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% http://angg.twu.net/LATEX/2016-1-C2-VR.pdf
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\begin{document}

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%   ____      _                    _ _           
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% | |   / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ 
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{\setlength{\parindent}{0em}
\footnotesize
\par CÃlculo 2
\par PURO-UFF - 2016.1
\par VR - 1$°$/ago/2016 - Eduardo Ochs
\par Links importantes:
\par \url{http://angg.twu.net/2016.1-C2.html} (pÃgina do curso)
\par \url{http://angg.twu.net/2016.1-C2/2016.1-C2.pdf} (quadros)
\par \url{http://angg.twu.net/LATEX/2016-1-C2-VR.pdf} (esta prova)
\par {\tt eduardoochs@gmail.com} (meu e-mail)
}

\setlength{\parindent}{0em}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B       (#1 pts){{\bf(#1 pts)}}

\bsk
\bsk



% (find-es "ipython" "2015.2-C2-P1")

1) \T(Total: 6.0 pts) Calcule:

$$a) \quad \B(2.0 pts) \quad \Intx 2 3 {x^4 \ln x}$$

$$b) \quad \B(2.0 pts) \quad \intx {x^3 \sqrt{1-x^2}}$$

$$c) \quad \B(2.0 pts) \quad \intx {\frac {x^2} {x^2 - x - 6}}$$

\bsk

2) \T(Total: 2.0 pts) Seja (*) esta EDO: $f'' - f' - 6f = 0$.

a) \B(1.0 pts) Encontre as soluÃÃes bÃsicas de $(*)$.

b) \B(1.0 pts) Encontre uma soluÃÃo de $(*)$ que obedeÃa $f(0)=0$ e $f(1)=1$.

\bsk

3) \T(Total: 2.0 pts) Seja (**) esta EDO: $f'(x) = -2(x+3) / f(x)$.

a) \B(1.0 pts) Encontre a soluÃÃo geral de (**).

b) \B(1.0 pts) Encontre uma soluÃÃo de (**) que obedeÃa $f(0)=10$.






\newpage



MÃtodo de Heaviside:

Se $f(x) = \frac{\aa}{x-a} + \frac{\bb}{x-b} + \frac{\cc}{x-c} = \frac{p(x)}{(x-a)(x-b)(x-c)}$,

entÃo $\lim_{x \to a} f(x)(x-a) = \aa = \frac{p(a)}{(a-b)(a-c)}$.

\bsk

SubstituiÃÃo:

\msk

$\begin{array}{l}
 \difx a b {g(h(x))} = \Intx a b {g'(h(x))\frac{d\,h(x)}{dx}} \\
 \phantom{mmm}|\,| \\
 \difu {h(a)} {h(b)} {g(u))} = \Intu {h(a)} {h(b)} {g'(u)} \\
 \end{array}
$

\msk

FÃrmulas:

\msk

$\begin{array}{l}
 \Intx   a b           {f(g(x))\frac{d\,g(x)}{dx}} \\
 = \Intx a b           {f(u)\frac{du}{dx}} \\
 = \Intu {g(a)} {g(b)} {f(u)}
 \end{array}
 \qquad
 \begin{array}{ll}
 \intx   {f(g(x))\frac{d\,g(x)}{dx}} \\
 = \intx {f(u)\frac{du}{dx}}         & \subst{u=g(x)} \\
 = \intu {f(u)}                      & \subst{u=g(x)}
 \end{array}
$

\bsk

SubstituiÃÃo inversa:

$\def\a{{hÂ(α)}}
 \def\b{{hÂ(β)}}
 \begin{array}{l}
 \Difx \a \b {g(h(x))} = \Intx \a \b {g'(h(x))\frac{d\,h(x)}{dx}} \\
 \phantom{mmm}|\,| \\
 \Difu {h(\a)} {h(\b)} {g(u))} = \Intu {h(\a)} {h(\b)} {g'(u)} \\
 \phantom{mmm}|\,| \\
 \Difu α β {g(u))} = \Intu α β {g'(u)} \\
 \end{array}
$

\msk

FÃrmulas:

\msk

$\def\a{{gÂ(α)}}
 \def\b{{gÂ(β)}}
 \begin{array}{l}
 \Intu   α β  {f(u)} \\
 = \Intx \a \b {f(u)\frac{du}{dx}} \\
 = \Intx \a \b {f(g(x))\frac{d\,g(x)}{dx}}
 \end{array}
 \qquad
 \begin{array}{ll}
 \intu   {f(u)} \\
 = \intx {f(u)\frac{du}{dx}}         & \subst{u=g(x)\\x=gÂ(u)} \\
 = \intx {f(g(x))\frac{d\,g(x)}{dx}} & \subst{x=gÂ(u)} \\
 \end{array}
$

\bsk

SubstituiÃÃo trigonomÃtrica:

\msk

$
 \begin{array}{ll}
 \Ints a b {F(s, \sqrt{1-s^2})} \\
 = \Intth {\arcsen a} {\arcsen b} {F(\senθ, \sqrt{1-\sen^2θ}) \frac{d\senθ}{dθ}} \\
 = \Intth {\arcsen a} {\arcsen b} {F(\senθ, \cosθ) \cosθ}                        \\
 \end{array}
 \qquad
 \begin{array}{ll}
 \ints    {F(s, \sqrt{1-s^2})} \\
 = \intth {F(s, \sqrt{1-s^2}) \frac{ds}{dθ}} & \subst{s=\senθ \\ θ=\arcsenθ} \\
 = \intth {F(s, c) c}                        & \subst{s=\senθ \\ c=\cosθ \\ θ=\arcsenθ} \\
 \end{array}
$

\msk

$
 \begin{array}{ll}
 \Intz a b {F(z, \sqrt{z^2-1})} \\
 = \Intth {\arcsec a} {\arcsec b} {F(\secθ, \sqrt{\sec^2θ-1}) \frac{d\secθ}{dθ}} \\
 = \Intth {\arcsec a} {\arcsec b} {F(\secθ, \tanθ) \secθ \tanθ}                   \\
 \end{array}
 \qquad
 \begin{array}{ll}
 \intz    {F(z, \sqrt{z^2-1})} \\
 = \intth {F(z, \sqrt{z^2-1}) \frac{dz}{dθ}} & \subst{z=\secθ \\ θ=\arcsec z} \\
 = \intth {F(z, t) zt}                       & \subst{z=\secθ \\ θ=\arcsec z \\ t=\tanθ} \\
 \end{array}
$

\msk

$
 \begin{array}{ll}
 \Intt a b {F(t, \sqrt{1+t^2})} \\
 = \Intth {\arctan a} {\arctan b} {F(\tanθ, \sqrt{1+\tan^2θ}) \frac{d\tanθ}{dθ}} \\
 = \Intth {\arctan a} {\arctan b} {F(\tanθ, \secθ) \sec^2θ}                      \\
 \end{array}
 \qquad
 \begin{array}{ll}
 \intt    {F(t, \sqrt{1+t^2})} \\
 = \intth {F(t, \sqrt{1+t^2}) \frac{dt}{dθ}} & \subst{t=\tanθ \\ θ=\arctan t} \\
 = \intth {F(t, z) z^2}                      & \subst{t=\tanθ \\ θ=\arctan t \\ z=\secθ} \\
 \end{array}
$





\newpage

Mini-gabarito:

\bsk

1a) $\intx{x^4 \ln x} = \frac{1}{25} x^5 (5 \ln x - 1)$

$\Intx 2 3 {x^4 \ln x}
  = \difx 2 3 {\frac{1}{25} x^5 (5 \ln x - 1)}
  = -\frac{211}{25} - \frac{32}{5} \ln 2 + \frac{243}{5} \ln 3
$

\msk

1b) $\intx {x^3 \sqrt{1-x^2}} = \frac{1}{15} (3x^4 - x^2 - 2) \sqrt{1-x^2}$

% Expr2 = (3 * x**4 - x**2 - 2) / 15 * sqrt(1 - x**2)

\msk

1c) $\frac {x^2} {x^2 - x - 6} = 1 - \frac 4 {5(x+2)} + \frac 9 {5(x-3)}$

$\intx {\frac {x^2} {x^2 - x - 6}} = x - \frac45 \ln |x+2| + \frac95 \ln |x-3|$

\bsk

2) $f'' - f' - 6f = (D^2 - D - 6)f = (D-3)(D+2)f$

2a) $f_1(x) = e^{3x}$, $f_2(x) = e^{-2x}$ 

2b) $f_3(x) = ae^{3x} + be^{-2x}$, onde $a = \frac {e^2} {e^5-1}$, $b = \frac {e^2} {1-e^5}$.

\bsk

3a) $f(x) = \pm \sqrt{2} \sqrt{C - (x+3)^2}$

3b) $f(x) =     \sqrt{2} \sqrt{59 - (x+3)^2}$



\end{document}

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