Question

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

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Sugestão: Utilize a simetria da função $f(x)=\dfrac{1}{1+\left(\sqrt{3}-\left|x-\sqrt{3}\right| \right )^{\!2}}$ sobre o intervalo de integração.
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Answer 1

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

Mostrando que $x=\sqrt{3}$ é eixo de simetria de $f$:

Avaliando f(√3 - δ):

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

Avaliando f(√3 + δ):

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

Como $f(\sqrt{3}-\delta)=f(\sqrt{3}+\delta)$ para δ apropriado (δ tal que √3 ± δ estejam no domínio de $f$), temos que $x=\sqrt{3}$ é eixo de simetria do gráfico de $f$

Além disso, temos que $f(1)=f(2\sqrt{3}-1)=\dfrac{1}{2}$:

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

e

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

Ou seja, $2\sqrt{3}-1$ e $1$ são simétricos em relação ao eixo $x=\sqrt{3}$
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Podemos integrar em 1 lado do eixo e encontrar a área desejada:

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

Nesse lado do eixo, a expressão para f é

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

Pois, nesse lado do eixo, temos $x~\textless~\sqrt{3}~\longrightarrow~|x-\sqrt{3}|=\sqrt{3}-x$

Então:

$\displaystyle\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=2\int\limits_{1}^{\sqrt{3}}\dfrac{1}{1+x^{2}}dx\\\\\\\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=2\int\limits_{1}^{\sqrt{3}}\dfrac{d}{dx}arctan(x)dx\\\\\\\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=2\bigg[ arctan(x)\bigg]_{1}^{\sqrt{3}}\\\\\\\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=2arctan(\sqrt{3})-2arctan(1)\\\\\\\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=\dfrac{2\pi}{3}-\dfrac{2\pi}{4}\\\\\\\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=\dfrac{8\pi-6\pi}{12}$

$\displaystyle\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=\dfrac{2\pi}{12}\\\\\\\boxed{\boxed{\int\limits_{1}^{2\sqrt{3}-1}f(x)dx=\dfrac{\pi}{6}}}$