Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 8.8.60

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 π to ∞ of ((1 + sin x) / x² dx)

검증된 단계별 안내
1
Identify the integral to be tested for convergence: \(\displaystyle \int_{\pi}^{\infty} \frac{1 + \sin x}{x^{2}} \, dx\).
Note that the integrand is \(\frac{1 + \sin x}{x^{2}}\). Since \(\sin x\) oscillates between \(-1\) and \(1\), the numerator \(1 + \sin x\) oscillates between \(0\) and \(2\).
To apply the Direct Comparison Test, find a simpler function that bounds the integrand from above and below. Since \(0 \leq 1 + \sin x \leq 2\), we have \(0 \leq \frac{1 + \sin x}{x^{2}} \leq \frac{2}{x^{2}}\) for \(x \geq \pi\).
Recall that the integral \(\int_{\pi}^{\infty} \frac{1}{x^{2}} \, dx\) converges because it is a p-integral with \(p=2 > 1\). Therefore, by the Direct Comparison Test, since \(\frac{1 + \sin x}{x^{2}} \leq \frac{2}{x^{2}}\) and the integral of \(\frac{2}{x^{2}}\) converges, the original integral converges.
Conclude that the integral \(\int_{\pi}^{\infty} \frac{1 + \sin x}{x^{2}} \, dx\) converges by the Direct Comparison Test.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
5m
도움이 되었나요?

주요 개념

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

Improper Integrals and Convergence

Improper integrals involve integration over an infinite interval or where the integrand is unbounded. To determine convergence, we evaluate the limit of the integral as the upper bound approaches infinity. If this limit exists and is finite, the integral converges; otherwise, it diverges.
추천 영상:
가이드 코스
11:11
Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test compares the given integral to a second integral with a known convergence behavior. If the integrand is less than or equal to a function with a convergent integral, the original integral also converges. Conversely, if it is greater than or equal to a function with a divergent integral, it diverges.
추천 영상:
가이드 코스
09:25
Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test involves taking the limit of the ratio of the given integrand to a simpler function with known convergence properties. If the limit is a positive finite number, both integrals share the same convergence behavior, allowing us to conclude about the original integral's convergence.
추천 영상:
가이드 코스
07:45
Limit Comparison Test