Respuesta :
Answer:
The improper integral diverges.
[tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \infty[/tex]
General Formulas and Concepts:
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Derivative Rule [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 Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Method: U-Substitution
Integration by Parts: [tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]
- [IBP] LIPET: Logs, Inverses, Polynomials, Exponentials, Trig
Improper Integral: [tex]\displaystyle \int\limits^{\infty}_a {f(x)} \, dx = \lim_{b \to \infty} \int\limits^b_a {f(x)} \, dx[/tex]
Step-by-step explanation:
*Note:
When using integration by parts, ignore the integration constant C when finding v.
Step 1: Define
Identify.
[tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx[/tex]
Step 2: Integrate Pt. 1
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \lim_{b \to \infty} \int\limits^{b}_1 {\ln |x|} \, dx[/tex]
Step 3: Integrate Pt. 2
Identify variables for integration by parts.
- Set u [LIPET]: [tex]\displaystyle u = \ln |x|[/tex]
- [u] Apply Logarithmic Differentiation [Derivative Rule - Chain Rule]: [tex]\displaystyle du = \frac{1}{|x|} \frac{d}{dx} |x| \ dx[/tex]
- [du] Apply Absolute Value Differentiation: [tex]\displaystyle du = \frac{1}{|x|} \frac{|x|}{x} \ dx[/tex]
- [du] Simplify: [tex]\displaystyle du = \frac{1}{x} \ dx[/tex]
- Set dv [LIPET]: [tex]\displaystyle dv = dx[/tex]
- [dv] Rewrite: [tex]\displaystyle \int {} \, dv = \int {} \, dx[/tex]
- [dv] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle v = x[/tex]
Step 4: Integrate Pt. 3
- [Integral] Apply Integration by Parts: [tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \lim_{b \to \infty} \Bigg[ \bigg( x \ln |x| \bigg) \bigg| \limits^b_1 - \int\limits^b_1 {\frac{x}{x}} \, dx \bigg][/tex]
- [Integral] Simplify: [tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \lim_{b \to \infty} \Bigg[ \bigg( x \ln |x| \bigg) \bigg| \limits^b_1 - \int\limits^b_1 {} \, dx \bigg][/tex]
- [Integral] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \lim_{b \to \infty} \Bigg[ \bigg( x \ln |x| \bigg) \bigg| \limits^b_1 - x \bigg| \limits^b_1 \bigg][/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \lim_{b \to \infty} \bigg[ b \ln |b| - \Big( b - 1 \Big) \bigg][/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \infty \ln | \infty | - \Big( \infty - 1 \Big)[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_1 {\ln |x|} \, dx = \infty[/tex]
∴ the improper integral tends to ∞ and is divergent.
---
Learn more about improper integrals: https://brainly.com/question/14413972
Learn more about calculus: https://brainly.com/question/23558817
---
Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
