% (find-LATEX "2024-1-C2-VRP1.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2024-1-C2-VRP1.tex" :end))
% (defun C () (interactive) (find-LATEXsh "lualatex 2024-1-C2-VRP1.tex" "Success!!!"))
% (defun D () (interactive) (find-pdf-page      "~/LATEX/2024-1-C2-VRP1.pdf"))
% (defun d () (interactive) (find-pdftools-page "~/LATEX/2024-1-C2-VRP1.pdf"))
% (defun e () (interactive) (find-LATEX "2024-1-C2-VRP1.tex"))
% (defun o () (interactive) (find-LATEX "2024-1-C2-P1.tex"))
% (defun u () (interactive) (find-latex-upload-links "2024-1-C2-VRP1"))
% (defun v () (interactive) (find-2a '(e) '(d)))
% (defun d0 () (interactive) (find-ebuffer "2024-1-C2-VRP1.pdf"))
% (defun cv () (interactive) (C) (ee-kill-this-buffer) (v) (g))
%          (code-eec-LATEX "2024-1-C2-VRP1")
% (find-pdf-page   "~/LATEX/2024-1-C2-VRP1.pdf")
% (find-sh0 "cp -v  ~/LATEX/2024-1-C2-VRP1.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2024-1-C2-VRP1.pdf /tmp/pen/")
%     (find-xournalpp "/tmp/2024-1-C2-VRP1.pdf")
%   file:///home/edrx/LATEX/2024-1-C2-VRP1.pdf
%               file:///tmp/2024-1-C2-VRP1.pdf
%           file:///tmp/pen/2024-1-C2-VRP1.pdf
%  http://anggtwu.net/LATEX/2024-1-C2-VRP1.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 "2024-1-C2-VRP1" "2" "c2m241vrp1" "c2vr1")

% «.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")
% «.title»		(to "title")
% «.links»		(to "links")
% «.questoes-12»	(to "questoes-12")



\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-dn6 "preamble6.lua" "preamble0")
%\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/2024-1-C2.pdf}
\def\drafturl{http://anggtwu.net/2024.1-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 "dednat6load.lua"}  % (find-LATEX "dednat6load.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

\sa  {[IP]}{\CFname{IP}{}}
\sa{[TFC2]}{\CFname{TFC2}{}}
\def\P   #1{\left(#1\right)}



%  _____ _ _   _                               
% |_   _(_) |_| | ___   _ __   __ _  __ _  ___ 
%   | | | | __| |/ _ \ | '_ \ / _` |/ _` |/ _ \
%   | | | | |_| |  __/ | |_) | (_| | (_| |  __/
%   |_| |_|\__|_|\___| | .__/ \__,_|\__, |\___|
%                      |_|          |___/      
%
% «title»  (to ".title")
% (c2m241vrp1p 1 "title")
% (c2m241vrp1a   "title")

\thispagestyle{empty}

\begin{center}

\vspace*{1.2cm}

{\bf \Large Cálculo 2 - 2024.1}

\bsk

Prova de reposição (VR)

pra quem perdeu a P1

\bsk

Eduardo Ochs - RCN/PURO/UFF

\url{http://anggtwu.net/2024.1-C2.html}

\end{center}

\newpage

% «links»  (to ".links")
% (c2m241vrp1p 2 "links")
% (c2m241vrp1a   "links")

{\bf Links}

\scalebox{0.6}{\def\colwidth{9cm}\firstcol{
}\anothercol{
}}

\newpage



% «questoes-12»  (to ".questoes-12")
% (c2m241vrp1p 3 "questoes-12")
% (c2m241vrp1a   "questoes-12")
% (c2m241p1p 2 "questoes-12")
% (c2m241p1a   "questoes-12")
% (c2m232p1p 2 "questoes-123")
% (c2m232p1a   "questoes-123")
% (c2m231p1p 2 "questoes-123")
% (c2m231p1a   "questoes-123")
% (c2m222p1p 1 "questao-1")
% (c2m222p1a   "questao-1")
% (c2m222p1p 2 "subst-trig")
% (c2m222p1a   "subst-trig")
% (c2m222mva "title")
% (c2m222mva "title" "Aula 10: Mudança de variáveis")
% (c2m222tudop 49)
% (c2m222striga "title")
% (c2m222striga "title" "Aulas 11 e 12: substituição trigonométrica")
% (find-es "maxima" "subst-trig-questions")
% (find-es "maxima" "subst-trig-questions" "F(2,1)")

%\vspace*{-0.4cm}



\scalebox{0.5}{\def\colwidth{10.5cm}\firstcol{

{\bf Questão 1}

\T(Total: 4.0 pts)

\msk

Resolva a integral abaixo por substituição trigonométrica e teste o
seu resultado. Dica: você vai ter que começar com a mudança de
variável $u=3x$.

$$\intx{x^3 {\sqrt{1-9x^2}}%^{\,3}
  }
  \;.
$$

\bsk


\bsk

{\bf Questão 2}

\T(Total: 4.0 pts)

\def\eqnp{\eqnpfull}

\msk

Lembre que:
%
$$\ga{[IP]} = \P{\intx{fg'} = fg - \intx{f'g}}$$

Complete as contas abaixo. Use uma folha inteira. Adicione mais linhas
se precisar. Vou dar algumas dicas no quadro.
%
$$\begin{array}{rclll}
  \intx{hm''} &\eqnp{1}& \Rq && \text{Por } \ga{[IP]} \bsm{f:=h \\ g:=m' \\ (...)} \\
  \intx{h'm'} &\eqnp{2}& \Rq && \text{Por } \ga{[IP]} \bsm{(...)} \\
  \intx{hm''} &\eqnp{3}& \Rq && \text{Cópia da (1)} \\
              &\eqnp{4}& \Rq && \text{Por (2)} \\
  \intx{hm''} &\eqnp{5}& \Rq && \text{Por (3) e (4)} \\
  \intx{(x-4)^2(100e^{10x})} &\eqnp{6}& \Rq && \text{Por (5)}\bsm{(...)} \\
  \end{array}
$$

%Calcule e teste o seu resultado:
%
%$$\intx{\frac{3x+2}{(x+4)(x-5)}}\;.$$

\bsk

% (setq eepitch-preprocess-regexp "^")
% (setq eepitch-preprocess-regexp "^%T ?")
% (find-es "maxima" "subst-trig-questions")
%
%T  (eepitch-maxima)
%T  (eepitch-kill)
%T  (eepitch-maxima)
%T
%T f(a,b) := x^a * sqrt(1 - x^2)^b;
%T F(a,b) := integrate(f(a,b), x);
%T f(3,1);
%T F(3,1);
%T
%T F : sin(2+sqrt(x));
%T diff(F, x);
%T
%T f : (2*x + 3) / ((x-4) * (x+5));
%T F : integrate(f, x);




}\anothercol{

% «questoes-12-dicas»  (to ".questoes-12-dicas")
% (c2m241p1p 3 "questoes-12-dicas")
% (c2m241p1a   "questoes-12-dicas")
% (c2m231p1p 2 "questoes-123-dicas")
% (c2m231p1a   "questoes-123-dicas")
{}

{\bf Dicas:}

\ssk

1) Nestas questões o que vai contar mais pontos é você organizar as
contas de modo que cada passo seja fácil de entender, de verificar, e
de justificar -- ``chegar no resultado certo'' vai valer relativamente
pouco.

\ssk

2) Recomendo que vocês usem o método das ``caixinhas de anotações''
nas mudanças de variável... numa caixinha de anotações a primeira
linha diz a relação entre a variável nova e a antiga, todas as outras
linhas são consequências da primeira, e dentro da caixinha de
anotações você pode usar as gambiarras com variáveis dependentes e
diferenciais, como isto aqui: $dx = 42\,du$...

\ssk

3) ...por exemplo:
%
$$\bmat{
  s = \sen θ \\
  \sqrt{1-s^2} = \cos θ \\
  \frac{ds}{dθ} = \cos θ \\
  ds = \cos θ \, dθ \\
  θ = \arcsen s \\
  }
$$

}}



\newpage

%  _  _            ____  
% | || |     ___  | ___| 
% | || |_   / _ \ |___ \ 
% |__   _| |  __/  ___) |
%    |_|    \___| |____/ 
%                        
% «questoes-45»  (to ".questoes-45")
% (c2m232p1p 3 "questoes-45")
% (c2m232p1a   "questoes-45")
% (c2m231p1p 3 "questoes-45")
% (c2m231p1a   "questoes-45")

\scalebox{0.6}{\def\colwidth{9cm}\firstcol{

{\bf Questão 3}

\T(Total: 1.0 pts)

\msk

No curso nós definimos que {\sl pra nós} a ``fórmula'' do TFC2 seria
esta aqui:
%
$$\ga{[TFC2]}
  \;=\;
  \left( \Intx{a}{b}{F'(x)} \;=\; \difx{a}{b}{F(x)} \right)
$$

Mostre que quando $a=1$, $b=3$ e
%
$$F(x) =
  \begin{cases}
     2x & \text{quando $x<2$}, \\
     x & \text{quando $x≥2$} \\
  \end{cases}
$$

a fórmula $\ga{[TFC2]}$ é falsa.

\msk

Dicas: o melhor modo de fazer isto é representando graficamente $F(x)$
e $F'(x)$ e calculando certas coisas a partir dos gráficos. Considere
que o leitor sabe calcular áreas de retângulos, triângulos e trapézios
no olhômetro quando as coordenadas deles são números simples, mas
complemente os seus gráficos com um pouquinho de português quando nem
tudo for óbvio só a partir dos gráficos.



}\anothercol{

{\bf Questão 4
}

\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.

}}

\newpage



% «questao-4-grids»  (to ".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,-1,0,-2,-1,0,1,2,0,1,2,0,-1,0,-2,1,-2}, Y0=-3} :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

% «questao-1-gab»  (to ".questao-1-gab")
% (c2m241p1p 5 "questao-1-gab")
% (c2m241p1a   "questao-1-gab")
% (find-es "maxima" "2024-1-C2-P1")

{\bf Questão 1: gabarito em Maxima}

%M (%i1) f : x^3 * sqrt(1-4*x^2);
%M (%o1) x^3\,\sqrt{1-4\,x^2}
%M (%i2) F1 : 'integrate(f, x);
%M (%o2) \int {x^3\,\sqrt{1-4\,x^2}}{\;dx}\big.
%M (%i3) F2 : changevar(F1, u=2*x, u, x);
%M (%o3) {\frac{\int {\sqrt{1-u}\,u^3\,\sqrt{u+1}}{\;du}\big.}{16}}
%M (%i4) F2 : rootscontract(F2);
%M (%o4) {\frac{\int {u^3\,\sqrt{1-u^2}}{\;du}\big.}{16}}
%M (%i5) F3 : changevar(F2, u=sin(th), th, u);
%M (%o5) {\frac{\int {\cos \theta \,\sqrt{1-\sin \theta }\,\left(\sin \theta \right)^3\,\sqrt{\sin \theta +1}}{\;d\theta }\big.}{16}}
%M (%i6) F3 : rootscontract(F3);
%M (%o6) {\frac{\int {\cos \theta \,\left(\sin \theta \right)^3\,\sqrt{1-\left(\sin \theta \right)^2}}{\;d\theta }\big.}{16}}
%M (%i7) F3 : subst([sqrt(1-sin(th)^2)=cos(th)], F3);
%M (%o7) {\frac{\int {\left(\cos \theta \right)^2\,\left(\sin \theta \right)^3}{\;d\theta }\big.}{16}}
%L maximahead:sa("gab q1", "")
\pu

%M (%i8) F4 : changevar(F3, c=cos(th), c, th);
%M (%o8) {\frac{\int {c^4-c^2}{\;dc}\big.}{16}}
%M (%i9) F5 : ev(F4, 'integrate);
%M (%o9) {\frac{{\frac{c^5}{5}}-{\frac{c^3}{3}}}{16}}
%M (%i10) F5 : expand(F5);
%M (%o10) {\frac{c^5}{80}}-{\frac{c^3}{48}}
%M (%i11) F6 : subst([c=cos(th)], F5);
%M (%o11) {\frac{\left(\cos \theta \right)^5}{80}}-{\frac{\left(\cos \theta \right)^3}{48}}
%M (%i12) F7 : subst([th=asin(u)], F6);
%M (%o12) {\frac{\left(1-u^2\right)^{{\frac{5}{2}}}}{80}}-{\frac{\left(1-u^2\right)^{{\frac{3}{2}}}}{48}}
%M (%i13) F8 : subst([u=2*x], F7);
%M (%o13) {\frac{\left(1-4\,x^2\right)^{{\frac{5}{2}}}}{80}}-{\frac{\left(1-4\,x^2\right)^{{\frac{3}{2}}}}{48}}
%L maximahead:sa("gab q1 b", "")
\pu

%M (%i14) align_eqs([F1, F2, F3, F4,
%M            F5, F6, F7, F8]);
%M (%o14) \begin{pmatrix}\int {x^3\,\sqrt{1-4\,x^2}}{\;dx}\big.&\mbox{ = }&{\frac{\int {u^3\,\sqrt{1-u^2}}{\;du}\big.}{16}}\cr &\mbox{ = }&{\frac{\int {\left(\cos \theta \right)^2\,\left(\sin \theta \right)^3}{\;d\theta }\big.}{16}}\cr &\mbox{ = }&{\frac{\int {c^4-c^2}{\;dc}\big.}{16}}\cr &\mbox{ = }&{\frac{c^5}{80}}-{\frac{c^3}{48}}\cr &\mbox{ = }&{\frac{\left(\cos \theta \right)^5}{80}}-{\frac{\left(\cos \theta \right)^3}{48}}\cr &\mbox{ = }&{\frac{\left(1-u^2\right)^{{\frac{5}{2}}}}{80}}-{\frac{\left(1-u^2\right)^{{\frac{3}{2}}}}{48}}\cr &\mbox{ = }&{\frac{\left(1-4\,x^2\right)^{{\frac{5}{2}}}}{80}}-{\frac{\left(1-4\,x^2\right)^{{\frac{3}{2}}}}{48}}\cr \end{pmatrix}
%M (%i15) 
%L maximahead:sa("gab q1 c", "")
\pu


\scalebox{0.35}{\def\colwidth{11cm}\firstcol{

\vspace*{0cm}
\def\hboxthreewidth {12cm}
\ga{gab q1}

}\def\colwidth{10cm}\anothercol{

\vspace*{0cm}
\def\hboxthreewidth {12cm}
\ga{gab q1 b}

}\anothercol{

\vspace*{0cm}
\def\hboxthreewidth {14cm}
\ga{gab q1 c}

}}






% (c2m231p1p 5 "questao-1-gab")
% (c2m231p1a   "questao-1-gab")
% (c2m222p1p 5 "questao-1-gab")
% (c2m222p1a   "questao-1-gab")

% (setq eepitch-preprocess-regexp "^")
% (setq eepitch-preprocess-regexp "^%T ")
%
%T  (eepitch-maxima)
%T  (eepitch-kill)
%T  (eepitch-maxima)
%T s : sqrt(1-4*x^2);
%T f : x^3 * s;
%T F : integrate(f, x);
%T G : (1/16) * (s^5/5 - s^3/3);
%T g : diff(G, x);
%T expand(rat(g*s));
%T expand(rat(f*s));

% {\bf Questão 1: gabarito}
% 
% \def\dds{\frac{d}{ds}}
% \def\sqs{\sqrt{1-s^2}}
% 
% 
% \scalebox{0.45}{\def\colwidth{20cm}\firstcol{
% 
% \vspace*{-0.5cm}
% 
% $$\begin{array}[t]{rcl}
%   \\
%   \ints{s^3 \sqrt{1-s^2}^3}
%     &=& \intth{(\senθ)^3 (\cosθ)^3(\cosθ)} \\
%     &=& \intth{(\cosθ)^4(\senθ)^2(\senθ)} \\
%     &=& \intth{(\cosθ)^4(1-(\cosθ)^2)(\senθ)} \\
%     &=& \intc {c^4(1-c^2)(-1)} \\
%     &=& \intc {c^4(c^2-1)} \\
%     &=& \intc {c^6-c^4} \\
%     &=& \frac17 c^7 - \frac15 c^5 \\
%     &=& \frac17 \sqrt{1-s^2}^7 - \frac15 \sqrt{1-s^2}^5 \\
%   \\
%   \dds\sqs &=& \dds(1-s^2)^{1/2} \\
%            &=& \frac12 (1-s^2)^{-1/2} \dds(1-s^2) \\
%            &=& \frac12 (1-s^2)^{-1/2} (-2s) \\
%            &=& - (1-s^2)^{-1/2} s \\
%            &=& -\sqs^{-1} s \\
%   \dds\sqs^k &=& (k\sqs^{k-1})(\dds\sqs) \\
%              &=& (k\sqs^{k-1})(-\sqs^{-1}s) \\
%              &=& -k\sqs^{k-2}s \\
%   \dds\sqs^7 &=& -7 \sqs^5 s \\
%   \dds\sqs^5 &=& -5 \sqs^3 s \\
%   \dds(\frac17 \sqs^7 - \frac15 \sqs^5)
%      &=& \frac17(-7\sqs^5 s) - \frac15(-5\sqs^3 s) \\
%      &=& -\sqs^5 s + \sqs^3 s \\
%      &=& (-\sqs^2+1) \sqs^3 s \\
%      &=& (-(1-s^2)+1) \sqs^3 s \\
%      &=& (-1+s^2+1) \sqs^3 s \\
%      &=& s^2 \sqs^3 s \\
%      &=& s^3 \sqs^3 \\
%   \end{array}
%   \hspace*{0cm}
%   \begin{array}[t]{l}
%     \\
%     \bsm{s = \senθ \\
%          s^2 = (\senθ)^2 \\
%          1-s^2 = (\cosθ)^2 \\
%          \sqrt{1-s^2} = \cosθ \\
%          \frac{ds}{dθ} = \cosθ \\
%          ds = \cosθ\,dθ \\
%         } \\
%     \\[-7pt]
%     \bsm{c = \cosθ \\
%          \frac{dc}{dθ} = -\senθ \\
%          dc = -\senθ\,dθ \\
%          (-1)dc = \senθ\,dθ \\
%          (\senθ)^2 = 1-c^2 \\
%         } \\
%     \\
%     \vspace*{6cm}
%   \end{array}
% $$
% 
% }\anothercol{
% }}


\newpage

% «questao-2-gab»  (to ".questao-2-gab")
% (c2m232p1p 6 "questao-2-gab")
% (c2m232p1a   "questao-2-gab")
% (c2m231p1p 6 "questao-2-gab")
% (c2m231p1a   "questao-2-gab")
% (c2m222p1p 6 "questao-2-gab")
% (c2m222p1a   "questao-2-gab")

{\bf Questão 2: gabarito}

\def\eqnp{\eqnpfull}

\scalebox{0.6}{\def\colwidth{9cm}\firstcol{

Lembre que:
%
$$\ga{[IP]} = \P{\intx{fg'} = fg - \intx{f'g}}$$

Então:
%
$$\begin{array}{rclll}
  \intx{hm''} &\eqnp{1}& hm' - \intx{h'm'}
               && \text{Por } \ga{[IP]} \bsm{f:=h \\ f':=h' \\ g:=m' \\ g':=m''} \\ \\[-10pt]
  \intx{h'm'} &\eqnp{2}& h'm - \intx{h''m}
               && \text{Por } \ga{[IP]} \bsm{f:=h' \\ f':=h'' \\ g:=m \\ g':=m'} \\ \\[-10pt]
  \intx{hm''} &\eqnp{3}& hm' - \intx{h'm'}
               && \text{Cópia da (1)} \\
              &\eqnp{4}&  hm' - \P{h'm - \intx{h''m}}
               && \text{Por (2)} \\
              &\eqnp{4b}& hm' - h'm + \intx{h''m} \\
  \intx{hm''} &\eqnp{5}& hm' - h'm + \intx{h''m}
               && \text{Por (3), (4) e (4b)} \\ \\[-10pt]
  \intx{(x-3)^2e^x} &\eqnp{6}& (x-3)^2e^x - 2(x-3)e^x + \intx{2e^x}
               && \text{Por (5)}
                  \bsm{ h:= (x-3)^2 \\
                        h':= 2(x-3) \\
                        h'':= 2 \\
                        m := e^x \\
                        m' := e^x \\
                        m'' := e^x \\
                      } \\
  \end{array}
$$

}\anothercol{
}}




% {\bf Questão 2: gabarito}
% 
% \scalebox{0.8}{\def\colwidth{14cm}\firstcol{
% 
% \vspace*{-0.5cm}
% 
% $$\begin{array}[t]{rcl}
%   \\
%   \intx{\frac{(\ln x)^3 \cos((\ln x)^4)}{x}}
%     &=& \intx{(\ln x)^3 \cos((\ln x)^4)\frac{1}{x}} \\
%     &=& \intu{u^3 \cos(u^4)} \\
%     &=& \intu{\cos(u^4)u^3} \\
%     &=& \intv{\cos v·\frac14} \\
%     &=& \frac14 \intv{\cos v} \\
%     &=& \frac14 \sen v \\
%     &=& \frac14 \sen(u^4) \\
%     &=& \frac14 \sen((\ln x)^4) \\
%   \\[-5pt]
%   \ddx (\frac14 \sen((\ln x)^4))
%     &=& \frac14 \cos((\ln x)^4) \, \ddx((\ln x)^4) \\
%     &=& \frac14 \cos((\ln x)^4) · 4(\ln x)^3 \ddx(\ln x) \\
%     &=& \frac14 \cos((\ln x)^4) · 4(\ln x)^3 \frac1x \\
%     &=& \cos((\ln x)^4) (\ln x)^3 \frac1x \\
%     &=& \frac{(\ln x)^3 \cos((\ln x)^4)}{x} \\
%   \end{array}
%   \hspace*{-0.5cm}
%   \begin{array}[t]{c}
%     \\
%     \subst{u \;=\; \ln x \\
%            \frac{du}{dx} \;=\; \frac1x \\
%            du \;=\; \frac1x dx \\
%           } \\
%     \\[-5pt]
%     \subst{v \;=\; u^4 \\
%            \frac{dv}{du} \;=\; 4u^3 \\
%            dv \;=\; 4u^3 du \\
%            \frac14 dv \;=\; u^3 du \\
%           } \\
%     \\
%     \vspace*{1.5cm}
%   \end{array}
% $$
% 
% }\anothercol{
% }}

\newpage

% (setq eepitch-preprocess-regexp "^")
% (setq eepitch-preprocess-regexp "^%T ?")
%
%T  (eepitch-maxima)
%T  (eepitch-kill)
%T  (eepitch-maxima)
%T linsolve ([A+B=4, 3*A-2*B=5], [A, B]);
%T linsolve ([A+B=2, 5*A-4*B=3], [A, B]);
%T
%T f : (4*x + 5) / ((x-2)*(x+3));
%T partfrac(f, x);
%T F : integrate(f, x);

\newpage

% «questao-3-gab»  (to ".questao-3-gab")
% (c2m232p1p 8 "questao-4-gab")
% (c2m232p1a   "questao-4-gab")

% (c2m231p1p 8 "questao-4-gab")
% (c2m231p1a   "questao-4-gab")
% (c2m222p1p 7 "questao-3-gab")
% (c2m222p1a   "questao-3-gab")
% (c2m221atisp 21 "1-then-2")
% (c2m221atisa    "1-then-2")

{\bf Questão 3: gabarito}

%L PictBounds.setbounds(v(0,-4), v(4,4))
%L spec = "(0,0)--(2,4)c (2,2)o--(4,4)"
%L pws = PwSpec.from(spec)
%L pws:topict():prethickness("1pt"):pgat("pgatc"):sa("F(x)"):output()
%L
%L PictBounds.setbounds(v(0,-4), v(4,4))
%L spec = "(0,2)--(2,2)o (2,1)o--(4,1)"
%L pws = PwSpec.from(spec)
%L pws:topict():prethickness("1pt"):pgat("pgatc"):sa("F'(x)"):output()
%L
%L spec = "(0,2)--(2,2)o (2,1)c--(4,1)"
%L pwsa = PwSpec.from(spec)
%L pf = Pict {
%L   pwsa:topwfunction():areaify(1, 3):Color("Orange"),
%L   pws:topict()
%L }
%L pf:pgat("pgatc"):sa("int F'(x)"):output()

\pu

\msk

\unitlength=5pt

$$F(x) = \ga{F(x)}
 \quad
 F'(x) = \ga{F'(x)}
 \quad
 \textstyle \Intx{1}{3}{F'(x)} = \ga{int F'(x)} = 3
$$

\def\und#1#2{\underbrace{#1}_{#2}}

$$\und{
  \und{\Intx{1}{3}{F'(x)}}{3} \;=\;
  \und{\und{\und{\difx{1}{3}{F(x)}}{F(3)-F(1)}}{3-2}}{1}
  }{\False}
$$

% (c2m221vsbp 6 "questao-1-gab")
% (c2m221vsba   "questao-1-gab")

\newpage

% «questao-4-gab»  (to ".questao-4-gab")
% (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 4: gabarito}

\unitlength=10pt

$$\begin{array}{r}
 f(x) \;=\; \ga{fig f}  \\ \\
 F(x) \;=\; \Intt{3}{x}{f(t)}
      \;=\; \ga{fig F}  \\
 \end{array}
$$


\newpage

% «erros»  (to ".erros")
% (c2m232p1p 10 "erros")
% (c2m232p1a    "erros")

% {\bf Erros que muitas pessoas cometeram}
% 
% \scalebox{0.9}{\def\colwidth{9cm}\firstcol{
% 
% \bsk
% 
% $\begin{array}{rl}
%   \standout{1a:}
%    & \D \intc{c^4(c^2-1)}
%     \;=\; \intc{c^8-c^4} \\ \\
%   \standout{3a:}
%    & \D \intx {\frac{3x+2}{(x+4)(x-5)}}
%     \;=\; \frac{A}{x+4} + \frac{B}{x-5} \\ \\
%   \standout{3b:}
%    & \D \int {\frac{3x+2}{(x+4)(x-5)}} 
%     \;=\; \int {\frac{10/9}{x+4} + \frac{17/9}{x-5}} \\ \\
%     % \;=\; \frac{10}{9} \int {\frac{1}{x+4}} + \frac{17}{9} \int {\frac{1}{x-5}} \\
%   \standout{3c:}
%    & \D
%      \frac{ \frac{10}{9}(x+4) + \frac{17}{9}(x-5) }
%              {(x+4)(x-5)}
%      \;=\;
%      \frac{ (\frac{10}{9}+\frac{17}{9})x + (\frac{68}{9}-\frac{50}{9}) }
%              {(x+4)(x-5)} \\
%   \end{array}
% $
% 
% }\anothercol{
% }}


\newpage

% «includegraphics»  (to ".includegraphics")
% (c2m241p1p 12 "includegraphics")
% (c2m241p1a    "includegraphics")




\GenericWarning{Success:}{Success!!!}  % Used by `M-x cv'

\end{document}

% (find-pdfpages2-links "~/LATEX/" "2024-1-C2-VRP1")
% (find-pdfpages2-links "~/LATEX/" "2024-1-C2-VRP1" "-pp" "pages=3-5,fitpaper,landscape=true")


% Local Variables:
% coding: utf-8-unix
% ee-tla: "c2vr1"
% ee-tla: "c2m241vrp1"
% End:
