Question

(50 PONTOS) Calcule a integral definida:
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$\displaystyle\int\limits_{2-\sqrt{3}}^{\sqrt{3}}{\dfrac{dx}{1+\left(1+\left|x-1\right| \right )^{2}}}$

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Sugestão: Utilize a simetria da função $f(x)=\dfrac{1}{1+\left(1+\left|x-1\right| \right )^{2}}$ sobre o intervalo de integração para simplificar os cálculos.
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Answer 1

Hola

$\displaystyle I=\int\limits_{2-\sqrt{3}}^{\sqrt{3}}{\dfrac{dx}{1+\left(1+\left|x-1\right| \right )^{2}}} \\ \\ \\ I=\int\limits_{2-\sqrt{3}}^{1}{\dfrac{dx}{1+\left(1+\left|x-1\right| \right )^{2}}} +\int\limits_{1}^{\sqrt{3}}{\dfrac{dx}{1+\left(1+\left|x-1\right| \right )^{2}}}$


$\displaystyle I=\int\limits_{2-\sqrt{3}}^{1}{\dfrac{dx}{1+\left(2-x\right)^{2}}} +\int\limits_{1}^{\sqrt{3}}{\dfrac{dx}{1+x^2}$


$\displaystyle I=\left.\left[-\arctan (2-x)\right]\right|_{2-\sqrt{3}}^{1}+(\arctan x)_{1}^{\sqrt3}\\ \\ \\ I=(\arctan \sqrt3-\arctan 1)+(\arctan \sqrt3-\arctan 1)\\ \\ \\ I=2(\arctan \sqrt3-\arctan 1)\\ \\ \\ I=2\left(\frac{\pi}{3}-\dfrac{\pi}{4}\right)\\ \\ \\ \boxed{I=\dfrac{\pi}{6}}$

Answer 2

Vou utilizar a sugestão!

$f(x)=\dfrac{1}{1+(1+|x-1|)^{2}}$

A função possui x = 1 como eixo de simetria. Veja:

$f(1+\delta)=\dfrac{1}{1+(1+|1+\delta-1|)^{2}}=\dfrac{1}{1+(1+|\delta|)^{2}}\\\\\\f(1-\delta)=\dfrac{1}{1+(1+|1-\delta-1|)^{2}}=\dfrac{1}{1+(1+|-\delta|)^{2}}=\dfrac{1}{1+(1+|\delta|)^{2}}$

Como $f(1+\delta)=f(1-\delta)$ para qualquer δ apropriado, temos que x = 1 é eixo de simetria do gráfico de $f$
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Avaliando f(2 - √3):

$f(2-\sqrt{3})=\dfrac{1}{1+(1+|2-\sqrt{3}-1|)^{2}}\\\\\\=\dfrac{1}{1+(1+|1-\sqrt{3}|)^{2}}=\dfrac{1}{1+(1-1+\sqrt{3})^{2}}=\dfrac{1}{4}$

Avaliando f(√3):

$f(\sqrt{3})=\dfrac{1}{1+(1+|\sqrt{3}-1|)^{2}}=\dfrac{1}{1+(1+\sqrt{3}-1)^{2}}\\\\\\=\dfrac{1}{1+(\sqrt{3})^{2}}=\dfrac{1}{4}$

Disso, tiramos que x = 2 - √3 é o simétrico de x = √3 com relação ao eixo de simetria x = 1 (verifique com δ = 1 - √3)

Portanto, só precisamos integrar de um lado do eixo:

$\displaystyle\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=2\int\limits_{1}^{\sqrt{3}}\dfrac{1}{1+(1+|x-1|)^{2}}dx$

Nesse lado do eixo, $x~\textgreater~1~~\longrightarrow~~x-1~\textgreater~0~~\longrightarrow~~|x-1|=x-1$

Então:

$\displaystyle\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=2\int\limits_{1}^{\sqrt{3}}\dfrac{1}{1+(1+x-1)^{2}}dx\\\\\\\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=2\int\limits_{1}^{\sqrt{3}}\dfrac{1}{1+x^{2}}dx\\\\\\\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=2\int\limits_{1}^{\sqrt{3}}\dfrac{d}{dx}tg^{-1}(x)dx\\\\\\\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=2\bigg[tg^{-1}(x)\bigg]_{1}^{\sqrt{3}}\\\\\\\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=2tg^{-1}(\sqrt{3})-2tg^{-1}(1)$

$\displaystyle\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=2\cdot\dfrac{\pi}{3}-2\cdot\dfrac{\pi}{4}\\\\\\\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=\dfrac{8\pi-6\pi}{12}\\\\\\\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=\dfrac{2\pi}{12}\\\\\\\boxed{\boxed{\int\limits_{2-\sqrt{3}}^{\sqrt{3}}f(x)dx=\dfrac{\pi}{6}}}$