Respuesta :

Answer:Convergent

Step-by-step explanation:

Given

Improper Integral I is given as

[tex]I=\int^{-4}_{-\infty}\frac{3}{x^4}dx[/tex]

integration of [tex]\frac{3}{x^4}[/tex]  is  -[tex]\frac{1}{x^3}[/tex]

[tex]I=-\left [ \frac{1}{x^3}\right ]^{-4}_{-\infty}[/tex]

[tex]I=-\left [ \frac{1}{(-4)^3}-\frac{1}{(-\infty)^3}\right ][/tex]

[tex]I=-\left [ -\frac{1}{4^3}\right ][/tex]

[tex]I=\frac{1}{4^3}[/tex]

Integral is convergent at   [tex]I=\frac{1}{64}[/tex]

 

Space

Answer:

The improper integral converges.

[tex]\displaystyle \int\limits^{-4}_{- \infty} {\frac{3}{x^4}} \, dx = \frac{1}{64}[/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]

Integration Property [Multiplied Constant]:                                                     [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(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^{-4}_{- \infty} {\frac{3}{x^4}} \, dx[/tex]

Step 2: Integrate

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                [tex]\displaystyle \int\limits^{-4}_{- \infty} {\frac{3}{x^4}} \, dx = 3 \int\limits^{-4}_{- \infty} {\frac{1}{x^4}} \, dx[/tex]
  2. [Integral] Rewrite [Improper Integral]:                                                         [tex]\displaystyle \int\limits^{-4}_{- \infty} {\frac{3}{x^4}} \, dx = \lim_{a \to - \infty} 3 \int\limits^{-4}_{a} {\frac{1}{x^4}} \, dx[/tex]
  3. [Integral] Apply Integration Rule [Reverse Power Rule]:                           [tex]\displaystyle \int\limits^{-4}_{- \infty} {\frac{3}{x^4}} \, dx = \lim_{a \to - \infty} 3 \bigg( \frac{-1}{3x^3} \bigg) \bigg| \limits^{-4}_{a}[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:         [tex]\displaystyle \int\limits^{-4}_{- \infty} {\frac{3}{x^4}} \, dx = \lim_{a \to - \infty} 3 \bigg( \frac{1}{3a^3} + \frac{1}{192} \bigg)[/tex]
  5. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:                       [tex]\displaystyle \int\limits^{-4}_{- \infty} {\frac{3}{x^4}} \, dx = 3 \bigg( \frac{1}{3(-\infty)^3} + \frac{1}{192} \bigg)[/tex]
  6. Simplify:                                                                                                        [tex]\displaystyle \int\limits^{-4}_{- \infty} {\frac{3}{x^4}} \, dx = \frac{1}{64}[/tex]

∴ the improper integral equals  [tex]\displaystyle \bold{\frac{1}{64}}[/tex]  and is convergent.

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Learn more about improper integrals: https://brainly.com/question/14412306

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

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

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