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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.8.56
Chapter 8, Problem 8.8.56

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 2 to ∞ of (dx / √(x² - 1))

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Identify the integral to be tested for convergence: \(\displaystyle \int_{2}^{\infty} \frac{dx}{\sqrt{x^{2} - 1}}\).
Recognize that the integral is an improper integral due to the infinite upper limit, so we consider the limit: \(\displaystyle \lim_{t \to \infty} \int_{2}^{t} \frac{dx}{\sqrt{x^{2} - 1}}\).
To test for convergence, compare the integrand \(\frac{1}{\sqrt{x^{2} - 1}}\) with a simpler function whose convergence behavior is known. For large \(x\), note that \(\sqrt{x^{2} - 1} \approx x\), so the integrand behaves like \(\frac{1}{x}\).
Use the Direct Comparison Test or Limit Comparison Test with the function \(\frac{1}{x}\), since \(\int_{2}^{\infty} \frac{1}{x} dx\) is a known divergent integral (harmonic integral).
Set up the Limit Comparison Test by computing \(\displaystyle \lim_{x \to \infty} \frac{\frac{1}{\sqrt{x^{2} - 1}}}{\frac{1}{x}}\) and analyze the limit to determine if the original integral converges or diverges.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, we consider limits approaching the problematic points, such as infinity, to determine if the integral converges to a finite value or diverges.
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Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test determines convergence by comparing the given integral's integrand to a simpler function with known behavior. If the integrand is smaller than a convergent function or larger than a divergent one, we can conclude about the integral's convergence accordingly.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test compares two functions by examining the limit of their ratio as the variable approaches infinity. If the limit is a positive finite number, both integrals either converge or diverge together, helping to analyze complex integrals by relating them to simpler ones.
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Limit Comparison Test
Related Practice
Textbook Question

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 e to e^e of (ln(ln x) dx)

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Textbook Question

[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.

y = x²/4, 0 ≤ x ≤ 2

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Textbook Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₋₈¹ dx / x^(1/3)

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Textbook Question

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.

∫ √(9 - w²) dw / w²

Textbook Question

Evaluate the integrals in Exercises 23–32.

∫₀^(π/6) √(1 + sin(x)) dx

(Hint: Multiply by √((1 - sin(x)) / (1 - sin(x))))

Textbook Question

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

86. Find the volume of the solid generated by revolving the region about the x-axis.

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