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}
$

}}





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