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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 8.8.48

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 0 to 2 of (dx / (1 - x))

검증된 단계별 안내
1
Identify the integral to be tested for convergence: \(\int_0^2 \frac{dx}{1 - x}\).
Notice that the integrand \(\frac{1}{1 - x}\) has a vertical asymptote at \(x = 1\) because the denominator becomes zero, which lies within the interval of integration \([0, 2]\). This means the integral is an improper integral.
Split the integral at the point of discontinuity to handle it as two improper integrals: \(\int_0^1 \frac{dx}{1 - x} + \int_1^2 \frac{dx}{1 - x}\).
For each integral, rewrite them as limits approaching the point of discontinuity: \(\lim_{t \to 1^-} \int_0^t \frac{dx}{1 - x}\) and \(\lim_{s \to 1^+} \int_s^2 \frac{dx}{1 - x}\).
Evaluate or analyze the behavior of these limits to determine if they converge or diverge. You may use comparison tests or direct integration to assess the convergence of each part.

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주요 개념

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Improper Integrals

Improper integrals involve integrals with infinite limits or integrands with discontinuities within the interval. In this problem, the integrand has a vertical asymptote at x = 1, making the integral improper. Understanding how to handle such integrals by taking limits is essential to determine convergence.
추천 영상:
11:11
Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test compares the given 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, conclusions about convergence or divergence can be drawn.
추천 영상:
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 known benchmark function as the variable approaches the problematic point. If the limit is a positive finite number, both integrals share the same convergence behavior, aiding in determining the original integral's convergence.
추천 영상:
07:45
Limit Comparison Test