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Warning: this is an htmlized version!
The original is here, and the conversion rules are here. |
% (find-LATEX "2025-1-C2-P1.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2025-1-C2-P1.tex" :end))
% (defun C () (interactive) (find-LATEXsh "lualatex 2025-1-C2-P1.tex" "Success!!!"))
% (defun D () (interactive) (find-pdf-page "~/LATEX/2025-1-C2-P1.pdf"))
% (defun d () (interactive) (find-pdftools-page "~/LATEX/2025-1-C2-P1.pdf"))
% (defun e () (interactive) (find-LATEX "2025-1-C2-P1.tex"))
% (defun o () (interactive) (find-LATEX "2024-2-C2-P1.tex"))
% (defun u () (interactive) (find-latex-upload-links "2025-1-C2-P1"))
% (defun v () (interactive) (find-2a '(e) '(d)))
% (defun d0 () (interactive) (find-ebuffer "2025-1-C2-P1.pdf"))
% (defun cv () (interactive) (C) (ee-kill-this-buffer) (v) (g))
% (defun oe () (interactive) (find-2a '(o) '(e)))
% (code-eec-LATEX "2025-1-C2-P1")
% (find-pdf-page "~/LATEX/2025-1-C2-P1.pdf")
% (find-sh0 "cp -v ~/LATEX/2025-1-C2-P1.pdf /tmp/")
% (find-sh0 "cp -v ~/LATEX/2025-1-C2-P1.pdf /tmp/pen/")
% (find-xournalpp "/tmp/2025-1-C2-P1.pdf")
% file:///home/edrx/LATEX/2025-1-C2-P1.pdf
% file:///tmp/2025-1-C2-P1.pdf
% file:///tmp/pen/2025-1-C2-P1.pdf
% http://anggtwu.net/LATEX/2025-1-C2-P1.pdf
% (find-LATEX "2019.mk")
% (find-Deps1-links "Caepro5 Piecewise2 Maxima2")
% (find-Deps1-cps "Caepro5 Piecewise2 Maxima2")
% (find-Deps1-anggs "Caepro5 Piecewise2 Maxima2")
% (find-MM-aula-links "2025-1-C2-P1" "2" "c2m251p1" "c2p1")
% «.defs» (to "defs")
% «.defs-T-and-B» (to "defs-T-and-B")
% «.defs-caepro» (to "defs-caepro")
% «.defs-pict2e» (to "defs-pict2e")
% «.defs-maxima» (to "defs-maxima")
% «.defs-V» (to "defs-V")
% «.defs-mvdefs» (to "defs-mvdefs")
% «.title» (to "title")
% «.links» (to "links")
% «.questao-1» (to "questao-1")
% «.questao-2» (to "questao-2")
% «.questao-3» (to "questao-3")
% «.questao-1-grids» (to "questao-1-grids")
% «.como-justificar-uma-MV» (to "como-justificar-uma-MV")
% «.gab-1» (to "gab-1")
% «.gab-2» (to "gab-2")
% «.gab-3» (to "gab-3")
% «.gab-3-cont» (to "gab-3-cont")
\documentclass[oneside,12pt]{article}
\usepackage[colorlinks,citecolor=DarkRed,urlcolor=DarkRed]{hyperref} % (find-es "tex" "hyperref")
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{pict2e}
\usepackage[x11names,svgnames]{xcolor} % (find-es "tex" "xcolor")
\usepackage{colorweb} % (find-es "tex" "colorweb")
%\usepackage{tikz}
%
% (find-LATEX "dednat7-test1.tex")
%\usepackage{proof} % For derivation trees ("%:" lines)
%\input diagxy % For 2D diagrams ("%D" lines)
%\xyoption{curve} % For the ".curve=" feature in 2D diagrams
%
\usepackage{edrx21} % (find-LATEX "edrx21.sty")
\input edrxaccents.tex % (find-LATEX "edrxaccents.tex")
\input edrx21chars.tex % (find-LATEX "edrx21chars.tex")
\input edrxheadfoot.tex % (find-LATEX "edrxheadfoot.tex")
\input edrxgac2.tex % (find-LATEX "edrxgac2.tex")
%
% (find-es "tex" "geometry")
\usepackage[a6paper, landscape,
top=1.5cm, bottom=.25cm, left=1cm, right=1cm, includefoot
]{geometry}
%
\begin{document}
% «defs» (to ".defs")
% (find-LATEX "edrx21defs.tex" "colors")
% (find-LATEX "edrx21.sty")
\def\drafturl{http://anggtwu.net/LATEX/2025-1-C2.pdf}
\def\drafturl{http://anggtwu.net/2025.1-C2.html}
\def\draftfooter{\tiny \href{\drafturl}{\jobname{}} \ColorBrown{\shorttoday{} \hours}}
% (find-LATEX "2024-1-C2-carro.tex" "defs-caepro")
% (find-LATEX "2024-1-C2-carro.tex" "defs-pict2e")
\catcode`\^^J=10
\directlua{dofile "dednat7load.lua"} % (find-LATEX "dednat7load.lua")
\directlua{dednat7preamble()} % (find-angg "LUA/DednatPreamble1.lua")
\directlua{dednat7oldheads()} % (find-angg "LUA/Dednat7oldheads.lua")
% «defs-T-and-B» (to ".defs-T-and-B")
\long\def\ColorDarkOrange#1{{\color{orange!90!black}#1}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){\ColorRed{\bf(Total: #1 pts)}}
\def\B (#1 pts){\ColorDarkOrange{\bf(#1 pts)}}
% «defs-caepro» (to ".defs-caepro")
%L dofile "Caepro5.lua" -- (find-angg "LUA/Caepro5.lua" "LaTeX")
\def\Caurl #1{\expr{Caurl("#1")}}
\def\Cahref#1#2{\href{\Caurl{#1}}{#2}}
\def\Ca #1{\Cahref{#1}{#1}}
% «defs-pict2e» (to ".defs-pict2e")
%L dofile "Piecewise2.lua" -- (find-LATEX "Piecewise2.lua")
%L dofile "Escadas1.lua" -- (find-LATEX "Escadas1.lua")
\def\pictgridstyle{\color{GrayPale}\linethickness{0.3pt}}
\def\pictaxesstyle{\linethickness{0.5pt}}
\def\pictnaxesstyle{\color{GrayPale}\linethickness{0.5pt}}
\celllower=2.5pt
% «defs-maxima» (to ".defs-maxima")
%L dofile "Maxima2.lua" -- (find-angg "LUA/Maxima2.lua")
\pu
% «defs-V» (to ".defs-V")
%L --- See: (find-angg "LUA/MiniV1.lua" "problem-with-V")
%L V = MiniV
%L v = V.fromab
\pu
% «defs-mvdefs» (to ".defs-mvdefs")
%\input 2024-1-C2-mv-defs.tex % (find-LATEX "2024-1-C2-mv-defs.tex")
\input 2025-1-C2-mv-defs.tex % (find-LATEX "2025-1-C2-mv-defs.tex")
\def\eqa{\overset{\scriptscriptstyle(a)}{=}}
\def\eqb{\overset{\scriptscriptstyle(b)}{=}}
\sa{gab 3a}{
\begin{array}[t]{ll}
\multicolumn{2}{l}{ \D \intx{\frac{4 \cos(\log x) (\sen(\log x))^3}{x}} } \\
=& \intx{4 \cos(\log x) (\sen(\log x))^3 ·\frac{1}{x}} \\
\eqa& \intu{4 \cos(u) (\sen(u))^3} \\
=& \intu{4 (\sen(u))^3 \cos(u)} \\
\eqb& \intv{4 v^3} \\
=& v^4 \\
=& (\sen u)^4 \\
=& (\sen(\log x))^4 \\
\end{array}
\hspace*{-0.1cm}
\begin{array}[t]{c}
\\[15pt]
\subst{u \;=\; \log x \\
\frac{du}{dx} \;=\; \frac1x \\
du \;=\; \frac1x dx \\
} \\
\\[-7pt]
\subst{v \;=\; \sen u \\
\frac{dv}{du} \;=\; \cos u \\
dv \;=\; \cos u \, du \\
} \\
\\
\vspace*{1.5cm}
\end{array}
}
\sa{[S3a]}{\CFname{S3a}{}}
\sa{[S4a]}{\CFname{S4a}{}}
\sa{[S3b]}{\CFname{S3b}{}}
\sa{[S4b]}{\CFname{S4b}{}}
\sa{[S2b]}{\CFname{S2b}{}}
\sa{S3a}{\bsm{g(x):=\log x \\ g'(x):=\frac1x \\ f'(u):=4\cos(x)(\sen(u))^3}}
\sa{S4a}{\bsm{g(x):=\log x \\ g'(x):=\frac1x \\ f'(u):=4\cos(x)(\sen(u))^3 \\ f(u):=(\sen u)^4}}
\sa{S2b}{\bsm{x:=u \\ u:=v}}
\sa{S3b}{\bsm{g(x):=sen(x) \\ g'(x):=cos(x) \\ f'(u):=4u^3}}
\sa{S4b}{\bsm{g(x):=sen(x) \\ g'(x):=cos(x) \\ f'(u):=4u^3 \\ f(u):=u^4}}
\sa {MVI1 _s_ S3a}{
\begin{array}{rcl}
\intx {4 \cos(\log x) (\sen(\log x))^3·\frac1x}
&=& \intu {4 \cos(u) (\sen(u))^3} \\
\end{array}
}
\sa {MVD4 _s_ S3a}{
\begin{array}{rcl}
\Intx {a} {b} {4 \cos(\log x) (\sen(\log x))^3·\frac1x}
&=& \Difx {a} {b} {f(\log x)} \\
&=& f(\log b) - f(\log a) \\
&=& \Difu {\log a} {\log b} {f(u)} \\
&=& \Intu {\log a} {\log b} {4 \cos(u) (\sen(u))^3} \\
\end{array}
}
\sa {MVD4 _s_ S4a}{
\begin{array}{rcl}
\Intx {a} {b} {4 \cos(\log x) (\sen(\log x))^3·\frac1x}
&=& \Difx {a} {b} {(\sen(\log x))^4} \\
&=& (\sen(\log b))^4 - (\sen(\log a))^4 \\
&=& \Difu {\log a} {\log b} {(\sen(u))^4} \\
&=& \Intu {\log a} {\log b} {4 \cos(u) (\sen(u))^3} \\
\end{array}
}
\sa {MVI1 _s_ S3b}{
\begin{array}{rcl}
\intx {4 (\sen(x))^3 \cos(x)}
&=& \intu {4 u^3} \\
\end{array}
}
\sa {MVI1 _s_ S3b _s_ S2b}{
\begin{array}{rcl}
\intu {4 (\sen(u))^3 \cos(u)}
&=& \intv {4 v^3} \\
\end{array}
}
\sa {MVD4 _s_ S3b _s_ S2b}{
\begin{array}{rcl}
\Intu {a} {b} {4 (\sen(u))^3 \cos(u)}
&=& \Difu {a} {b} {f(\sen(u))} \\
&=& f(\sen(b)) - f(\sen(a)) \\
&=& \Difv {\sen(a)} {\sen(b)} {f(v)} \\
&=& \Intv {\sen(a)} {\sen(b)} {4 v^3} \\
\end{array}
}
\sa {MVD4 _s_ S4b _s_ S2b}{
\begin{array}{rcl}
\Intu {a} {b} {4 (\sen(u))^3 \cos(u)}
&=& \Difu {a} {b} {(\sen(u))^4} \\
&=& (\sen(b))^4 - (\sen(a))^4 \\
&=& \Difv {\sen(a)} {\sen(b)} {v^4} \\
&=& \Intv {\sen(a)} {\sen(b)} {4 v^3} \\
\end{array}
}
\newpage
% _____ _ _ _
% |_ _(_) |_| | ___ _ __ __ _ __ _ ___
% | | | | __| |/ _ \ | '_ \ / _` |/ _` |/ _ \
% | | | | |_| | __/ | |_) | (_| | (_| | __/
% |_| |_|\__|_|\___| | .__/ \__,_|\__, |\___|
% |_| |___/
%
% «title» (to ".title")
% (c2m251p1p 1 "title")
% (c2m251p1a "title")
\thispagestyle{empty}
\begin{center}
\vspace*{1.2cm}
{\bf \Large Cálculo 2 - 2025.1}
\bsk
P1 (Primeira prova)
\bsk
Eduardo Ochs - RCN/PURO/UFF
\url{http://anggtwu.net/2025.1-C2.html}
\end{center}
\newpage
% «links» (to ".links")
% (c2m251p1p 2 "links")
% (c2m251p1a "links")
% {\bf Links}
% \scalebox{0.6}{\def\colwidth{16cm}\firstcol{
% }\anothercol{
% }}
\newpage
% «questao-1» (to ".questao-1")
% 2jT223: (c2m242p1p 6 "questao-4")
% (c2m242p1a "questao-4")
\scalebox{0.55}{\def\colwidth{9cm}\firstcol{
{\bf Questão 1}
\T(Total: 1.0 pts)
\msk
Seja $f(t)$ a função no topo da página seguinte.
Seja
%
$$F(x) \;=\; \Intt{3}{x}{f(t)}.$$
Desenhe o gráfico de $F(x)$ em algum dos grids vazios da próxima
página. Indique claramente qual é a versão final e quais desenhos são
rascunhos.
\bsk
\bsk
% «questao-2» (to ".questao-2")
% (c2m251p1p 2 "questao-2")
% (c2m251p1a "questao-2")
{\bf Questão 2}
\T(Total: 4.0 pts)
\msk
Resolva esta integral:
%
$$\intx{\frac{2x^3 - 6x^2 - 15x + 55}{x^2 - 2x - 15}}.$$
Lembre que eu vou corrigir a sua solução usando os critérios de
correção que eu expliquei no PDFzinho de ``Dicas para a P1'', mas as
bancas de revisão de prova costumam ignorar os meus critérios de
correção e costumam usar outros critérios -- que nunca são explicados
direito.
% ⌠ 3 2
% ⎮ 2 x - 6 x - 15 x + 55
% (%o18) ⎮ ─────────────────────── dx
% ⎮ 2
% ⌡ x - 2 x - 15
}\def\colwidth{10.5cm}\anothercol{
% «questao-3» (to ".questao-3")
% (c2m251p1p 2 "questao-3")
% (c2m251p1a "questao-3")
% (c2m251dip 12 "chutar-e-testar")
% (c2m251dia "chutar-e-testar")
{\bf Questão 3}
\T(Total: 5.0 pts)
\msk
Seja:
%
$$F(x) \;=\; \intx{\frac{4 \cos(\log x) (\sen(\log x))^3}{x}}.$$
% ⌠ 3
% ⎮ 4 cos(log(x)) sin (log(x))
% (%o9) ⎮ ────────────────────────── dx
% ⎮ x
% ⌡
a) \B (0.5 pts) Integre $F(x)$ pelo ``método rápido'' dos anexos --
use duas mudanças de variável, cada uma com uma caixinha de anotações,
e siga exatamente o modelo -- alinhe os sinais de `$=$', etc.
\ssk
{\sl Chame a igualdade da primeira mudança de variável de `$\eqa$' e a
da segunda mudança de variável de `$\eqb$'.}
\msk
b) \B (2.0 pts) Imagine que alguém te diz ``eu não acredito no método
rápido, você pode me mostrar justificativas pras igualdades `$\eqa$' e
`$\eqb$' usando casos particulares da \ga{[MVI1]}?''
\ssk
{\sl Traduza as suas justificativas do item (a) pra justificativas que
satifaçam a pessoa do item (b).}
\msk
c) \B (2.5 pts) Imagine que alguém te diz ``eu não acredito na
\ga{[MVI1]}, você pode me mostrar justificativas pras igualdades
`$\eqa$' e `$\eqb$' usando casos particulares da \ga{[MVD4]}?''
\ssk
{\sl Traduza as suas justificativas do item (b) pra justificativas que
satifaçam a pessoa do item (c).}
}}
\newpage
% «questao-1-grids» (to ".questao-1-grids")
% (c2m242p1p 5 "questao-4-grids")
% (c2m242p1a "questao-4-grids")
% (c2m232p1p 4 "questao-5-grids")
% (c2m232p1a "questao-5-grids")
% (c2m231p1p 4 "questao-5-grids")
% (c2m231p1a "questao-5-grids")
% (c2m222p1p 4 "questao-5-grids")
% (c2m222p1a "questao-5-grids")
%L -- (find-angg "LUA/Pict2e1-1.lua" "FromYs")
%L fry = FromYs.from {ys={0,-1,1,-2,2,-3,3,-3,2,-2,1,-1,0}, Y0=0} :setall()
%L fry = FromYs.from {ys={0,-1,-3,3,1,0,1,2,1,0,-1,-2,-1,0}, Y0=0} :setall()
%L fry = FromYs.from {ys={2,1,0,1,2,-2,1,-2,0,-2,0,1,2,1,0,-1,-2,-1,0}, Y0=-3} :setall()
%L fry = FromYs.from {ys={2,-1,1,-2,-1,0,1,2,0,1,2,0,-1,0,-2,1,-2}, Y0=-2} :setall()
%L Pict {
%L fry:ypict() :sa("fig f"),
%L fry:ypict() :prethickness("1pt"):sa("fig f"),
%L fry:Ypict() :sa("fig F"),
%L fry:grid(-5,5):sa("grid F"),
%L } :output()
\pu
\unitlength=8pt
$\scalebox{0.9}{$
\begin{array}{ll}
\ga{fig f} \phantom{mm} & \ga{fig f} \\ \\
\ga{grid F} & \ga{grid F} \\ \\
\ga{grid F} & \ga{grid F} \\
\end{array}
$}
$
\newpage
% (c2m251dicasp1p 4 "mais-sobre-o-modo-rapido")
% (c2m251dicasp1a "mais-sobre-o-modo-rapido")
% (c2m251dicasp1p 4 "mais-sobre-o-modo-rapido")
% (c2m251dicasp1a "mais-sobre-o-modo-rapido")
% 2hT191: (c2m232p1p 6 "questao-2-gab")
% (c2m232p1a "questao-2-gab")
% 2gT46: (c2m231mvp 6 "caixinhas")
% (c2m231mva "caixinhas")
{\bf Anexo: método rápido, \ga{[MVI1]}, \ga{[MVD4]}}
\scalebox{0.55}{\def\colwidth{10cm}\firstcol{
Lembre que o ``método rápido'' tem essa cara aqui:
%
$$\begin{array}[t]{ll}
\\
\multicolumn{2}{l}{ \D \intx{\frac{(\ln x)^3 \cos((\ln x)^4)}{x}} } \\
=& \intx{(\ln x)^3 \cos((\ln x)^4)\frac{1}{x}} \\
=& \intu{u^3 \cos(u^4)} \\
=& \intu{\cos(u^4)u^3} \\
=& \intv{\cos v·\frac14} \\
=& \frac14 \intv{\cos v} \\
=& \frac14 \sen v \\
=& \frac14 \sen(u^4) \\
=& \frac14 \sen((\ln x)^4) \\
\end{array}
\hspace*{-0.1cm}
\begin{array}[t]{c}
\\
\subst{u \;=\; \ln x \\
\frac{du}{dx} \;=\; \frac1x \\
du \;=\; \frac1x dx \\
} \\
\\[-5pt]
\subst{v \;=\; u^4 \\
\frac{dv}{du} \;=\; 4u^3 \\
dv \;=\; 4u^3 du \\
\frac14 dv \;=\; u^3 du \\
} \\
\\
\vspace*{1.5cm}
\end{array}
$$
e em cada caixinha de anotações a) a primeira linha diz a relação
entre a variável antiga e a variável nova, b) todas as outras linhas
da caixinha são consequências dessa primeira, e c) dentro da caixinha
a gente permite gambiarras com diferenciais.
}\def\colwidth{10cm}\anothercol{
\aligneqswide
\mvdefaults
E lembre que:
%
$$\begin{array}{rcl}
\ga{[MVI1]} &=& \ga{(MVI1)} \\
\ga{[MVD4]} &=& \ga{(MVD4)} \\
\end{array}
$$
}}
\newpage
% «como-justificar-uma-MV» (to ".como-justificar-uma-MV")
% 2kT110: (c2m251dip 17 "como-justificar-uma-MV")
% (c2m251dia "como-justificar-uma-MV")
{\bf Anexo: como justificar uma MV de cabeça}
\def\dudx{\frac{du}{dx}}
\def\und#1#2{\underbrace{\mathstrut #1}_{#2}}
\def\sfrac#1#2{{\textstyle\frac{#1}{#2}}}
\sa{anot1}{\bmat{u=x^3 \\ \dudx=\ddx u=\ddx x^3 = 3x^2 }}
\sa{anot2}{\bmat{u=x^3 \\ \dudx=3x^2 }}
\sa{intx 2}{\int { \cos(\und{x^3}{ u }) · \sfrac13 · \und{\und{3x^2}{\dudx}\,dx}{du}}}
\sa{intu 2}{\intu{ \cos( u ) · \sfrac13}}
\sa{intx 3}{\intx{\und{\cos(\und{x^3}{g(x)}) · \sfrac13}{f'(g(x))} · \und{3x^2}{g'(x)} }}
\sa{intu 3}{\intu{\und{\cos( u ) · \sfrac13}{f'(u)}}}
\scalebox{0.45}{\def\colwidth{16cm}\firstcol{
{}
Por exemplo...
$$\begin{array}{rclc}
\D \intt{t^2 \cos(t^3)} &=& \Rq \\
\D \intx{x^2 \cos(x^3)} &=& \Rq & \ga{anot1} \\ \\[-11pt]
\D \ga{intx 2} &=& \D \ga{intu 2} & \ga{anot2} \\ \\[-11pt]
\D \ga{intx 3} &=& \D \ga{intu 3} \\
\end{array}
$$
\mvdefaults
\def\und#1#2{\underbrace{\mathstrut #1}_{#2}}
\def\und#1#2{\underbrace{\mathstrut #1}_{\textstyle #2}}
\sa{[MVI1] sp}{\phantom{mmmmmm} \ga{[MVI1]} \phantom{mmmmmm}}
\sa {mvi1 1}{\intx{f'(g(x))g'(x)} = \intu{f'(u)}}
\sa {mvi1 2}{\intx{\cos(x^3)·\sfrac13·3x^2} = \intu{\cos(u)·\sfrac13}}
\sa {s1}{\bmat{g(x):=x^3 \\ g'(x):=3x^2 \\ f'(u):=\cos(u)·\sfrac13}}
\sa {s2}{\bmat{x:=t \\ u:=w}}
\sa {s1 b}{\bmat{g(x):=x^3 \\ g'(x):=3x^2 \\ f'(u):=\sfrac13\cos(u)}}
\sa {mvi1 3}{\intt{\cos(t^3)·\sfrac13·3t^2} = \intw{\cos(w)·\sfrac13}}
$$\begin{array}{rclc}
\und{\und{\und{\ga{[MVI1] sp}}
{\ga{mvi1 1}}
\ga{s1}
}{\ga{mvi1 2}}
\ga{s2}
}{\ga{mvi1 3}}
\\
\end{array}
$$
$$\begin{array}{rclc}
\D \intt{t^2 \cos(t^3)}
&=& \D \intw{\sfrac13 \, \cos(w)}
& \;\; \text{Por $\ga{[MVI1]} \ga{s1 b} \ga{s2}$} \\
\end{array}
$$
}\def\colwidth{9cm}\anothercol{
{\bf Dicas:}
Repare que no exemplo à esquerda o problema original era este,
%
$$ \D \intt{t^2 \cos(t^3)} = \Rq $$
e eu resolvi ele nesta ordem: 1) eu mudei a variável dele pra $x$ pra
ficar com algo mais parecido com a $\ga{[MVI1]}$, 2) eu escolhi a
mudança de variável certa, que era $u=x^3$, 3) eu calculei o $\dudx$,
4) eu rearrumei o problema original pro $\dudx$ ficar colado no $dx$,
5) eu fiz a mudança de variável pelo método rápido, 6) eu reescrevi as
anotações do método rápido pra obter $g(x)$, $g'(x)$ e $f'(u)$, 7) eu
transformei essas $g(x)$, $g'(x)$ e $f'(u)$ numa substituição, 8) eu
calculei os resultados parciais dessa substituição e da
$\bsm{x:=t \\ u:=w}$, 9) eu reescrevi a substituição que eu tinha
obtido e testado pra fingir que eu primeiro tinha resolvido o problema
original de cabeça e depois eu escrevi a justificativa porque alguém
me perguntou como eu tinha chegado naquele resultado.
}}
\newpage
% «gab-1» (to ".gab-1")
% (c2m251p1p 8 "gab-1")
% (c2m251p1a "gab-1")
% 2jT228: (c2m242p1p 11 "gab-4")
% (c2m242p1a "gab-4")
% (c2m241p1p 5 "questao-4-gab")
% (c2m241p1a "questao-4-gab")
% (c2m231p1p 9 "questao-5-gab")
% (c2m231p1a "questao-5-gab")
% (c2m222p1p 9 "questao-5-gab")
% (c2m222p1a "questao-5-gab")
{\bf Questão 1: gabarito}
\unitlength=10pt
$$\begin{array}{r}
f(x) \;=\; \ga{fig f} \\ \\
F(x) \;=\; \Intt{3}{x}{f(t)}
\;=\; \ga{fig F} \\
\end{array}
$$
\newpage
% «gab-2» (to ".gab-2")
% (c2m251p1p 7 "gab-2")
% (c2m251p1a "gab-2")
{\bf Questão 2: gabarito}
% (find-angg "MAXIMA/2024-2-C2-partfrac.mac" "test-solve2")
\scalebox{0.38}{\def\colwidth{16cm}\firstcol{
Temos:
%
$$\begin{array}{rcl}
\D \frac{2 x^3 - 6 x^2 - 15 x + 55}{x^2 - 2 x - 15}
&=& \D \frac{(2 x - 2) (x^2 - 2 x - 15) + 11 x + 25}{x^2 - 2 x - 15} \\
&=& \D 2x - 2 + \frac{11 x + 25}{x^2 - 2 x - 15} \\
&=& \D 2x - 2 + \frac{11 x + 25}{(x - 5) (x + 3)} \\
\end{array}
$$
Queremos encontrar $A$ e $B$ tais que:
%
$$\begin{array}{rcl}
\D \frac{11 x + 25}{(x - 5) (x + 3)}
&=& \D \frac{A}{x - 5} + \frac{B}{x + 3} \\
&=& \D \frac{A(x+3) + B(x-5)}{(x - 5) (x + 3)} \\
&=& \D \frac{(A+B)x + (3A-5B)}{(x - 5) (x + 3)} \\
\end{array}
$$
Vamos resolver um problema um pouco mais simples:
%
$$\begin{array}{rcl}
11 x + 25 &=& (A+B)x + (3A-5B) \\
A+B &=& 11 \\
3A-5B &=& 25 \\
B &=& 11-A \\
3A-5(11-A) &=& 25 \\
3A-55+5A &=& 25 \\
8A &=& 80 \\
A &=& 10 \\
B &=& 11-10 \\
&=& 1 \\
\end{array}
$$
Temos:
%
$$\begin{array}{rcl}
\D \intx{\frac{2 x^3 - 6 x^2 - 15 x + 55}{x^2 - 2 x - 15}}
&=& \D \intx{2x - 2 + \frac{11 x + 25}{(x - 5) (x + 3)}} \\
&=& \D \intx{2x - 2 + \frac{A}{x - 5} + \frac{B}{x + 3}} \\
&=& \D \intx{2x - 2 + \frac{10}{x - 5} + \frac{1}{x + 3}} \\
&=& x^2 - 2x + 10\log(x-5) + \log(x+3) \\
\end{array}
$$
% ⌠ 3 2 ⌠
% ⎮ 2 x - 6 x - 15 x + 55 ⎮ 1 10
% ⎮ ─────────────────────── dx = ⎮ (───── + 2 x + ───── - 2) dx
% ⎮ 2 ⎮ x + 3 x - 5
% ⌡ x - 2 x - 15 ⌡
%
% 2
% = log(x + 3) + x - 2 x + 10 log(x - 5)
%
}\anothercol{
}}
\newpage
% «gab-3» (to ".gab-3")
% (c2m251p1p 8 "gab-3")
% (c2m251p1a "gab-3")
{\bf Questão 3: mini-gabarito}
\scalebox{0.52}{\def\colwidth{25cm}\firstcol{
\vspace*{0cm}
$\ga{gab 3a}$
\vspace*{-0.5cm}
$\begin{array}{lcl}
\ga{[S3a]} &=& \ga {S3a} \\
\ga{[S4a]} &=& \ga {S4a} \\
\ga{[MVI1]} \ga{[S3a]} &=& \left( \ga {MVI1 _s_ S3a} \right) \\
\ga{[MVD4]} \ga{[S3a]} &=& \left( \ga {MVD4 _s_ S3a} \right) \\
\ga{[MVD4]} \ga{[S4a]} &=& \left( \ga {MVD4 _s_ S4a} \right) \\
\end{array}
$
}}
\newpage
% «gab-3-cont» (to ".gab-3-cont")
% (c2m251p1p 8 "gab-3-cont")
% (c2m251p1a "gab-3-cont")
{\bf Questão 3: mini-gabarito (cont.)}
\scalebox{0.48}{\def\colwidth{20cm}\firstcol{
\vspace*{0cm}
$\ga{gab 3a}$
\vspace*{-0.5cm}
$\begin{array}{lcl}
\ga{[S2b]} &=& \ga {S2b} \\
\ga{[S3b]} &=& \ga {S3b} \\
\ga{[S4b]} &=& \ga {S4b} \\
\ga{[MVI1]} \ga{[S3b]} &=& \left( \ga {MVI1 _s_ S3b} \right) \\
\ga{[MVI1]} \ga{[S3b]} \ga{[S2b]} &=& \left( \ga {MVI1 _s_ S3b _s_ S2b} \right) \\
\ga{[MVD4]} \ga{[S3b]} \ga{[S2b]} &=& \left( \ga {MVD4 _s_ S3b _s_ S2b} \right) \\
\ga{[MVD4]} \ga{[S4b]} \ga{[S2b]} &=& \left( \ga {MVD4 _s_ S4b _s_ S2b} \right) \\
\end{array}
$
}}
\GenericWarning{Success:}{Success!!!} % Used by `M-x cv'
\end{document}
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% coding: utf-8-unix
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