Respuesta :
Answer:
[tex]\displaystyle AL = \int\limits^{\frac{\pi}{2}}_0 {\sqrt{1+ 4cos^2(2x)}} \, dx = 2.63518[/tex]
General Formulas and Concepts:
Algebra I
Functions
- Function Notation
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(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 = sin(2x)
Interval [0, π/2]
Step 2: Find Arc Length
- [Function] Differentiate [Trigonometric Differentiation, Chain Rule]: [tex]\displaystyle \frac{dy}{dx} = 2cos(2x)[/tex]
- Substitute in variables [Arc Length Formula - Rectangular]: [tex]\displaystyle AL = \int\limits^{\frac{\pi}{2}}_0 {\sqrt{1+ [2cos(2x)]^2}} \, dx[/tex]
- [Integrand] Simplify: [tex]\displaystyle AL = \int\limits^{\frac{\pi}{2}}_0 {\sqrt{1+ 4cos^2(2x)}} \, dx[/tex]
- [Integral] Evaluate: [tex]\displaystyle AL = 2.63518[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Applications of Integration
Book: College Calculus 10e