Respuesta :

Answer:

Since the [tex]\lim_{x\to -\infty} \frac{1}{(x-2)^2}=0[/tex] then we have this:

[tex]\int_{-\infty}^{-1} (x-2)^{-3} dx =-\frac{1}{18}[/tex]

And we see that our integral on this case converged to -1/18.

Step-by-step explanation:

For this case we need to determine if the following integral converges or not:

[tex] \int_{-\infty}^{-1} (x-2)^{-3} dx[/tex]

We can rewrite the integral like this:

[tex] \int_{-\infty}^{-1} \frac{1}{(x-2)^3} dx[/tex]

Then we can use the substitution [tex] u = x-2[/tex] and then [tex] du = dx[/tex] and we have this:

[tex] \int_{-\infty}^{-3} \frac{1}{u^3} du=\int_{-\infty}^{-3} u^{-3} du [/tex]

If we solve the integral we got:

[tex] =\frac{u^{-3+1}}{-3+1} =-\frac{u^{-2}}{2}=-\frac{1}{2}\frac{1}{u^2}[/tex]

And then the integral would be equal to:

[tex]\int_{-\infty}^{-1} (x-2)^{-3} dx = -\frac{1}{2(x-2)^2}\Big|_{-\infty}^{-1}[/tex]

And if we replace and using the fundamental theorem of calculus we got:

[tex] = -\frac{1}{2} [\frac{1}{(-1-2)^2} -\lim_{x\to -\infty} \frac{1}{(x-2)^2}][/tex]

Since the [tex]\lim_{x\to -\infty} \frac{1}{(x-2)^2}=0[/tex] then we have this:

[tex]\int_{-\infty}^{-1} (x-2)^{-3} dx =-\frac{1}{18}[/tex]

And we see that our integral on this case converged to -1/18.

Space

Answer:

The improper integral converges.

[tex]\displaystyle \int\limits^{-1}_{- \infty} {(x - 2)^{-3}} \, dx = \frac{-1}{18}[/tex]

General Formulas and Concepts:
Calculus

Limit

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]

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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 Method: U-Substitution

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 - 2)^{-3}} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite [Improper Integral]:                                                     [tex]\displaystyle \int\limits^{-1}_{- \infty} {(x - 2)^{-3}} \, dx = \lim_{a \to - \infty} \int\limits^{-1}_{a} {(x - 2)^{-3}} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:                                                                                                         [tex]\displaystyle u = x - 2[/tex]
  2. [u] Differentiate [Derivative Properties and Rules]:                                 [tex]\displaystyle du = dx[/tex]
  3. [Bounds] Swap:                                                                                         [tex]\displaystyle \left \{ {{x = -1 \rightarrow u = -3} \atop {x = a \rightarrow u = a - 2}} \right.[/tex]

Step 4: Integrate Pt. 3

  1. [Integral] Apply Integration Method [U-Substitution]:                             [tex]\displaystyle \int\limits^{-1}_{- \infty} {(x - 2)^{-3}} \, dx = \lim_{a \to - \infty} \int\limits^{-3}_{a - 2} {u^{-3}} \, du[/tex]
  2. [Integral] Apply Integration Rule [Reverse Power Rule]:                       [tex]\displaystyle \int\limits^{-1}_{- \infty} {(x - 2)^{-3}} \, dx = \lim_{a \to - \infty} \frac{-1}{2u^2} \bigg| \limits^{-3}_{a - 2}[/tex]
  3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:       [tex]\displaystyle \int\limits^{-1}_{- \infty} {(x - 2)^{-3}} \, dx = \lim_{a \to - \infty} \bigg[ \frac{1}{2(a - 2)^2} - \frac{1}{18} \bigg][/tex]
  4. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:                     [tex]\displaystyle \int\limits^{-1}_{- \infty} {(x - 2)^{-3}} \, dx = \frac{1}{2(\infty - 2)^2} - \frac{1}{18}[/tex]
  5. Simplify:                                                                                                     [tex]\displaystyle \int\limits^{-1}_{- \infty} {(x - 2)^{-3}} \, dx = - \frac{1}{18}[/tex]

∴ the improper integral equals  [tex]\displaystyle \bold{\frac{-1}{18}}[/tex]  and is convergent.

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Learn more about improper integrals: https://brainly.com/question/14413082

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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