Math Problem Statement
Qual integral dá o comprimento do arco da curva \[y=\cos(x)\] no intervalo \[[0, \pi]\]? Escolha 1 resposta: Escolha 1 resposta: (Escolha A) \[\displaystyle \int_{0}^{\pi} \sqrt{1+\operatorname{sen} x}\,dx\] A \[\displaystyle \int_{0}^{\pi} \sqrt{1+\operatorname{sen} x}\,dx\] (Escolha B) \[\displaystyle \int_{0}^{\pi} \sqrt{1+\operatorname{sen}^2x}\,dx\] B \[\displaystyle \int_{0}^{\pi} \sqrt{1+\operatorname{sen}^2x}\,dx\] (Escolha C) \[\displaystyle \int_{0}^{\pi} \sqrt{1+\cos x}\,dx\] C \[\displaystyle \int_{0}^{\pi} \sqrt{1+\cos x}\,dx\] (Escolha D) \[\displaystyle \int_{0}^{\pi} \sqrt{1+\cos^2x}\,dx\] D \[\displaystyle \int_{0}^{\pi} \sqrt{1+\cos^2x}\,dx\]
Solution
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Math Problem Analysis
Mathematical Concepts
Integral calculus
Arc length of curves
Trigonometric functions
Formulas
Length of arc formula: L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
Theorems
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Suitable Grade Level
Advanced Mathematics
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