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%   ____      _                    _ _           
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% | |   / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ 
% | |__| (_| | |_) |  __/ (_| (_| | | | | | (_) |
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{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2016.1
\par VR - 3/ago/2016 - Eduardo Ochs
\par Links importantes:
\par \url{http://angg.twu.net/2016.1-C2.html} (página 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" "2016.1-C2-P2")
% (find-books "__analysis/__analysis.el" "stewart")
% (find-stewart7page 1376       "Table of integrals")
% (find-angg ".emacs.papers" "hernandez")

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

$$a) \quad \B(2.0 pts) \quad \intx {\frac {x^3} {x^2 - 4}}$$

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

$$c) \quad \B(2.0 pts) \quad \Intx {-2} {2} {|x^2-1|}$$


\bsk

2) \T(Total: 2.0 pts) Seja (*) esta EDO: $f'' + 2f' + 5f = 0$.

a) \B(1.0 pts) Encontre as soluções básicas reais de $(*)$.

b) \B(1.0 pts) Encontre uma solução de $(*)$ que obedeça $f(0)=0$ e $f'(0)=1$.

\bsk

3) \T(Total: 1.0 pts) Seja (**) esta EDO: $f'(x) = (2x-2)/e^{f(x)}$.

Encontre a solução geral de (**).

% a) \B(1.0 pts)



\bsk

4) \T(Total: 1.5 pts)

a) \B(0.5 pts) Teste a sua solução da 1a.

a) \B(0.5 pts) Teste a sua solução da 1b.

a) \B(0.5 pts) Teste a sua solução da 3.



% b) \B(1.0 pts) Encontre uma solução de (**) que obedeça $f(0)=10$.





\newpage



Método 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

Substituição:

\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

Fórmulas:

\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

Substituição inversa:

$\def\a{{h¹(α)}}
 \def\b{{h¹(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

Fórmulas:

\msk

$\def\a{{g¹(α)}}
 \def\b{{g¹(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=g¹(u)} \\
 = \intx {f(g(x))\frac{d\,g(x)}{dx}} & \subst{x=g¹(u)} \\
 \end{array}
$

\bsk

Substituição trigonométrica:

\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:

(não revisado, contém erros)

\bsk

% (find-es "ipython" "2016.1-C2-VS")

1a) $\frac {x^3} {x^2 - 4} = x + \frac{2}{x+2} + \frac{2}{x-2}$

$\intx {\frac {x^3} {x^2 - 4}} = \intx {x + \frac{2}{x+2} + \frac{2}{x-2}}
  = \frac{x^2}{2} + 2 \ln|x+2| + 2 \ln|x-2|$

\msk

1b) $\intx {\frac 1 {x\sqrt{x^2-1}}} = \arcsec x$

\msk

1c) $\Intx {-2} {2} {|x^2-1|}
    = \Intx {-2} {-1} {(x^2-1)} + \Intx {-1} {1} {(1-x^2)} + \Intx {1} {2} {(x^2-1)}
    $

    $= \frac{4}{3} + \frac{4}{3} + \frac{4}{3}
    = 4
    $


\bsk



2a) $f_1(x) = e^{-x}\cos 2x$, $f_2(x) = e^{-x}\sen 2x$.

2b) $f_3(x) = \frac12 e^{-x}\sen 2x$.

\bsk

3) $\frac{dy}{dx} = \frac {2x-2} {e^y}$

$e^{-y} dy = (2x-2) dx$

$\inty {e^{-y}} = \intx {2x-2} + C_1$

$- {e^{-y}} = x^2 - 2x + C_1$

${e^{-y}} = 2x - x^2 + C_2$

$-y = \ln(2x - x^2 + C_2)$

$y = -\ln(2x - x^2 + C_2)$







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