Vaiporra

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\title{Cálculo II - Exercícios}
\author{Grupo 5}
\date{}
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\textbf{Página 255 ex 20}

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 \begin{equation}
\int x^2\cdotln{x}\,dx \\ \bigg\{\begin{matrix}u = ln{x} & du = \frac{1}{x}\\ dv = x^2 & v = \frac{x^3}{3}\end{matrix}
\\ \begin{matrix} \\ \Rightarrow I = ln{x} \cdot\frac{x^3}{3} - \int \frac{x^3}{3}\cdot \frac{1}{x}\, dx \Rightarrow I = ln{x}\cdot \frac{x^3}{3} - \int \frac{x^2}{3}\,dx \\ \Rightarrow I = ln{x}\cdot \frac{x^3}{3} - \frac{1}{3}\int x^2\, dx \\ \Rightarrow I = ln{x}\cdot \frac{x^3}{3} - \frac{1}{3}\cdot \frac{x^3}{3} \\ \Rightarrow I = \frac {ln{x}\cdot x^3}{3} - \frac{x^3}{9} + c \end{matrix}
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\textbf{Página 278 ex 16}
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Seja y = tanx , x = 0 e y = 1. Obter área.
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 \begin{equation}
\bigg\{\begin{matrix} \tan{x}=1\Rightarrow x= \frac{\Pi}{4}, & \mbox{intersecção com y = 1} \\ \tan{x}=0\Rightarrow x = 0, & \mbox{intersecção com x = 0} \end{matrix}
 \\ \begin{matrix} A = \int_{o}^{\frac{\Pi}{4}}1\,dx - \int_{o}^{\frac{\Pi}{4}}\tan{x}\,dx
\Rightarrow A = \frac{\Pi}{4} - (- ln{|\cos{\frac{\Pi}{4}}|} - (- ln{|\cos{0}}|})) \\ \Rightarrow A = \frac{\Pi}{4} - (- ln{\frac{\sqrt{2}}{2}} + ln{1}) \\ \Rightarrow  A = \frac{\Pi}{4} + ln{\frac{\sqrt{2}}{2}} \\ \Rightarrow A = \frac{\Pi}{4} + ln{{\sqrt{2}} - ln{2} \\ \Rightarrow A = \frac{\Pi}{4} - \frac{1}{2}\cdot ln{2}\end{matrix} \right
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\textbf{Página 291 ex 12a)}
\begin{equation}
	\int_{-\infty}^{0}e^x\,dx = \lim_{x\rightarrow -\infty}\int_{n}^{0}e^x\,dx = \lim_{x\rightarrow -\infty} [e^0 - e^n] = \lim_{x\rightarrow -\infty}[1-e^n]= 1 - 0 = 1
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Sendo assim, convergente.
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