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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 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))

검증된 단계별 안내
1
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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
11:11
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.
추천 영상:
09:25
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.
추천 영상:
07:45
Limit Comparison Test