Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.8.52
Capitolo 8, Problema 8.8.52

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from 4 to ∞ of (dx / (√x - 1))

Guida verificata passo dopo passo
1
Identify the integral to be tested for convergence: \(\displaystyle \int_{4}^{\infty} \frac{dx}{\sqrt{x} - 1}\).
Analyze the behavior of the integrand as \(x \to \infty\). Since \(\sqrt{x}\) grows without bound, \(\sqrt{x} - 1\) behaves like \(\sqrt{x}\) for large \(x\), so the integrand behaves like \(\frac{1}{\sqrt{x}}\).
Choose a comparison function to apply the Direct Comparison Test or Limit Comparison Test. A natural choice is \(\frac{1}{\sqrt{x}}\), because it simplifies the integrand's behavior at infinity.
Recall that the integral \(\int_{4}^{\infty} \frac{1}{\sqrt{x}} \, dx\) is a p-integral with \(p = \frac{1}{2}\), which diverges since \(p \leq 1\).
Use the Direct Comparison Test or Limit Comparison Test by comparing \(\frac{1}{\sqrt{x} - 1}\) with \(\frac{1}{\sqrt{x}}\) to determine if the original integral converges or diverges.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
4m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, we take limits to handle the infinite bounds or singularities, determining whether the integral converges (has a finite value) or diverges.
Video consigliato:
Percorso guidato
11:11
Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test compares the given integral's integrand to a simpler function whose convergence behavior is known. If the integrand is smaller than a convergent function or larger than a divergent function on the interval, we can conclude about the original integral's convergence.
Video consigliato:
Percorso guidato
09:25
Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test uses the limit of the ratio of two functions as the variable approaches infinity. If this limit is a positive finite number, both integrals either converge or diverge together, allowing us to infer the behavior of a complicated integral by comparing it to a simpler one.
Video consigliato:
Percorso guidato
07:45
Limit Comparison Test
Pratica correlata
Domanda del libro di testo

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ θ cos(πθ) dθ

12
views
Domanda del libro di testo

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.

∫ √(1 - (ln x)²) / (x ln x) dx

32
views
Domanda del libro di testo

Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e^(-x) and the x-axis.

85. Find the volume of the solid generated by revolving the region about the y-axis.

19
views
Domanda del libro di testo

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.

∫ (csc t sin 3t dt)

23
views
Domanda del libro di testo

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (8x² + 8x + 2) / (4x² + 1)² dx

31
views
Domanda del libro di testo

Volume: Find the volume of the solid generated by revolving the region in Exercise 45 about the x-axis.

58
views