Warning: this is an htmlized version!
The original is across this link,
and the conversion rules are here.
% (find-angg "LATEX/2009-1-C2-prova-1-gab.tex")
% (find-angg "LATEX/2009-1-C2-prova-1.tex")
% (find-dn4ex "edrx08.sty")
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% (defun c () (interactive) (find-zsh "cd ~/LATEX/ && ~/dednat4/dednat41 2009-1-C2-prova-1-gab.tex && pdflatex 2009-1-C2-prova-1-gab.tex"))
% (eev "cd ~/LATEX/ && Scp 2009-1-C2-prova-1-gab.{dvi,pdf} edrx@angg.twu.net:slow_html/LATEX/")
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\documentclass[oneside]{book}
\usepackage[latin1]{inputenc}
\usepackage{edrx08}       % (find-dn4ex "edrx08.sty")
%L process "edrx08.sty"  -- (find-dn4ex "edrx08.sty")
\input edrxheadfoot.tex   % (find-dn4ex "edrxheadfoot.tex")
\begin{document}

\input 2009-1-C2-prova-1-gab.dnt

%*
% (eedn4-51-bounded)

\def\ddx{\frac{d}{dx}}
\def\ddth{\frac{d}{d\theta}}
\def\sen{\operatorname{sen}}
\def\sec{\operatorname{sec}}
\def\ln{\operatorname{ln}}

% To update the list of slides uncomment this line:
%\makelos{tmp.los}
% then rerun LaTeX on this file, and insert the contents of "tmp.los"
% below, by hand (i.e., with "insert-file"):
% (find-fline "tmp.los")
% (insert-file "tmp.los")

{\setlength{\parindent}{0pt}

Cálculo Diferencial e Integral II

PURO-UFF - 2009.1

Turma: A1/RCT00017

Professor: Eduardo Ochs

{Gabarito (parcial?) da primeira prova}

Versão: 2009jun05

}

\bsk
\bsk

\def\Area{\text{Área}}


1a) (1.5 pontos):

$$\begin{array}{rcl}
  \Area(A) &=& Å_{-4}^{-2} f(x)dx \\
  \Area(B) &=& Å_{-3}^{-0} g(x)dx - Å_{-3}^{-2} f(x)dx - Å_{-2}^{0} h(x)dx \\
  \Area(C) &=& Å_{-2}^{-0} h(x)dx + Å_{0}^{2} k(x)dx \\ 
  \Area(D) &=& Å_{0}^{4}   p(x)dx - Å_{0}^{4} k(x)dx \\ 
  \end{array}
$$

1b) (1.0 pontos):

$$\begin{array}{rclcrcl}
  f(x) &=& 1-(x+3)^2       && Åf(x)dx &=& x - \frac13 (x+3)^3 \\ 
  g(x) &=& 2 \sqrt{x+4}-1  && Åg(x)dx &=&     \frac43 (x+4)^{3/2} - x \\
  h(x) &=& ((x+4)/2)^2-1   && Åh(x)dx &=&  \frac1{12} (x+4)^3     - x \\ 
  k(x) &=& ((4-x)/2)^2-1   && Åk(x)dx &=& -\frac1{12} (4-x)^3     - x \\ 
  p(x) &=& 2 \sqrt{4-x}-1  && Åp(x)dx &=&   - \frac43 (4-x)^{3/2} - x \\ 
  \end{array}
$$

% F = function (x) return x - 1/3 * (x+3)^3         end
% G = function (x) return     4/3 * (x+4)^(3/2) - x end
% H = function (x) return    1/12 * (x+4)^3     - x end
% K = function (x) return  - 1/12 * (4-x)^3     - x end
% P = function (x) return   - 4/3 * (4-x)^(3/2) - x end

$\Area(B) = 3$

$\Area(D) = 16/3$

% (find-angg      "LATEX/2009-1-C2-prova-1.tex")
% (find-dvipage "~/LATEX/2009-1-C2-prova-1.dvi")

\bsk\bsk

2a) (0.5 pontos)
%
$$\begin{array}{rcl}
  Åx^2 \ln x\,dx &=& (\frac13 x^3)(\ln x) - Å(\frac13 x^3)\frac1x \, dx \\
                 &=& (\frac13 x^3)(\ln x) - \frac13 Å x^2 \, dx \\
                 &=& (\frac13 x^3)(\ln x) - \frac19 x^3 \, dx \\
  \end{array}
$$

2b) (1.0 pontos)

Eu fiz uma conta errada quando bolei esta questão... Temos primitivas
para $Å(\sen x)x^n\,dx$ e para $Åx^n \ln x\,dx$, mas não para $Å(\sen
x)\ln x\,dx$ --- quando tentamos calcular $Å(\sen x)\ln x\,dx$ por
integração por partes caímos em $Å \frac{\sen x}{x}\,dx$, que não tem
uma primitiva que possa ser expressa por funções elementares... Veja
\url{http://en.wikipedia.org/wiki/Differential_Galois_theory} para
detalhes.

{\sl Vou dobrar o valor desta questão, e dar até 2.0 pontos pra quem tiver
feito várias boas tentativas incompletas.}

\newpage

3a) (0.5 pontos)
%
$$\begin{array}{rcl}
  Å\frac{1}{(x-2)(x-3)} dx &=& Å\frac{a}{x-2} + \frac{b}{x-3}    \,dx \\
                           &=& Å\frac{a(x-3)+b(x-2)}{(x-2)(x-3)} \,dx \\
                           &=& Å\frac{-(x-3)+(x-2)}{(x-2)(x-3)}  \,dx \\
                           &=& Å\frac{-1}{x-2} + \frac{1}{x-3}   \,dx \\
                           &=& -\ln|x-2| + \ln|x-3| \\
  \end{array}
$$

3b) (0.5 pontos)
%
$$\begin{array}{rcl}
  Å\frac{x+1}{(x-1)^2} dx &=& Å\frac{a}{(x-1)} + \frac{b}{(x-1)^2} \,dx \\
                          &=& Å\frac{a(x-1)+b}{(x-1)^2}            \,dx \\
                          &=& Å\frac{(x-1)+2}{(x-1)^2}             \,dx \\
                          &=& Å\frac{1}{(x-1)} + \frac{2}{(x-1)^2} \,dx \\
                          &=& \ln|x-1| - \frac{2}{x-1} \\
  \end{array}
$$

\bsk
\bsk

4a) (0.8 pontos)

Se $u=x-10$ então $x = u+10$ e $x^2 - 20x + 1000 = (u+10)^2 - 20(u+10)
+ 1000 = (u^2 + 20u + 100) - (20u + 200) + 1000 = u^2 + 900$.
%
$$\begin{array}{rcl}
  Å_{x=2}^{x=3}       \frac{x}{\sqrt{x^2 - 20x + 1000}} \, dx &=&
  Å_{u=2-10}^{u=3-10} \frac{u+10}{\sqrt{u^2 + 900}} \, du
  \end{array}
$$

4b) (0.2 pontos)
%
$$\begin{array}{rcl}
      Å \frac{x^2+1}{x\sqrt{4-x^2}}\,dx
  &=& Å \frac{x^2+1}{x\sqrt{4(1-\frac14 x^2)}}\,dx \\
  &=& Å \frac{x^2+1}{2x\sqrt{(1-\frac14 x^2)}}\,dx \\
  \end{array}
$$

% $$\begin{array}{rcl}
%       Å \frac{x^2+1}{x\sqrt{4-x^2}}\,dx
%   &=& Å \frac{x^2+1}{x\sqrt{4(1-\frac14 x^2)}}\,dx \\
%   &=& Å \frac{x^2+1}{x\sqrt{4(1-(\frac12 x)^2)}}\,dx \\
%   &=& Å \frac{4(\frac12 x)^2+1}{2(\frac12 x)2\sqrt{1-(\frac12 x)^2}}\,dx \\
%   &=& Å \frac{(\frac12 x)^2+1}{(\frac12 x)\sqrt{1-(\frac12 x)^2}}\,dx \\
%   &=& Å \frac{u^2+1}{u\sqrt{1-u^2}}\,2\,du \\
%   \end{array}
% $$

4c) (0.5 pontos)

Se $u=3x$ então $x=\frac13 u$ e $dx = \frac13\,du$. Aí:
%
$$\begin{array}{rcl}
      Å_{x=1/9}^{x=1/6} \frac{\sqrt{1-9x^2}}{3-x}\,dx
  &=& Å_{u=1/3}^{u=1/3} \frac{\sqrt{1-u^2}}{3-\frac13 u}\,\frac13 du \\
  \end{array}
$$


4d) (1.0 pontos)
%
$$\begin{array}{rcl}
  Å \frac{t^2}{(\sqrt{1+t^2})^5}\,dt &=& Å\frac{t^2}{z^5}z^2\,d \\
                                     &=& Å\frac{t^2}{z^3}\,d \\
                                     &=& Å t^2 z^{-3}\,d \\
                                     &=& Å \frac{s^2}{c^2} c^3\,d \\
                                     &=& Å s^2 c\,d \\
                                     &=& Å s^2 \,ds \\
                                     &=& \frac13 s^3 \\
  \end{array}
$$

\newpage

4e) (1.0 pontos)
%
$$\begin{array}{rcl}
      Å_{t=a}^{t=b} \frac{t^2}{(\sqrt{1+t^2})^5}\,dt
  &=& Å_{=\arctan a}^{=\arctan b} \frac{t^2}{z^5}z^2\,d \\
  &=& Å_{=\arctan a}^{=\arctan b} s^2 c\,d \\
  &=& Å_{s=\sen \arctan a}^{s=\sen \arctan b} s^2 \,ds \\
  &=& \frac13 s^3 \big|_{s=\sen \arctan a}^{s=\sen \arctan b} \\
  &=& \frac13 (\sen \arctan t)^3 \big|_{t=a}^{t= b} \\
  \end{array}
$$

\bsk
\bsk

5a) (0.4 pontos)

Altura: $2f(x_0) = 2 \sqrt{1-(x_0-3)^2}$

Largura: $\ee$

Raio (interno) da casca cilíndrica: $x_0$

\msk

5b) (0.4 pontos)

$A_i = (x_{i+1}-x_i)2f(x_i) = 2f(x_i)(x_{i+1}-x_i)$

$V_i = \pi (2(x_{i+1}-x_i)x_i)2f(x_i) = 4\pi x_i f(x_i) (x_{i+1}-x_i)$

\msk

5c) (0.6 pontos)

Aproximações por somatórios:

Área total: $\sum_{i=0,\ldots,n-1} A_i = \sum_{i=0,\ldots,n-1} 2f(x_i)(x_{i+1}-x_i)$

Volume total: $\sum_{i=0,\ldots,n-1} V_i = \sum_{i=0,\ldots,n-1} 4\pi x_i f(x_i) (x_{i+1}-x_i)$

\msk

5d) (0.6 pontos)

Área total:
%
$$\begin{array}{rcl}
  \sum_{i=0,\ldots,n-1} 2f(x_i)(x_{i+1}-x_i)
    &\approx& Å_{x=x_0}^{x=x_n} 2f(x)\,dx          \\
          &=& Å_{x=2}^{x=4} 4\sqrt{1-(x-3)^2} \,dx \\
  \end{array}
$$

Volume total:
%
$$\begin{array}{rcl}
  \sum_{i=0,\ldots,n-1} 4\pi x_i f(x_i) (x_{i+1}-x_i)
    &\approx& Å_{x=x_0}^{x=x_n} 4\pi x f(x) \,dx       \\
          &=& Å_{x=2}^{x=4} 4\pi \sqrt{1-(x-3)^2} \,dx \\
  \end{array}
$$






%*

\end{document}

* (eepitch-lua51)
* (eepitch-kill)
* (eepitch-lua51)
sqrt = math.sqrt

-- Funções do problema 1:
f = function (x) return 1-(x+3)^2 end
g = function (x) return 2*sqrt(x+4)-1 end
h = function (x) return ((x+4)/2)^2-1 end
k = function (x) return ((4-x)/2)^2-1 end
p = function (x) return 2*sqrt(4-x)-1 end
print(f(-4), f(-3), f(-2))
print(g(-4), g(-3), g(0))
print(h(-4), h(-2), h(0))
print(k(0),  k(2),  k(4))
print(p(0),  p(3),  p(4))

-- Primitivas:
F = function (x) return x - 1/3 * (x+3)^3         end
G = function (x) return     4/3 * (x+4)^(3/2) - x end
H = function (x) return    1/12 * (x+4)^3     - x end
K = function (x) return  - 1/12 * (4-x)^3     - x end
P = function (x) return   - 4/3 * (4-x)^(3/2) - x end
pri = function (str, ...)
    local spl = split(str)
    print(str .. ": ", _G[spl[1]](spl[3]) - _G[spl[1]](spl[2]), ...)
  end
pri("F -4 -3", 2/3)
pri("F -3 -2", 2/3)
pri("G -4 -3", 4/3 - 1) 
pri("G -3 -0", 9 - 8/3) 
pri("H -4 -2", 2/3 - 2) 
pri("H -2 -0", 8/3) 
pri("K  0  2", 8/3)
pri("K  2  4", 2/3 - 2)
pri("P  0  3", 9 - 8/3)
pri("P  3  4", 4/3 - 1)


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