contestada

Determine whether the improper integral converges or diverges, and find the value of each that converges.
∫^[infinity]_2 1/x ln x dx

Respuesta :

Answer:

Diverges

Step-by-step explanation:

We can solve this by using integral by parts:

Let

[tex]u=lnx[/tex]            [tex]dv=(1/x)dx[/tex]

[tex]du=(1/x)dx[/tex]    [tex]v=lnx[/tex]

[tex]\int\limits {lnx/x} \, dx = (lnx)^2-\int\limits {lnx/x} \, dx[/tex]

We can add

[tex]\int\limits {lnx/x} \, dx[/tex] to both sides

[tex]\int\limits{lnx/x} \, dx = (1/2)\cdot{(lnx)^2}[/tex]

We can evaluate the limit between 2 and infinity.

If x tends to infinity the limit will be infinity and therefore the integral diverges to ∞

Space

Answer:

The improper integral diverges.

[tex]\displaystyle \int\limits^{\infty}_2 {\frac{1}{x} \ln x} \, dx = \infty[/tex]

General Formulas and Concepts:
Calculus

Limits

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

Differentiation

  • Derivatives
  • Derivative Notation

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 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^{\infty}_2 {\frac{1}{x} \ln x} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite [Improper Integral]:                                                         [tex]\displaystyle \int\limits^{\infty}_2 {\frac{1}{x} \ln x} \, dx = \lim_{b \to \infty} \int\limits^{b}_2 {\frac{1}{x} \ln x} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:                                                                                                             [tex]\displaystyle u = \ln x[/tex]
  2. [u] Apply Logarithmic Differentiation:                                                         [tex]\displaystyle du = \frac{1}{x} \ dx[/tex]
  3. [Bounds] Swap:                                                                                            [tex]\displaystyle \left \{ {{x = b \rightarrow u = \ln b} \atop {x = 2 \rightarrow u = \ln 2}} \right.[/tex]

Step 4: Integrate Pt. 3

  1. [Integral] Apply Integration Method [U-Substitution]:                                [tex]\displaystyle \int\limits^{\infty}_2 {\frac{1}{x} \ln x} \, dx = \lim_{b \to \infty} \int\limits^{\ln b}_{\ln 2} {u} \, du[/tex]
  2. [Integral] Apply Integration Rule [Reverse Power Rule]:                           [tex]\displaystyle \int\limits^{\infty}_2 {\frac{1}{x} \ln x} \, dx = \lim_{b \to \infty} \frac{u^2}{2} \bigg| \limits^{\ln b}_{\ln 2}[/tex]
  3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:          [tex]\displaystyle \int\limits^{\infty}_2 {\frac{1}{x} \ln x} \, dx = \lim_{b \to \infty} \frac{\ln (b)^2 - \ln (2)^2}{2}[/tex]
  4. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:                       [tex]\displaystyle \int\limits^{\infty}_2 {\frac{1}{x} \ln x} \, dx = \frac{\ln (\infty)^2 - \ln (2)^2}{2}[/tex]
  5. Simplify:                                                                                                        [tex]\displaystyle \int\limits^{\infty}_2 {\frac{1}{x} \ln x} \, dx = \infty[/tex]

∴ the improper integral tends to ∞ and is divergent.

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Learn more about improper integrals: https://brainly.com/question/14412893

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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