contestada

Determine whether the improper integral converges or diverges, and find the value of each that converges.
∫^-10_-[infinity] x^-2 dx

Respuesta :

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Answer:

The improper integral converges.

[tex]\displaystyle \int\limits^{-10}_{- \infty} {x^{-2}} \, dx = \frac{1}{10}[/tex]

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:                                                         [tex]\displaystyle \lim_{x \to c} x = c[/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]

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:

Step 1: Define

Identify.

[tex]\displaystyle \int\limits^{-10}_{- \infty} {x^{-2}} \, dx[/tex]

Step 2: Integrate

  1. [Integral] Rewrite [Improper Integral]:                                                     [tex]\displaystyle \int\limits^{-10}_{- \infty} {x^{-2}} \, dx = \lim_{a \to - \infty} \int\limits^{-10}_{a} {x^{-2}} \, dx[/tex]
  2. [Integral] Apply Integration Rule [Reverse Power Rule]:                       [tex]\displaystyle \int\limits^{-10}_{- \infty} {x^{-2}} \, dx = \lim_{a \to - \infty} \frac{-1}{x} \bigg| \limits^{-10}_{a}[/tex]
  3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:       [tex]\displaystyle \int\limits^{-10}_{- \infty} {x^{-2}} \, dx = \lim_{a \to - \infty} \bigg( \frac{1}{10} + \frac{1}{a} \bigg)[/tex]
  4. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:                    [tex]\displaystyle \int\limits^{-10}_{- \infty} {x^{-2}} \, dx = \frac{1}{10} + \frac{1}{- \infty}[/tex]
  5. Simplify:                                                                                                     [tex]\displaystyle \int\limits^{-10}_{- \infty} {x^{-2}} \, dx = \frac{1}{10}[/tex]

∴ the improper integral is equal to 0.10 and is convergent.

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Learn more about improper integrals: https://brainly.com/question/14413972

Learn more about calculus: https://brainly.com/question/20197752

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration