|
Warning: this is an htmlized version!
The original is here, and the conversion rules are here. |
% (find-LATEX "2023-1-C2-P1.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2023-1-C2-P1.tex" :end))
% (defun C () (interactive) (find-LATEXsh "lualatex 2023-1-C2-P1.tex" "Success!!!"))
% (defun D () (interactive) (find-pdf-page "~/LATEX/2023-1-C2-P1.pdf"))
% (defun d () (interactive) (find-pdftools-page "~/LATEX/2023-1-C2-P1.pdf"))
% (defun e () (interactive) (find-LATEX "2023-1-C2-P1.tex"))
% (defun o () (interactive) (find-LATEX "2022-2-C2-P1.tex"))
% (defun u () (interactive) (find-latex-upload-links "2023-1-C2-P1"))
% (defun v () (interactive) (find-2a '(e) '(d)))
% (defun d0 () (interactive) (find-ebuffer "2023-1-C2-P1.pdf"))
% (defun cv () (interactive) (C) (ee-kill-this-buffer) (v) (g))
% (code-eec-LATEX "2023-1-C2-P1")
% (find-pdf-page "~/LATEX/2023-1-C2-P1.pdf")
% (find-sh0 "cp -v ~/LATEX/2023-1-C2-P1.pdf /tmp/")
% (find-sh0 "cp -v ~/LATEX/2023-1-C2-P1.pdf /tmp/pen/")
% (find-xournalpp "/tmp/2023-1-C2-P1.pdf")
% file:///home/edrx/LATEX/2023-1-C2-P1.pdf
% file:///tmp/2023-1-C2-P1.pdf
% file:///tmp/pen/2023-1-C2-P1.pdf
% http://anggtwu.net/LATEX/2023-1-C2-P1.pdf
% (find-LATEX "2019.mk")
% (find-sh0 "cd ~/LUA/; cp -v Pict2e1.lua Pict2e1-1.lua Piecewise1.lua ~/LATEX/")
% (find-sh0 "cd ~/LUA/; cp -v Pict2e1.lua Pict2e1-1.lua Pict3D1.lua ~/LATEX/")
% (find-sh0 "cd ~/LUA/; cp -v C2Subst1.lua C2Formulas1.lua ~/LATEX/")
% (find-sh0 "cd ~/LUA/; cp -v Gram2.lua Tree1.lua Caepro5.lua ~/LATEX/")
% (find-MM-aula-links "2023-1-C2-P1" "C2" "c2m231p1" "c2p1")
% «.defs» (to "defs")
% «.defs-caepro» (to "defs-caepro")
% «.defs-pict2e» (to "defs-pict2e")
% «.defs-T-and-B» (to "defs-T-and-B")
% «.title» (to "title")
% «.questoes-123» (to "questoes-123")
% «.questoes-123-dicas» (to "questoes-123-dicas")
% «.questoes-45» (to "questoes-45")
% «.questao-5-grids» (to "questao-5-grids")
% «.questao-1-gab» (to "questao-1-gab")
% «.questao-2-gab» (to "questao-2-gab")
% «.questao-3-gab» (to "questao-3-gab")
% «.questao-4-gab» (to "questao-4-gab")
% «.questao-5-gab» (to "questao-5-gab")
% «.links» (to "links")
%
% «.djvuize» (to "djvuize")
% <videos>
% Video (not yet):
% (find-ssr-links "c2m231p1" "2023-1-C2-P1")
% (code-eevvideo "c2m231p1" "2023-1-C2-P1")
% (code-eevlinksvideo "c2m231p1" "2023-1-C2-P1")
% (find-c2m231p1video "0:00")
\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")
%\usepackage{emaxima} % (find-LATEX "emaxima.sty")
%
% (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/2023-1-C2.pdf}
\def\drafturl{http://anggtwu.net/2023.1-C2.html}
\def\draftfooter{\tiny \href{\drafturl}{\jobname{}} \ColorBrown{\shorttoday{} \hours}}
% (find-LATEX "2023-1-C2-carro.tex" "defs-caepro")
% (find-LATEX "2023-1-C2-carro.tex" "defs-pict2e")
\catcode`\^^J=10
\directlua{dofile "dednat6load.lua"} % (find-LATEX "dednat6load.lua")
% «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 V = nil -- (find-angg "LUA/Pict2e1.lua" "MiniV")
%L dofile "Piecewise1.lua" -- (find-LATEX "Piecewise1.lua")
%L Pict2e.__index.suffix = "%"
\def\pictgridstyle{\color{GrayPale}\linethickness{0.3pt}}
\def\pictaxesstyle{\linethickness{0.5pt}}
\def\pictnaxesstyle{\color{GrayPale}\linethickness{0.5pt}}
\celllower=2.5pt
\pu
\sa{[IP]}{\CFname{IP}{}}
\sa{[TFC2]}{\CFname{TFC2}{}}
% «defs-T-and-B» (to ".defs-T-and-B")
\long\def\ColorOrange#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){\ColorOrange{\bf(#1 pts)}}
% _____ _ _ _
% |_ _(_) |_| | ___ _ __ __ _ __ _ ___
% | | | | __| |/ _ \ | '_ \ / _` |/ _` |/ _ \
% | | | | |_| | __/ | |_) | (_| | (_| | __/
% |_| |_|\__|_|\___| | .__/ \__,_|\__, |\___|
% |_| |___/
%
% «title» (to ".title")
% (c2m231p1p 1 "title")
% (c2m231p1a "title")
\thispagestyle{empty}
\begin{center}
\vspace*{1.2cm}
{\bf \Large Cálculo 2 - 2023.1}
\bsk
P1 (Primeira prova)
\bsk
Eduardo Ochs - RCN/PURO/UFF
\url{http://anggtwu.net/2023.1-C2.html}
\end{center}
\newpage
% «questoes-123» (to ".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{0.6}{\def\colwidth{9cm}\firstcol{
{\bf Questão 1}
\T(Total: 2.5 pts)
\msk
Calcule:
$$\ints{s^3 \sqrt{1-s^2}}\;.$$
\bsk
{\bf Questão 2}
\T(Total: 2.5 pts)
\msk
Calcule a integral abaixo fazendo pelo menos duas mudanças de variável
e teste o seu resultado:
$$\intx{\frac{\cos(2+\sqrt x)}{2 \sqrt x}}.$$
\bsk
{\bf Questão 3}
\T(Total: 2.5 pts)
\msk
Calcule e teste o seu resultado:
$$\intx{\frac{2x+3}{(x-4)(x+5)}}\;.$$
\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-123-dicas» (to ".questoes-123-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 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")
% (c2m231p1p 3 "questoes-45")
% (c2m231p1a "questoes-45")
\scalebox{0.6}{\def\colwidth{9cm}\firstcol{
{\bf Questão 4}
\T(Total: 1.5 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
%
$$F(x) =
\begin{cases}
x & \text{quando $x<2$}, \\
-x & \text{quando $x≥2$} \\
\end{cases}
$$
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 5}
\T(Total: 1.0 pts)
\msk
Seja $f(t)$ a função no topo da página seguinte.
Seja
%
$$F(x) \;=\; \Intt{5}{x}{f(t)}.$$
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-5-grids» (to ".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 fryF = FromYs.fromys({0,-1,1,-2,2,-3,3,-3,2,-2,1,-1,0}):getYs(1)
%L fryF = FromYs.fromys({0,-1,-3,3,1,0,1,2,1,0,-1,-2,-1,0}):getYs(0)
%L fryF:getypict():pgat("pgatc"):sa("fig f"):output()
%L fryF:getYpict():pgat("pgatc"):sa("fig F"):output()
%L fryF:getYgrid(-4,4):
%L pgat("pgatc"):sa("grid F"):output()
\pu
\unitlength=8pt
$\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")
% (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}
$$\scalebox{0.6}{$
\begin{array}{rcl}
\ints{s^3 \sqrt{1-s^2}}
%&=& \intu{\frac18 u^3 \sqrt{1-u^2}·\frac12} \\
%&=& \frac1{16}\intu{u^3 \sqrt{1-u^2}} \\
&=& \intth{(\senθ)^3 (\cosθ)(\cosθ)} \\
&=& \intth{(\cosθ)^2 (\senθ)^3} \\
&=& \intth{(\cosθ)^2 (\senθ)^2 (\senθ)} \\
&=& \intc{c^2 (1-c^2)(-1)} \\
&=& \intc{c^2 (c^2-1)} \\
&=& \intc{c^4 - c^2} \\
&=& \frac{c^5}{5} - \frac{c^3}{3} \\
&=& \frac{(\cosθ)^5}{5} - \frac{(\cosθ)^3}{3} \\
&=& \frac{\sqrt{1-s^2}^5}{5} - \frac{\sqrt{1-s^2}^3}{3} \\
\\
\frac{d}{ds}(\frac{\sqrt{1-s^2}^5}{5} - \frac{\sqrt{1-s^2}^3}{3})
&=& \frac15 \frac{d}{ds} \sqrt{1-s^2}^5 - \frac13 \frac{d}{ds} \sqrt{1-s^2}^3 \\
&=& \frac15 \frac{d}{ds} (1-s^2)^{5/2} - \frac13 \frac{d}{ds} (1-s^2)^{3/2} \\
&=& \frac15 \frac52 (1-s^2)^{3/2} \frac{d}{ds}(1-s^2)
- \frac13 \frac32 (1-s^2)^{1/2} \frac{d}{ds}(1-s^2) \\
&=& \frac15 \frac52 (1-s^2)^{3/2} (-2s)
- \frac13 \frac32 (1-s^2)^{1/2} (-2s) \\
&=& \frac15 \frac52 (-2)s (1-s^2)^{3/2}
- \frac13 \frac32 (-2)s (1-s^2)^{1/2} \\
&=& - s (1-s^2)^{3/2}
+ s (1-s^2)^{1/2} \\
&=& - s (1-s^2)^{2/2} (1-s^2)^{1/2}
+ s (1-s^2)^{1/2} \\
&=& - s (1-s^2) (1-s^2)^{1/2}
+ s (1-s^2)^{1/2} \\
&=& (- s (1-s^2) + s) (1-s^2)^{1/2} \\
&=& (- s + s^3 + s) (1-s^2)^{1/2} \\
&=& s^3 \sqrt{1-s^2} \\
\end{array}
\hspace*{-3cm}
\begin{array}{l}
% \bsm{u = 2x \\
% u^2 = 4x^2 \\
% x = u/2 \\
% x^3 = u^3/8 \\
% du = 2dx \\
% dx = \frac12 du \\
% } \\
% \\[-7pt]
\bsm{s = \senθ \\
s^2 = (\senθ)^2 \\
1-s^2 = (\cosθ)^2 \\
\sqrt{1-s^2} = \cosθ \\
\frac{ds}{dθ} = \cosθ \\
ds = \cosθ\,dθ \\
} \\
\\[-7pt]
\bsm{c = \cosθ \\
\frac{dc}{dθ} = -\senθ \\
dc = -\senθ\,dθ \\
(-1)dc = \senθ\,dθ \\
(\senθ)^2 = 1-c^2 \\
} \\
\\
\vspace*{6cm}
\end{array}
$}
$$
\newpage
% «questao-2-gab» (to ".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}
$$\begin{array}{rcl}
\intx{\frac{\cos(2+\sqrt x)}{2 \sqrt x}}
&=& \intu{\cos(2+u)} \\
&=& \intv{\cos v} \\
&=& \sen v \\
&=& \sen(2+u) \\
&=& \sen(2+\sqrt{x}) \\
\\[-5pt]
\ddx \sen(2+\sqrt{x})
&=& \cos(2+\sqrt{x}) \, \ddx(2+\sqrt{x}) \\
&=& \cos(2+\sqrt{x}) \, \ddx x^{1/2} \\
&=& \cos(2+\sqrt{x}) \, \frac{1}{2} x^{-1/2} \\
&=& \cos(2+\sqrt{x}) \, \frac{1}{2\sqrt{x}} \\
&=& \frac{\cos(2+\sqrt{x})}{2\sqrt{x}} \\
\end{array}
%\qquad
\begin{array}{c}
\subst{u \;=\; \sqrt{x} \;=\; x^{1/2} \\
\frac{du}{dx} \;=\; \frac12 x^{-1/2} \;=\; \frac{1}{2\sqrt{x}} \\
du \;=\; \frac{1}{2\sqrt{x}} dx \\
u^2 \;=\; x \\
x \;=\; u^2 \\
} \\
\\[-5pt]
\subst{v \;=\; 2+u \\
dv \;=\; du \\
v-2 \;=\; u \\
u \;=\; v-2 \\
} \\
\\
\vspace*{1.5cm}
\end{array}
$$
\newpage
% «questao-3-gab» (to ".questao-3-gab")
% (c2m231p1p 7 "questao-3-gab")
% (c2m231p1a "questao-3-gab")
% (c2m222p1p 8 "questao-4-gab")
% (c2m222p1a "questao-4-gab")
{\bf Questão 3: gabarito}
$$\scalebox{0.6}{$
\begin{array}{rcl}
\frac{2x+3}{(x-4)(x+5)} &=& \frac{A}{x-4} + \frac{B}{x+5} \\
&=& \frac{A(x+5)}{(x-4)(x+5)} + \frac{B(x-4)}{(x-4)(x+5)} \\
&=& \frac{A(x+5)+B(x-4)}{(x-4)(x+5)} \\
&=& \frac{Ax+5A+Bx-4B}{(x-4)(x+5)} \\
&=& \frac{(A+B)x+(5A-4B)}{(x-4)(x+5)} \\
\\[-5pt]
2x+3 &=& (A+B)x+(5A-4B) \\
A+B &=& 2 \\
5A-4B &=& 3 \\
A &=& 11/9 \\
B &=& 7/9 \\
\\[-5pt]
\frac{2x+3}{(x-4)(x+5)} &=& \frac{11/9}{x-4} + \frac{7/9}{x+5} \\
\intx{\frac{2x+3}{(x-4)(x+5)}} &=& \intx{\frac{11/9}{x-4} + \frac{7/9}{x+5}} \\
&=& \frac{11}{9}\intx{\frac{1}{x-4}}
+ \frac{7}{9}\intx{\frac{1}{x+5}} \\
&=& \frac{11}{9} \ln |x-4|
+ \frac{7}{9} \ln |x+5| \\
\\[-5pt]
\ddx(\frac{11}{9} \ln |x-4| + \frac{7}{9} \ln |x+5|)
&=& \frac{11}{9} \frac{1}{x-4} + \frac{7}{9} \frac{1}{x+5} \\
&=& \frac{\frac{11}{9}(x+5) + \frac{7}{9}(x-4)}{(x-4)(x+5)} \\
&=& \frac{(\frac{11}{9}+ \frac{7}{9})x + (\frac{55}{9} - \frac{28}{9})}{(x-4)(x+5)} \\
&=& \frac{2x+3}{(x-4)(x+5)} \\
\end{array}
$}
$$
% (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-4-gab» (to ".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 4: gabarito}
%L Pict2e.bounds = PictBounds.new(v(0,-4), v(4,4))
%L spec = "(0,0)--(2,2)c (2,-2)o--(4,-4)"
%L pws = PwSpec.from(spec)
%L pws:topict():prethickness("1pt"):pgat("pgatc"):sa("F(x)"):output()
%L
%L Pict2e.bounds = PictBounds.new(v(0,-4), v(4,4))
%L spec = "(0,1)--(2,1)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 = "(0,1)--(2,1)o (2,-1)c--(4,-1)"
%L pwsa = PwSpec.from(spec)
%L pf = PictList{
%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
$$F(x) = \ga{F(x)}
\quad
F'(x) = \ga{F'(x)}
\quad
\textstyle \Intx{1}{3}{F'(x)} = \ga{int F'(x)} = 0
$$
\def\und#1#2{\underbrace{#1}_{#2}}
$$\und{
\und{\Intx{1}{3}{F'(x)}}{0} \;=\;
\und{\und{\und{\difx{1}{3}{F(x)}}{F(3)-F(1)}}{(-3)-1}}{-4}
}{\False}
$$
% (c2m221vsbp 6 "questao-1-gab")
% (c2m221vsba "questao-1-gab")
\newpage
% «questao-5-gab» (to ".questao-5-gab")
% (c2m231p1p 9 "questao-5-gab")
% (c2m231p1a "questao-5-gab")
% (c2m222p1p 9 "questao-5-gab")
% (c2m222p1a "questao-5-gab")
{\bf Questão 5: gabarito}
\unitlength=10pt
$$\begin{array}{r}
f(x) \;=\; \ga{fig f} \\ \\
F(x) \;=\; \Intt{5}{x}{f(t)}
\;=\; \ga{fig F} \\
\end{array}
$$
% «links» (to ".links")
\GenericWarning{Success:}{Success!!!} % Used by `M-x cv'
\end{document}
% Local Variables:
% coding: utf-8-unix
% ee-tla: "c2p1"
% ee-tla: "c2m231p1"
% End: