Respuesta :
Answer:
[tex]\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = \ln(3) - 1[/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]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Integration
- Integrals
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
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 by Parts: [tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]
- [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^1_0 {\ln(3x)} \, dx[/tex]
Step 2: Integrate Pt. 1
Identify variables for integration by parts using LIPET.
- Set u: [tex]\displaystyle u = \ln(3x)[/tex]
- [u] Logarithmic Differentiation [Derivative Rule - Chain Rule]: [tex]\displaystyle du = \frac{(3x)'}{3x} \ dx[/tex]
- [du] Basic Power Rule [Derivative Property - Multiplied Constant]: [tex]\displaystyle du = \frac{3}{3x} \ dx[/tex]
- [du] Simplify: [tex]\displaystyle du = \frac{1}{x} \ dx[/tex]
- Set dv: [tex]\displaystyle dv = 1 \ dx[/tex]
- [dv] Integration Rule [Reverse Power Rule]: [tex]\displaystyle v = x[/tex]
Step 3: Integrate Pt. 2
- [Integral] Integration by Parts: [tex]\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = x \ln(3x) \bigg| \limits^1_0 - \int\limits^1_0 {1} \, dx[/tex]
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = x \ln(3x) \bigg| \limits^1_0 - \int\limits^1_0 {} \, dx[/tex]
- [Integral] Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = x \ln(3x) \bigg| \limits^1_0 - x \bigg| \limits^1_0[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = \ln(3) - 1[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration