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Write and simplify the integral that gives the arc length of the following curve on the given interval b. If necessary, use technology to evaluate or approximate the integral. y:31n x, for 2sxs5 a. The integral that gives the arc length of the curve is L dx &The arc length of the curve is approximately (Round to three decimal places as needed.)

Respuesta :

Space

Answer:

a.  [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex]

b.  [tex]\displaystyle AL = 4.10322[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Integration

  • Integrals
  • Definite Integrals
  • Integration Constant C

Integration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

U-Substitution

Arc Length Formula [Rectangular]:                                                                       [tex]\displaystyle AL = \int\limits^b_a {\sqrt{1+ [f'(x)]^2}} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

Identify

y = 3ln(x)

Interval [2, 5]

Step 2: Find Arc Length

  1. [Function] Differentiate [Logarithmic Differentiation]:                                 [tex]\displaystyle \frac{dy}{dx} = \frac{3}{x}[/tex]
  2. Substitute in variables [Arc Length Formula - Rectangular]:                     [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ [\frac{3}{x}]^2}} \, dx[/tex]
  3. [Integrand] Simplify:                                                                                       [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex]
  4. [Integral] Evaluate:                                                                                         [tex]\displaystyle AL = 4.10322[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Applications of Integration

Book: College Calculus 10e

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