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{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2017.2
\par P1 - 13/nov/2017 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrônicos.

}

\bsk
\bsk

\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)}}
% Usage:
% 1) \T(Total: 2.34 pts) Foo
% a) \B(0.45 pts) Bar


% \bsk
% \bsk

% (c2q)

1) \T(Total: 2.0 pts) Calcule $$\intx {(\cos x)^6}.$$

\bsk

2) \T(Total: 3.0 pts) Calcule $$\intx {\sqrt{1-x^2}}.$$

\bsk

3) \T(Total: 2.0 pts) Calcule $$\intx {\frac{x^3}{x^2 + 3x - 10}}.$$

\bsk

4) \T(Total: 2.0 pts) Sejam $F(a,b) = \Intx{a}{b}{|1-x^4|}$ e $G(b) = F(0,b)$.

a) \B(0.5 pts) Calcule $F(0,2)$.

b) \B(1.5 pts) Dê uma definição por casos para $G(b)$.

\bsk

5) \T(Total: 1.5 pts) Sejam $(SD)$ e $(SI)$ as fórmulas para
integração por substituição na integral definida e na integral
indefinida, e $(WI)$ uma integração por substituição que demonstramos
no curso usando ``bloquinhos de substituições'':
%
$$\begin{array}{ccccc}
  {(SD)} && \left( \Intx {a}{b}{f(g(x))g'(x)} = \Intu {g(a)}{g(b)}{f(u)} \right) \\
  {(SI)} && \left( \intx       {f(g(x))g'(x)} = \intu             {f(u)} \right) \\
  {(WI)} && \left( \intth {(\senθ)^3(\cosθ)^2\cosθ} = \ints {s^3(1-s^2)} \right)
         && \bsm{s=\senθ \\ \senθ→s \\ \frac{ds}{dθ}=\cosθ \\ \cosθ\,dθ→ds \\ (\cosθ)^2→(1-s^2)} \\
  \end{array}
$$

a) \B(0.1 pts) O que é $(SD)\bsm{f(u):=\sen u \\ g(x):=x^2 \\ a:=3 \\ b:=4}$?

b) \B(0.4 pts) Adicione limites de integração na fórmula $(WI)$ para
obter uma fórmula ``$(WD)$'' que calcula (ou melhor, ``simplifica'') a
integral definida $(WDL)$ abaixo; note que falta pôr os limites de
integração no ``lado direito'' (``$(WDR)$'').
%
$$\begin{array}{ccc}
  (WDL) && \Intth{a}{b} {(\senθ)^3(\cosθ)^2\cosθ}.
  \end{array}
$$

c) \B(1.0 pts) Encontre uma substituição simultânea que aplicada a
$(SD)$ demonstra $(WD)$.



\newpage

{\bf Mini-gabarito (não revisado):}

\bsk

1) $(\cosθ)^6 = (\frac{E+E¹}{2})^6 = \frac{1}{64} (E^6 + 6 E^4 + 15
E^2 + 20 + 15 E^{-2} + 6 E^{-4} + E^{-6})$

$= \frac{1}{32} \frac{1}{2} (E^6 + E^{-6} + 6 (E^4 + E^{-4}) + 15 (E^2
+ E^{-2}) + 20)$

$= \frac{1}{32} (\cos6θ + 6\cos4θ + 15\cos2θ + 10)$

$\intx{(\cos x)^6} = \frac{1}{32} \intx {(\cos6x + 6\cos4x + 15\cos2x + 10)}$

$= \frac{1}{32} (\frac{\sen6x}{6} + \frac{6\sen4x}{4} + \frac{15\sen2x}{2} + 10x)$

\bsk

2) $\intx {\sqrt{1-x^2}} = \frac{1}{2}(\arcsen(x) + x\sqrt{1-x^2})$

\bsk

3) $\intx {\frac{x^3}{x^2 + 3x - 10}} =
    \frac{x^2}{2} - 3x + \frac{8}{7} \ln|x-2| + \frac{125}{7}\ln|x+5|$

\bsk

4) Sejam $f(x)=|1-x^4|$, $h(x)=1-x^4$. Então $f(x)=h(x)$ em $[-1,1]$,
$f(x)=-h(x)$ em $(-∞,-1]∪[1,∞)$. Seja $H(x)=∫h(x)\,dx =
    x-\frac{x^5}{5}$; $H(x)$ é uma primitiva de $f(x)$ em $[-1,1]$, e
    no resto da reta podemos usar $-H(x)$ como primitiva de $f(x)$.

4a) $F(0,2) = \Intx{0}{1}{f(x)} + \Intx{1}{2}{f(x)} =
\difx{0}{1}{H(x)} + \difx{1}{2}{(-H(x))}$

$= (H(1)-H(0))+(-H(2)+H(1)) = -H(0) + 2H(1) -H(2)$

$= 2(1 - \frac15) - (2-\frac{32}5) = \frac{33}5$

4b)
$G(b)= \begin{cases}
    \difx{0}{-1}{H(x)} + \difx{-1}{b}{(-H(x))} & \text{se $b≤-1$} \\
    \difx{0}{b}{H(x)}                          & \text{se $1≤b≤-1$} \\
    \difx{0}{1}{H(x)} + \difx{1}{b}{(-H(x))}   & \text{se $1≤b$} \\
    \end{cases}
$

$= \begin{cases}
    (H(-1)-H(0)) + (-H(b)+H(-1)) & \text{se $b≤-1$} \\
    H(b)-H(0)                     & \text{se $1≤b≤-1$} \\
    (H(1)-H(0)) + (-H(b)+H(1))    & \text{se $1≤b$} \\
    \end{cases}
$

$= \begin{cases}
    2H(-1)-H(b) & \text{se $b≤-1$} \\
    H(b)         & \text{se $1≤b≤-1$} \\
    2H(1)-H(b)   & \text{se $1≤b$} \\
    \end{cases}
$
\quad
$= \begin{cases}
    -\frac85-b+\frac{b^5}{5} & \text{se $b≤-1$} \\
    b-\frac{b^5}{5}          & \text{se $1≤b≤-1$} \\
    \frac85-b+\frac{b^5}{5}  & \text{se $1≤b$} \\
    \end{cases}
$

\bsk

5a) $\left( \Intx {a}{b}{f(g(x))g'(x)} = \Intu {g(a)}{g(b)}{f(u)} \right)
     \bsm{f(u):=\sen u \\ g(x):=x^2 \\ a:=3 \\ b:=4}$

$= \left( \Intx {3}{4}{\sen(x^2)·2x} = \Intu {3^2}{4^2}{\sen(u)} \right)$

5b) $\Intth{a}{b} {(\senθ)^3(\cosθ)^2\cosθ} = \Ints{\sen a}{\sen b} {s^3(1-s^2)}$

\msk

5c) $\left( \Intth {a}{b}{f(g(θ))g'(θ)} = \Ints {g(a)}{g(b)}{f(s)} \right)
     \bsm{f(s):=s^3(1-s^2) \\ g(θ):=\senθ \\}$

$= \left( \Intth{a}{b} {(\senθ)^3(1-\sen^2θ)\cosθ} = \Ints{\sen a}{\sen b} {s^3(1-s^2)} \right)$





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