \tan θ \\
\sqrt{1+t^2} = \sec θ \\
dt = \sec^2 θ \, dθ \\
θ = \arctan t}
\int F(\tan θ, \sec θ) \sec^2 θ \, dθ$
$\int F(z, \sqrt{z^2 - 1})\,dz =
\subst{z = \sec θ \\
\sqrt{z^2-1} = \tan θ \\
dz = \tan θ \sec θ\, dθ \\
θ = \arcsec z}
\int F(\sec θ, \tan θ) \tan θ \sec θ \, dθ$
\ssk
$\int\sqrt{1-x^2}\,dx = \frac{\arcsen x}{2} + \frac{x\,\sqrt{1-x^2}}{2}$
\msk
Método de Heaviside:
Se $f(x) = \frac{\aa}{x-a} + \frac{\bb}{x-b} + \frac{\cc}{x-c} = \frac{p(x)}{(x-a)(x-b)(x-c)}$,
então $\lim_{x \to a} f(x)(x-a) = \aa = \frac{p(a)}{(a-b)(a-c)}$.
% \end{document}
\newpage
Funções usadas nas aulas de 30/nov e 2/dez/2015:
\msk
% (find-djvupage "~/2015.2-C2/2015.2-C2.djvu")
$f_1(x) = \tikzp{[scale=0.3,auto]
\mygrid (-1,-2) (5,2);
\drawdots{ (-2,1)--(2,1)c (2,-1)o--(3,-1)o (3,0)c--(6,0) };
}
$
% (find-fline "~/2015.2-C2/20151130_C23.jpg")
$f_2(x) = \tikzp{[scale=0.3,auto]
\mygrid (-1,-2) (4,3);
\drawdots{ (-3,0)--(0,0)o (0,2)c--(1,0)c (1,-1)o--((5,-1) };
}
$
% (find-fline "~/2015.2-C2/20151130_C23.jpg")
$f_3(x) = \tikzp{[scale=0.3,auto]
\mygrid (-1,-2) (5,3);
\drawdots{ (-2,0)--(1,0)c--(2,2)c--(6,2) };
}
$
$f_4(x) = \tikzp{[scale=0.3,auto]
\mygrid (-1,-1) (6,5);
\drawdots{ (-2,0)--(0,0)c--(1,3)c--(2,4)c--(3,3)c--(4,0)c--(7,0) };
}
$
$f_6(x) = \tikzp{[scale=0.3,auto]
\mygrid (-1,-2) (8,3);
\drawdots{ (-2,0)--(1,0)o (1,1)c--(2,1)c (2,2)o--(3,2)o (3,1)c--(4,1)c
(4,0)o--(5,0)o (5,-1)c--(6,-1)c (6,0)o--(10,0) };
}
$
\end{document}
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% coding: utf-8-unix
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