• Matéria: Matemática
  • Autor: Lukyo
  • Perguntado 9 anos atrás

(30 PONTOS) Calcule a integral definida:
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\displaystyle\int\limits_{0}^{2\pi/3}{\mathrm{sen}\!\left(\dfrac{\pi}{3}-\left|x-\dfrac{\pi}{3} \right| \right )dx}
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Sugestão: Utilize a simetria da função f(x)=\mathrm{sen}\!\left(\dfrac{\pi}{3}-\left|x-\dfrac{\pi}{3}\right| \right ).
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Lukyo: Caso alguém queira, confirmar a resposta é 1.

Respostas

respondido por: carlosmath
2
\displaystyle
I=\int\limits_0^{2\pi/3}\sin\left(\frac{\pi}{3}-\left|x-\frac{\pi}{3} \right| \right )dx\\ \\ \\
I=\int\limits_0^{\pi/3}\sin\left(\frac{\pi}{3}-\left|x-\frac{\pi}{3} \right| \right )dx+\int\limits_{\pi/3}^{2\pi/3}\sin\left(\frac{\pi}{3}-\left|x-\frac{\pi}{3} \right| \right )dx\\ \\ \\
I=\int\limits_0^{\pi/3}\sin x\,dx+\int\limits_{\pi/3}^{2\pi/3}\sin\left(\frac{2\pi}{3}-x \right )dx


\displaystyle
I=\left.-\cos x\right|_{0}^{\pi/3}}+\left.\cos\left(\frac{2\pi}{3}-x \right)\right|_{\pi/3}^{2\pi/3}\\ \\ \\
I=2\left(\cos 0 - \cos \dfrac{\pi}{3}\right)\\ \\ \\
I=2\left(1-\dfrac{1}{2}\right)\\ \\ \\
\boxed{I=1}

Lukyo: Obrigado!
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