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% (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") \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") % (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: