Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx = 1[/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 [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals
  • Improper 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]

U-Substitution

Area of a Region Formula:                                                                                     [tex]\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle f(x) = \frac{1}{(x - 1)^2} \\\left[ -\infty ,\ 0 \right][/tex]

Step 2: Integrate Pt. 1

  1. Substitute in variables [Area of a Region Formula]:                                   [tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx[/tex]
  2. [Integral] Rewrite [Improper Integral]:                                                         [tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx = \lim_{a \to -\infty} \int\limits^{0}_{a} {\frac{1}{(x - 1)^2}} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:                                                                                                             [tex]\dipslaystyle u = x - 1[/tex]
  2. [u] Basic Power Rule [Derivative Property - Addition/Subtraction]:           [tex]\dipslaystyle du = dx[/tex]
  3. [Limits] Switch:                                                                                               [tex]\displaystyle \left \{ {{x = 0 ,\ u = 0 - 1 = -1} \atop {x = a ,\ u = a - 1}} \right.[/tex]

Step 4: Integrate Pt. 3

  1. [Integral] U-Substitution:                                                                               [tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx = \lim_{a \to -\infty} \int\limits^{-1}_{a - 1} {\frac{1}{u^2}} \, du[/tex]
  2. [Integral] Integration Rule [Reverse Power Rule]:                                       [tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx = \lim_{a \to -\infty} \frac{-1}{x} \bigg| \limits^{-1}_{a - 1}[/tex]
  3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx = \lim_{a \to -\infty} \bigg( \frac{1}{a - 1} + 1 \bigg)[/tex]
  4. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           [tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx = 0 + 1[/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle \int\limits^{0}_{- \infty} {\frac{1}{(x - 1)^2}} \, dx = 1[/tex]

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

Unit: Advanced Integration Techniques

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