Respuesta :
Answer:
4.25
Step-by-step explanation:
Given:
[tex]f(x) = \frac{x^3}{12} + \frac{1}{x} \\\\Arc Length = \int\limits^a_b {\sqrt{1 + (f'(x))^2} } \, dx \\f '(x) = \frac{x^2}{6} - \frac{1}{x^2}\\\\f'(x)^2 = (\frac{x^2}{6} - \frac{1}{x^2})^2 = \frac{x^4}{36} + \frac{1}{x^4} - \frac{1}{3}\\\\1 + f'(x)^2 = \frac{x^4}{36} + \frac{1}{x^4} + \frac{2}{3}\\\\\= \frac{x^8 + 36 + 12x^4}{36x^4}\\\\= \frac{(x^4 + 6)^2}{36x^4}\\\\=\sqrt{1 + f'(x)^2} = \sqrt{ \frac{(x^4 + 6)^2}{36x^4}}\\\\= \frac{x^2}{6} + \frac{1}{x^2} \\\\[/tex]
[tex]ArcLength = \int\limits^4_1 {\frac{x^2}{6} + \frac{1}{x^2} } \, dx \\= (\frac{x^3}{18} - \frac{1}{x})\limits^4_1\\\\= (\frac{64}{18} - \frac{1}{4}) - (\frac{1}{18} - 1)\\\\= \frac{17}{4}= 4.25[/tex]
