Respuesta :

Space

Answer:

The improper integral converges.

[tex]\displaystyle \int\limits^{-1}_{- \infty} {x^\Big{\frac{-8}{3}}} \, dx = \frac{3}{5}[/tex]

General Formulas and Concepts:
Calculus

Limit

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]

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^{-1}_{- \infty} {x^\Big{\frac{-8}{3}}} \, dx[/tex]

Step 2: Integrate

  1. [Integral] Rewrite [Improper Integral]:                                                     [tex]\displaystyle \int\limits^{-1}_{- \infty} {x^\Big{\frac{-8}{3}}} \, dx = \lim_{a \to - \infty} \int\limits^{-1}_{a} {x^\Big{\frac{-8}{3}}} \, dx[/tex]
  2. [Integral] Apply Integration Rule [Reverse Power Rule]:                       [tex]\displaystyle \int\limits^{-1}_{- \infty} {x^\Big{\frac{-8}{3}}} \, dx = \lim_{a \to - \infty} \frac{-3}{5x^\big{\frac{5}{3}}} \bigg| \limits^{-1}_{a}[/tex]
  3. Evaluate [Integration Rules - Fundamental Theorem of Calculus 1]:     [tex]\displaystyle \int\limits^{-1}_{- \infty} {x^\Big{\frac{-8}{3}}} \, dx = \lim_{a \to - \infty} \bigg( \frac{3}{5a^\big{\frac{5}{3}}} + \frac{3}{5} \bigg)[/tex]
  4. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:                     [tex]\displaystyle \int\limits^{-1}_{- \infty} {x^\Big{\frac{-8}{3}}} \, dx = \frac{3}{5(-\infty)^\big{\frac{5}{3}}} + \frac{3}{5}[/tex]
  5. Simplify:                                                                                                     [tex]\displaystyle \int\limits^{-1}_{- \infty} {x^\Big{\frac{-8}{3}}} \, dx = \frac{3}{5}[/tex]

∴ the improper integral equals  [tex]\displaystyle \bold{\frac{3}{5}}[/tex]  and is convergent.

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Learn more about improper integrals: https://brainly.com/question/14412693

Learn more about calculus: https://brainly.com/question/23558817

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration