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\begin{document}

\centerline{\bf MAT 0146 - C\'alculo 1 para Economia} 

\centerline{\bf 3$^a$ Prova - 20 de junho de 2016}

\vskip1cm


\noindent {\bf Quest\ao\ 1} (1 pt) {\em Seja ${\displaystyle f(x)=\int_2^{x^3}\frac{1}{\sqrt{\ln t}}\,dt}$, $x>1$. Calcule} $f'(x)$. 

\noindent {\tt Solu\c{c}\~ao}\\ Usando o Teorema Fundamental do C\'alculo e a Regra da Cadeia, obtemos:
\[
f'(x)=\frac{1}{\sqrt{\ln x^3}}\cdot3x^2\]
para todo $x>1$.


\vskip1cm 

\noindent {\bf Quest\ao\ 2} (2 pts) {\em Calcule} ${\displaystyle \int_0^\pi x^2\cos x\,dx}$. 

\noindent {\tt Solu\c{c}\~ao}\\ Integrando por partes com $u=x^2$ e $dv=\cos x\,dx$, obtemos:
\[
\left.\int_0^\pi x^2\cos x\,dx=\left.x^2\sen x\right.\right|_0^\pi-\int_0^\pi2x\sen x\,dx=-2\int_0^\pi x\sen x\,dx.
\]
Integrando por partes com $u=x$ e $dv=\sen x\,dx$, obtemos:
\[
\left.\int_0^\pi x\sen x\,dx=-\left.x\cos x\right.\right|_0^\pi+\left.\int_0^\pi\cos x\,dx=\pi+\left.\sen x\right.\right|_0^\pi=\pi.
\]
Da\'i:
\[
\int_0^\pi x^2\cos x\,dx=-2\pi
\]

\vskip1cm

\noindent {\bf Quest\ao\ 3} (2 pts) {\em Mostre que} ${\displaystyle \int_1^\infty\frac{1}{(x^2+1)^{3/2}}\,dx<\frac{1}{2}}$.

%\noindent{\tt Dica}: Sai por compara\cao. Quem fizer calculando o valor exato, ganha 1 ponto de b\^onus.

\noindent {\tt Solu\c{c}\~ao por compara\c{c}\~ao} \\Para todo $x>1$, temos $x^2+1>x^2$, logo $(x^2+1)^{1/2}>(x^2)^{1/2}=x$, logo
$(x^2+1)^{3/2}>x^3>0$, logo
\[
0<\frac{1}{(x^2+1)^{3/2}}<\frac{1}{x^3},
\]
logo
\[
\left.\int_1^\infty\!\frac{1}{(x^2+1)^{3/2}}\,dx<\!\int_1^\infty\frac{1}{x^3}\,dx=\lim_{M\to\infty}\int_1^Mx^{-3}\,dx=\lim_{M\to\infty}\frac{x^{-2}}{-2}\right|_1^M\!\!\!=
-\lim_{M\to\infty}\!\left(\frac{1}{2M^2}-\frac{1}{2}\right)=\frac{1}{2}.
\]

\newpage
\noindent {\tt C\'alculo do valor exato}
\\Substituindo na integral indefinida
\[
x=\tan\theta,\ \ dx=\sec^2\theta\,d\theta,\ \
(1+x^2)^{3/2}=(\sqrt{1+\tan^2\theta})^3=\sec^3\theta,
\]
segue que:
\[
\int\frac{1}{(x^2+1)^{3/2}}\,dx=\int\frac{\sec^2\theta}{\sec^3\theta}\,d\theta=\int\cos\theta\,d\theta=\sen\theta+C.
\]
Para voltar para a vari\'avel $x$, observamos que s\'o estamos interessados em valores positivos de $x$ (pois a integral que queremos calcular \'e para
$x\geq 1$). Podemos portanto supor que $0<\theta<\frac{\pi}{2}$, intervalo em que todas as fun\coes\ trigonom\'etricas que aparecem aqui s\ao\ positivas.
Da\'i, obtemos
\[
\sen\theta\!=\!\sqrt{1-\cos^2\theta}=\sqrt{1-\frac{1}{\sec^2\theta}}=\sqrt{1-\frac{1}{1+\tan^2\theta}}=\sqrt{1-\frac{1}{1+x^2}}=
\sqrt{\frac{x^2}{x^2+1}}=\frac{x}{\sqrt{1+x^2}}
\]
e, da\'i,
\[
\left.\int_1^M\!\frac{1}{(x^2+1)^{3/2}}\,dx
=\frac{x}{\sqrt{x^2+1}}\right|_1^M=\frac{M}{\sqrt{M^2+1}}-\frac{1}{\sqrt{2}}
\]
e, da\'i,
\[
\int_1^\infty\!\frac{1}{(x^2+1)^{3/2}}\,dx=\lim_{M\to\infty}\int_1^M\!\frac{1}{(x^2+1)^{3/2}}\,dx=%\lim_{M\to\infty}\left(\frac{M}{\sqrt{M^2+1}}-\frac{1}{\sqrt{2}}\right)=
\lim_{M\to\infty}\left(\frac{1}{\sqrt{1+\frac{1}{M^2}}}-\frac{1}{\sqrt{2}}\right)=1-\frac{1}{\sqrt{2}}<\frac{3}{10}<\frac{1}{2}.
\]
\vskip1cm

\noindent {\bf Quest\ao\ 4} (3 pts) (a) {\em Calcule} ${\displaystyle \int_0^{\frac{\sqrt{3}}{2}}\sqrt{1-x^2}\,dx}$. (b) {\em Calcule,
ilustrando seu racioc\'inio com uma figura, a \'area da regi\ao\ do plano que consiste dos pontos simultaneamente
interiores aos c\'irculos de raio 1 centrados nos pontos $(0,0)$ e $(0,1)$.}

\noindent {\tt Solu\c{c}\~ao}\\ (a) Fazendo $x=\sen\theta$, $dx=\cos\theta\,d\theta$, 
notando que
\[
0\leq\theta\leq\frac{\pi}{3}\ \ \Longrightarrow\ \  0\leq \sen\theta\leq\frac{\sqrt{3}}{2}\ \mbox{e}\ 
\sqrt{1-x^2}=\sqrt{1-\sen\!^2\,\theta}=\cos\theta,
\]
vem:
\[
\int_0^{\frac{\sqrt{3}}{2}}\sqrt{1-x^2}\,dx=\int_0^{\frac{\pi}{3}}\cos^2\theta\,d\theta=
\left.\int_0^{\frac{\pi}{3}}\frac{1+\cos(2\theta)}{2}\,d\theta=\frac{1}{2}\left[\theta+\frac{\sen(2\theta)}{2}\right]\right|_{0}^{\frac{\pi}{3}}=
\frac{\pi}{6}+\frac{\sqrt{3}}{8}.
\]

\noindent
(b) Um quarto da \'area $A$ pedida \'e igual ao valor da integral menos a \'area do ret\^angulo de lados $\frac{1}{2}$ e $\frac{\sqrt{3}}{2}$ (isto
pode ser entendido com a ajuda de uma figura). Logo
\[
A=4\cdot\left(\frac{\pi}{6}+\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{4}\right)=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}.
\]


\vskip1cm

\noindent {\bf Quest\ao\ 5} (2 pts) {\em Calcule} ${\displaystyle \lim_{n\to\infty}\frac{1+2^{10}+3^{10}+\cdots+n^{10}}{n^{11}}}$.

%\noindent {\tt Sugest\ao}: Use a defini\cao\ de integral definida e o teorema fundamental do c\'alculo para uma fun\cao\ e um intervalo convenientemente escolhidos.

\noindent {\tt Solu\c{c}\~ao}: Seja $f(x)=x^{10}$. Pelo teorema fundamental do c\'alculo,
\[
\left.\int_0^1f(x)\,dx=\frac{x^{11}}{11}\right|_0^1=\frac{1}{11}.
\]
Por outro lado, essa integral \'e o limite, quando $n\to\infty$, das somas de Riemann
\[
\sum_{j=1}^{n} f(x_j)\Delta x,\ \  x_j=\frac{j}{n},\ j=1,\cdots,n,\ \Delta x=\frac{1}{n}.
\]
Logo,
\[
 \lim_{n\to\infty}\frac{1+2^{10}+3^{10}+\cdots+n^{10}}{n^{11}}=\frac{1}{n}\sum_{j=1}^n\left(\frac{j}{x}\right)^{10}=\lim_{n\to\infty}\sum_{j=1}^{n} f(x_j)\Delta x=\int_0^1f(x)\,dx=\frac{1}{11}.
\]

\end{document}
