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Resolução do simulado - Integrais impróprias

1. a. Resolvendo a integral indefinida,

$$\begin{align*} \int x \cdot e^{-x^2} \ dx =&\left[ \begin{array}{lcl} v = -x^2 \\dv = -2x \ dx\\x \cdot dx = - \frac{dv}{2}\end{array} \right] = \\=& \int e^v \cdot -\frac{dv}{2} = -\frac{1}{2} e^v + C = \\=& -\frac{1}{2} e^{-x^2}+ C \end{align*}$$

Sendo assim,

$$\begin{align*} \int_{0}^{ \infty} x \cdot e^{-x^2} \ dx =& \lim_{a \to \infty} x \cdot e^{-x^2} = \\ =& \lim_{a \to \infty} \left( - \frac{1}{2} e^{-x^2} \right) \Bigg\vert_0^a = \\=& 0 - \left( \frac{1}{2} \right) e^{-2x} + C \end{align*}$$

b. Resolvendo a integral indefinida,

$$\begin{align*} \int x \ln x \ dx =& \left[ \begin{array}{lcl} v = \ln x & dv= x \ dx \\dv = \frac{1}{x} \ dx & v = \frac{x^2}{2}\end{array} \right] = \\=& \frac{x^2 \cdot \ln x}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \ dx = \\=& \frac{x^2 \cdot \ln x}{2} - \frac{1}{4} x^2 + C \end{align*}$$

Sendo assim,

$$\begin{align*} \therefore \lim_{a \to 0^+} \int_{a}^{e} x \ln x \ dx =& \lim_{a \to 0^+} \left( \frac{x^2 \ln x}{2} - \frac{1}{4} x^2 \right) \Bigg\vert_a^e = \\=& \lim_{a \to 0^+} \left[ \underbrace{\left( \frac{e^2 \ln e}{2} - \frac{e^2}{4} \right)}_{\frac{e^2}{2} - \frac{e^2}{4} = \frac{e^2}{4}} - \left( \frac{a^2 \ln a}{2} - \underbrace{\frac{a^2}{4}}_{\to 0} \right) \right] = \\=& \frac{e^2}{4} - \frac{1}{2} \cdot \lim_{a \to 0^+} \left( a^2 \ln a \right) =\frac{e^2}{4} - \frac{1}{2} \cdot \lim_{a \to 0^+} \frac{\ln a}{\frac{1}{a^2}} = \\\stackrel{\text{L’H}}{=}& \ \frac{e^2}{4} - \frac{1}{2} \cdot \lim_{a \to 0^+} \frac{\frac{1}{a}}{- \frac{2}{a^3}} = \frac{e^2}{4} - \frac{1}{2} \cdot \underbrace{\lim_{a \to 0^+} \left( - \frac{a^2}{2} \right)}_{\to 0} = \\=& \frac{e^2}{4}\end{align*}$$

c. Sabendo que $\displaystyle 2x^2 +3x+1$ pode ser fatorado como $\displaystyle \left( 2x+1 \right) \left( x+1 \right) $, temos a seguinte integral indefinida:

$$ \begin{align*} \int \frac{1}{2x^2 +3x+1} \ dx =& \int \frac{1}{ \left( 2x+1 \right) \left( x+1 \right)} \ dx = \\=& \int \left( \frac{A}{ 2x+1} + \frac{B}{x+1} \right) \ dx \end{align*}$$

Cálculo das constantes $A$ e $B$:

$$ A = \frac{1}{x+1} \Bigg\vert_{x=- \frac{1}{2}} = \frac{1}{-\frac{1}{2}+1} = 2$$

$$ B = \frac{1}{2x+1} \Bigg\vert_{x=-1} = \frac{1}{-2+1} = -1$$

$$ \begin{align*} \therefore \int \frac{1}{2x^2 +3x+1} \ dx =&\int \left( \frac{2}{ 2x+1} + \frac{-1}{x+1} \right) \ dx = \\=& 2 \int \frac{dx}{2x+1} -  \int \frac{dx}{x+1} = \\=& \ln \Big\vert 2x+1 \Big\vert - \ln \Big\vert x+1 \Big\vert + C = \\=& \ln \Bigg\vert \frac{2x+1}{x+1} \Bigg\vert +C\end{align*}$$

Sendo assim,

$$ \int_{- \infty}^{-2} \frac{1}{2x^2 +3x+1} \ dx = \lim_{a \to - \infty} \int_{a}^{-2} \frac{1}{2x^2 +3x+1} \ dx = $$

$$= \lim_{a \to - \infty} \ln \Bigg\vert \frac{2x+1}{x+1} \Bigg\vert \ \Bigg\vert_{a}^{-2} = \lim_{a \to - \infty} \left( \ln \Big\vert -3 \Big\vert - \ln \Bigg\vert \frac{2a+1}{a+1} \Bigg\vert \right) =$$

$$ = \ln 3 - \lim_{a \to - \infty} \ln \Bigg\vert \frac{2a+1}{a+1} \Bigg\vert \ = \ln 3 - \ln \left( \lim_{a \to - \infty} \frac{2a+1}{a+1} \right) = $$

$$ \stackrel{\text{L’H}}{=} \ln 3 - \ln \left( \lim_{a \to - \infty} \frac{2}{1} \right) = \ln 3 - \ln 2 = $$

$$= \ln \left( \frac{3}{2} \right) \approx 0,4055 $$

2. a. $$ f(x) = \frac{x-2}{x^3 +2x} < g(x) = \frac{x}{x^3} = \frac{1}{x^2} $$

Dado que a integral $\displaystyle \int_{1}^{ \infty}  \frac{1}{x^2} \ dx$ é convergente e que $f(x) < g(x)$, a integral $\displaystyle \int_{1}^{ \infty} \frac{x-2}{x^3 +2x} \ dx$ também é convergente.

b. Resolvendo a integral indefinida,

$$ f(x) = \frac{1}{\left( 1+x^3 \right)^{\frac{1}{3}}} \quad > \quad g(x) = \frac{1}{\left( x^3 + x^3 \right)^{\frac{1}{3}}} = \frac{1}{x \cdot \sqrt[3]{2}} $$

$$g(x) < f(x) $$

$$\int_{0}^{ \infty} \frac{dx}{\sqrt[3]{2} \cdot x} = \int_{0}^{1} \frac{dx}{\sqrt[3]{2} \cdot x} + \int_{1}^{ \infty} \frac{dx}{\sqrt[3]{2} \cdot x} = $$

$$= \underbrace{\frac{1}{\sqrt[3]{2}} \int_{0}^{1} \frac{dx}{x}}_{\text{divergente}} +  \underbrace{\frac{1}{\sqrt[3]{2}} \int_{1}^{\infty} \frac{dx}{x}}_{\text{divergente}} $$

$$\therefore \int_{0}^{\infty} \frac{dx}{\left( 1+x^3 \right)^{\frac{1}{3}}} \text{é também divergente.}$$

3. Começamos pelo cálculo da integral indefinida,

$$\int e^{-st} \cdot \sin \left( 4t \right) \ dt =  \left[ \begin{array}{lcl} u = \sin 4t & dv= e^{-st} \ dt \\du = 4 \cos 4t \ dt & v = - \frac{e^{-st}}{s} \\\frac{du}{4} = \cos 4t \ dt &\end{array} \right] = $$

$$= - \frac{ \sin 4t \cdot e^{-st}}{s} + \frac{4}{s} \int e^{-st} \cdot \cos 4t \ dt = \left[ \begin{array}{lcl} u = \cos 4t & dv= e^{-st} \ dt \\du = -4 \sin 4t \ dt & v = - \frac{e^{-st}}{s} \\- \frac{du}{4} = \sin 4t \ dt &\end{array} \right] = $$

$$= - \frac{ \sin 4t \cdot e^{-st}}{s} + \frac{4}{s} \left( - \frac{e^{-st} \cdot \cos 4t}{s} - \frac{4}{s} \int e^{-st} \cdot \sin 4t \ dt \right) = $$

$$= - \frac{ \sin 4t \cdot e^{-st}}{s} - \frac{ 4 e^{-st} \cdot \cos 4t}{s^2} - \frac{16}{s^2} \int e^{-st} \cdot \sin 4t \ dt $$

Sendo assim,

$$\int e^{-st} \cdot \sin \left( 4t \right) \ dt = -\frac{\sin 4t \cdot e^{-st}}{s} - \frac{4 e^{-st} \cdot \cos 4t}{s^2} - \frac{16}{s^2} \int e^{-st} \cdot \sin 4t \ dt $$

$$ \left( 1+ \frac{16}{s^2} \right)  \int e^{-st} \cdot \sin 4t \ dt = -\frac{\sin 4t \cdot e^{-st}}{s} - \frac{4 e^{-st} \cdot \cos 4t}{s^2} $$

$$ \left( \frac{s^2 +16}{s^2} \right)  \int e^{-st} \cdot \sin 4t \ dt = -\frac{\sin 4t \cdot e^{-st}}{s} - \frac{4 e^{-st} \cdot \cos 4t}{s^2} $$

$$\therefore \int e^{-st} \cdot \sin 4t \ dt = \frac{s^2}{s^2 +16} \left( -\frac{\sin 4t \cdot e^{-st}}{s} - \frac{4 e^{-st} \cdot \cos 4t}{s^2} \right) $$

$$ \int_{0}^{\infty} e^{-st} \cdot \sin 4t \ dt = \lim_{0 \to \infty} \int_{0}^{a} e^{-st} \cdot \sin 4t \ dt = $$

$$= \lim_{a \to \infty} \frac{s^2}{s^2 +16} \cdot \left( - \frac{ \sin 4t \cdot e^{-st}}{s} - \frac{4 e^{-st} \cdot \cos 4t}{s^2} \right) \Bigg\vert_{0}^{a} =$$

$$= \frac{s^2}{s^2 +16} \cdot  \lim_{a \to \infty} \left[ \left( - \frac{ \sin 4a \cdot e^{-sa}}{s} - \frac{4 e^{-sa} \cdot \cos 4a}{s^2} \right) - \left( -0 - \frac{4}{s^2} \right) \right] =$$

$$= \frac{s^2}{s^2 +16} \cdot  \lim_{a \to \infty} \left( - \underbrace{ \frac{ \sin 4a}{s \cdot e^{sa}}}_{\to 0} - \underbrace{\frac{4 \cos 4a}{s^2 \cdot e^{sa}}}_{\to 0} + \frac{4}{s^2} \right) =$$

$$ = \frac{s^2}{s^2 +16} \cdot \frac{4}{s^2} = \frac{4}{s^2 +16} $$