Answer:
Area: 16
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Calculus
Derivatives
Derivative Notation
Integrals - Area under the curve
Trig Integration
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
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
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle f(x) = 8sin(x) + sin(8x)[/tex]
[tex]\displaystyle y = 0[/tex]
Bounds of Integration: 0 ≤ x ≤ π
Step 2: Find Area Pt. 1
- Set up integral: [tex]\displaystyle A = \int\limits^{\pi}_0 {[8sin(x) + sin(8x)]} \, dx[/tex]
- Rewrite integral [Integration Property - Addition/Subtraction]: [tex]\displaystyle A = \int\limits^{\pi}_0 {8sin(x)} \, dx + \int\limits^{\pi}_0 {sin(8x)} \, dx[/tex]
- [1st Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle A = 8\int\limits^{\pi}_0 {sin(x)} \, dx + \int\limits^{\pi}_0 {sin(8x)} \, dx[/tex]
- [1st Integral] Integrate [Trig Integration]: [tex]\displaystyle A = 8[-cos(x)] \bigg| \limits^{\pi}_0 + \int\limits^{\pi}_0 {sin(8x)} \, dx[/tex]
- [1st Integral] Evaluate [Integration Rule - FTC 1]: [tex]\displaystyle A = 8(2) + \int\limits^{\pi}_0 {sin(8x)} \, dx[/tex]
- Multiply: [tex]\displaystyle A = 16 + \int\limits^{\pi}_0 {sin(8x)} \, dx[/tex]
Step 3: Identify Variables
Identify variables for u-substitution.
u = 8x
du = 8dx
Step 4: Find Area Pt. 2
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle A = 16 + \frac{1}{8}\int\limits^{\pi}_0 {8sin(8x)} \, dx[/tex]
- [Integral] U-Substitution: [tex]\displaystyle A = 16 + \frac{1}{8}\int\limits^{8\pi}_0 {sin(u)} \, du[/tex]
- [Integral] Integrate [Trig Integration]: [tex]\displaystyle A = 16 + \frac{1}{8}[-cos(u)] \bigg| \limits^{8\pi}_0[/tex]
- [Integral] Evaluate [Integration Rule - FTC 1]: [tex]\displaystyle A = 16 + \frac{1}{8}(0)[/tex]
- Simplify: [tex]\displaystyle A = 16[/tex]
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Integration - Area under the curve
Book: College Calculus 10e