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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.9.81

77–86. Comparison Test Determine whether the following integrals converge or diverge.
81. ∫(from 1 to ∞) (sin²x) / x² dx

Guida verificata passo dopo passo
1
Identify the integral to analyze: \(\int_1^{\infty} \frac{\sin^2 x}{x^2} \, dx\) and note that it is an improper integral because the upper limit is infinity.
Recall the Comparison Test for improper integrals: if \(0 \leq f(x) \leq g(x)\) for all \(x\) in \([1, \infty)\) and \(\int_1^{\infty} g(x) \, dx\) converges, then \(\int_1^{\infty} f(x) \, dx\) also converges.
Find a suitable function \(g(x)\) to compare with \(f(x) = \frac{\sin^2 x}{x^2}\). Since \(\sin^2 x\) is always between 0 and 1, we have \(0 \leq \sin^2 x \leq 1\), so \(0 \leq \frac{\sin^2 x}{x^2} \leq \frac{1}{x^2}\).
Check the convergence of the comparison integral \(\int_1^{\infty} \frac{1}{x^2} \, dx\). This is a p-integral with \(p=2 > 1\), which is known to converge.
Conclude by the Comparison Test that since \(\int_1^{\infty} \frac{1}{x^2} \, dx\) converges and \(\frac{\sin^2 x}{x^2} \leq \frac{1}{x^2}\), the original integral \(\int_1^{\infty} \frac{\sin^2 x}{x^2} \, dx\) also converges.

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Improper integrals involve integration over an infinite interval or where the integrand has an infinite discontinuity. To evaluate convergence, we consider the limit of the integral as the upper bound approaches infinity. Determining whether such integrals converge or diverge is essential for understanding the behavior of functions over unbounded domains.
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The Comparison Test helps determine convergence by comparing the given integral to a simpler integral with known behavior. If the integrand is positive and less than or equal to a function whose integral converges, then the original integral also converges. Conversely, if it is greater than a divergent integral, it diverges.
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