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Warning: this is an htmlized version!
The original is here, and the conversion rules are here. |
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% http://angg.twu.net/LATEX/2016-1-C2-P1.pdf
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%
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\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B (#1 pts){{\bf(#1 pts)}}
% (find-LATEX "2015-2-C2-P1.tex")
\def\ddx{\frac{d}{dx}}
\def\ddth{\frac{d}{d\theta}}
\def\arcsen{\operatorname{arcsen}}
\def\sen{\operatorname{sen}}
\def\sec{\operatorname{sec}}
\def\ln{\operatorname{ln}}
\def\subst#1{\left[\sm{#1}\right]}
\def\prims #1{∫#1\,ds}
\def\primth#1{∫#1\,dθ}
\def\difs #1#2#3{\left. #3 \right|_{s=#1}^{s=#2}}
\def\difu #1#2#3{\left. #3 \right|_{u=#1}^{u=#2}}
\def\difx #1#2#3{\left. #3 \right|_{x=#1}^{x=#2}}
\def\difth#1#2#3{\left. #3 \right|_{θ=#1}^{θ=#2}}
% Indefinite integrals
\def\ints #1{∫#1\,ds}
\def\intc #1{∫#1\,dc}
\def\intt #1{∫#1\,dt}
\def\intu #1{∫#1\,du}
\def\intx #1{∫#1\,dx}
\def\intz #1{∫#1\,dz}
\def\intth#1{∫#1\,dθ}
% Definite integrals
\def\Ints #1#2#3{∫_{s=#1}^{s=#2}#3\,ds}
\def\Intt #1#2#3{∫_{t=#1}^{t=#2}#3\,dt}
\def\Intu #1#2#3{∫_{u=#1}^{u=#2}#3\,du}
\def\Intx #1#2#3{∫_{x=#1}^{x=#2}#3\,dx}
\def\Intz #1#2#3{∫_{z=#1}^{z=#2}#3\,dz}
\def\Intth#1#2#3{∫_{θ=#1}^{θ=#2}#3\,dθ}
% Difference
\def\Difs #1#2#3{\left. #3 \right|_{s=#1}^{s=#2}}
\def\Difu #1#2#3{\left. #3 \right|_{u=#1}^{u=#2}}
\def\Difx #1#2#3{\left. #3 \right|_{x=#1}^{x=#2}}
\def\Difth#1#2#3{\left. #3 \right|_{θ=#1}^{θ=#2}}
% ____ _ _ _
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% | |__| (_| | |_) | __/ (_| (_| | | | | | (_) |
% \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/
%
{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2016.1
\par P1 - 4/jul/2016 - Eduardo Ochs
% \par Versão: 14/mar/2016
\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-material.pdf}
\par {\tt eduardoochs@gmail.com} (meu e-mail)
}
\bsk
\bsk
% (c2q 7 "integral de Riemann")
% (c2q 9 "integral de Riemann sem partição especificada")
% (c2q 11 "TFC")
% (c2q 13 "TFC 2")
% (c2q 15 "Substituição")
% (c2q 16 "Diferenciais")
% (c2q 22 "G(x,y) = x^2 + y^2")
% (c2q 24 "Derivada da função inversa")
% (c2q 27 "Integrando funções racionais")
% (c2q 30 "Método de Heaviside")
% (c2q 32 "Integrando funções racionais impróprias")
% (c2q 34 "Integração por partes")
% (c2q 35 "Truque do `onde'")
% (c2q 36 "Tabelas de integrais")
% (c2q 37 "Substituição trigonométrica")
% (c2q 43 "Série de Taylor")
% (c2q 45 "Plano complexo")
% (c2q 48 "Grande truque: E")
% (c2q 50 "Substituição trigonométrica")
% (c2q 51 "EDOs")
% (c2q 53 "EDOs: D")
% (c2q 55 "EDOs: sen e cos vezes exp")
% (find-es "ipython" "2015.2-C2-P1")
1) \T(Total: 1.0 pts) Calcule $\intx {\tan x}$.
\bsk
2) \T(Total: 1.0 pts) Calcule $\Intx {-1} {2} {x^{-4}}$.
\bsk
3) \T(Total: 1.5 pts) Calcule $\intx {\cos^4 x}$.
\bsk
4) \T(Total: 1.5 pts) Calcule $\intx {\frac {x^2} {x^2+x-2}}$.
\bsk
5) \T(Total: 1.5 pts) Calcule $\intx {x \, e^x \cos x}$.
\bsk
6) \T(Total: 2.0 pts)
6a) \B(1.5 pts) Calcule $\Intx 0 1 {\sqrt{4-x^2}}$.
6b) \B(0.1 pts) Represente $\Intx 0 1 {\sqrt{4-x^2}}$ graficamente.
6c) \B(0.4 pts) Mostre como calcular $\Intx 0 1 {\sqrt{4-x^2}}$ pelo gráfico.
\bsk
7) \T(Total: 2.5 pts) Calcule $\Intx {-1} {2} {|e^x-1|}$.
\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
% ____ _ _ _
% / ___| __ _| |__ __ _ _ __(_) |_ ___
% | | _ / _` | '_ \ / _` | '__| | __/ _ \
% | |_| | (_| | |_) | (_| | | | | || (_) |
% \____|\__,_|_.__/ \__,_|_| |_|\__\___/
%
Gabarito (não revisado, contém erros de vários tipos!):
\bsk
% _
% / |
% | |
% | |
% |_|
%
1)
%
$\begin{array}[t]{rcl}
\intth {\tanθ} &=& \intth {\frac{s}{c}} \\
&=& \intth {\frac{1}{c} s} \\
&=& - \intc {\frac{1}{c}} \\
&=& - \ln |c| \\
&=& - \ln |\cosθ| \\
\end{array}
$
% ____
% |___ \
% __) |
% / __/
% |_____|
%
{
\def\lima{\lim\limits_{a→0^-}}
\def\limb{\lim\limits_{b→0^+}}
\def\lima{\lim_{a→0^-}}
\def\limb{\lim_{b→0^+}}
\def\limap#1{\lim_{a→0^-} \left(#1\right)}
\def\limbp#1{\lim_{b→0^+} \left(#1\right)}
\def\p#1{\left(#1\right)}
\bsk
2)
%
$\begin{array}[t]{rcl}
\intx {x^{-4}} &=& \frac{x^{-3}}{-3} \\
\Intx{-1}{2} {x^{-4}} &=& \Intx{-1}{0} {x^{-4}} + \Intx{0}{2} {x^{-4}} \\
&=& \lima \Intx{-1}{a} {x^{-4}} +
\limb \Intx{b}{2} {x^{-4}} \\
&=& \limap {\difx{-1}{a} {\frac{x^{-3}}{-3}}} +
\limbp {\difx{b}{2} {\frac{x^{-3}}{-3}}} \\
&=& \limap {\frac{a^{-3}}{-3} - \frac{(-1)^{-3}}{-3}} +
\limbp {\frac{2^{-3}}{-3} - \frac{b^{-3}}{-3}} \\
&=& \p {\frac{-∞}{-3} - \frac{-1}{-3}} +
\p {\frac{1/8}{-3} - \frac{+∞}{-3}} \\
\end{array}
$
}
\bsk
% _____
% |___ /
% |_ \
% ___) |
% |____/
%
3) Seja $E = e^{iθ}$. Então
$\begin{array}[t]{rcl}
\cosθ &=& \frac {E + E¹} {2}, \\
(\cosθ)^2 &=& \frac {E^2 + 2 + E^{-2}} {4}, \\
(\cosθ)^4 &=& \frac {E^4 + 4E^2 + 6 + 4E^{-2} + E^{-4}} {16} \\
&=& \frac 1 8 \frac{E^4 + E^{-4}}{2} + \frac 1 2 \frac{E^2 + E^{-2}}{2} + \frac 3 8 \\
&=& \frac 1 8 \cos 4θ + \frac 1 2 \cos 2θ + \frac 3 8 \\
\intth {(\cosθ)^4} &=& \intth {\frac 1 8 \cos 4θ + \frac 1 2 \cos 2θ + \frac 3 8} \\
&=& \frac 1 8 \frac{\sen 4θ}{4} + \frac 1 2 \frac{\sen 2θ}{2} + \frac 3 8 θ \\
&=& \frac{\sen 4θ}{32} + \frac{\sen 2θ}{4} + \frac 3 8 θ \\
\end{array}
$
\bsk
% _ _
% | || |
% | || |_
% |__ _|
% |_|
%
4) $\frac {x^2} {x^2 + x - 2} =
\frac {(x^2 + x - 2) - x + 2} {x^2 + x - 2} =
1 + \frac {- x + 2} {x^2 + x - 2} =
1 + \frac {- x + 2} {(x+2)(x-1)}
$
Se $\frac {- x + 2} {(x+2)(x-1)} = \frac A {x+2} + \frac B {x-1}$ então
$\lim_{x→-2} \frac {-x+2} {x-1} = A = \frac 4 {-3}$ e
$\lim_{x→1} \frac {-x+2} {x+2} = B = \frac 1 3$.
Conferindo:
$\begin{array}[t]{rcl}
\frac A {x+2} + \frac B {x-1} &=& \frac {-4/3} {x+2} + \frac {1/3} {x-1} \\
&=& \frac {- \frac 4 3 (x-1) + \frac 1 3 (x+2)} {(x+2)(x-1)} \\
&=& \frac {-x+2} {(x+2)(x-1)} \\
\end{array}
$
Daí:
$\begin{array}[t]{rcl}
\intx {\frac {x^2} {x^2 + x - 2}} &=& \intx {1 + {-4/3} {x+2} + \frac {1/3} {x-1}} \\
&=& x - \frac 4 3 \ln |x+2| + \frac 1 3 \ln |x-1| \\
\end{array}
$
\newpage
% ____
% | ___|
% |___ \
% ___) |
% |____/
%
\def\uf #1{\underbrace {#1}_{f}}
\def\uff#1{\underbrace {#1}_{f'}}
\def\ug #1{\underbrace {#1}_{g}}
\def\ugg#1{\underbrace {#1}_{g'}}
5) Sejam $c = \cos x$, $s = \sen x$. Então
$\intx {\uf{e^x} \ugg{c}} = \uf{e^x} \ug{s} - \intx {\uff{e^x} \ug{s}}$
$\intx {\uf{e^x} \ugg{s}} = \uf{e^x} \ug{(-c)} - \intx {\uff{e^x} \ug{(-c)}} = -e^x c + \intx {e^x c}$
\msk
$\intx {e^x c} = e^x s - \intx {e^x s} = e^x s - (-e^x c + \intx {e^x c})
= e^x s + e^x c - \intx {e^x c}
$
$2 \intx {e^x c} = e^x s + e^x c$
$\intx {e^x c} = (e^x s + e^x c)/2$
\msk
$\intx {e^x s} = -e^x c + \intx {e^x c} = -e^x c + (e^x s + e^x c)/2 = (e^x s - e^x c)/2$
\msk
$\begin{array}{rcl}
\intx {\uf{x} \ugg{e^x c}}
&=& \uf{x} \ug{(e^x s + e^x c)/2} - \intx {\uff{1} \ug{(e^x s + e^x c)/2}} \\
&=& \frac12 (x e^x s + x e^x c) - \frac12 \intx {e^x s} - \frac12 \intx{e^x c} \\
&=& \frac12 (x e^x s + x e^x c) - \frac14 (e^x s - e^x c) - \frac14 (e^x s + e^x c) \\
&=& \frac12 (x e^x s + x e^x c) - \frac12 e^x s \\
&=& \frac12 (x e^x s + x e^x c - e^x s) \\
\end{array}
$
\bsk
% __
% / /_
% | '_ \
% | (_) |
% \___/
%
6) $\begin{array}[t]{rclcl}
\sqrt{4 - x^2} &=& \sqrt{4 - 4(x/2)^2} \\
&=& \sqrt{4(1 - (x/2)^2)} \\
&=& 2 \sqrt{1 - (x/2)^2} \\
\intx {\sqrt{4 - x^2}} &=& \intx {2 \sqrt{1 - (x/2)^2}} && \bsm{s = x/2 \\ x = 2s \\ dx = 2ds} \\
&=& \ints {2 \sqrt{1 - s^2} · 2} \\
&=& 4 \ints {\sqrt{1 - s^2}} &&
\bsm{s = \senθ \\ θ = \arcsen s \\ ds = \cosθ dθ} \\
&=& 4 \intth {\sqrt{1 - (\senθ)^2} \cosθ} \\
&=& 4 \intth {(\cosθ)^2} \\
&=& 4 \intth {\frac {1 + \cos 2θ} 2 } \\
&=& 4 ({\frac θ2 + \frac {\sen 2θ} 4 }) \\
&=& 2θ + \sen 2θ \\
\end{array}
$
\bsk
% _____
% |___ |
% / /
% / /
% /_/
%
7) $\begin{array}[t]{rclcl}
\Intx{-1}{2} {|e^x-1|} &=& \Intx{-1}{0} {|e^x-1|} + \Intx{0}{2} {|e^x-1|} \\
&=& \Intx{-1}{0} {1-e^x} + \Intx{0}{2} {e^x-1} \\
&=& \difx{-1}{0} {(x-e^x)} + \difx{0}{2} {(e^x-x)} \\
&=& (0-e^0) - (-1-e^{-1}) + (e^2-2) - (e^0-0) \\
&=& -1 + 1 + e^{-1} + e^2 - 2 - 1 \\
&=& -3 + e^{-1} + e^2 \\
\end{array}
$
\end{document}
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