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7. 1 Suppose f (x) is a continuous function and 0 < f (x) < ². Does ₁° f (x) dx converge or diverge? Justify your solution with the appropriate calculus and written explanation.

7. 1 Suppose f (x) is a continuous function and 0 < f (x) < ². Does ₁° f (x) dx converge or diverge? Justify your solution with the appropriate calculus and written explanation.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please solve the following (use calc 2 questions) 

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**Problem 7:** Suppose \( f(x) \) is a continuous function and \( 0 < f(x) < \frac{1}{x^{2\pi}} \). Does \( \int_{1}^{\infty} f(x) \, dx \) converge or diverge? Justify your solution with the appropriate calculus and written explanation. --- In this problem, you need to analyze the convergence of the improper integral \( \int_{1}^{\infty} f(x) \, dx \), where \( f(x) \) is a continuous function bounded by \( 0 < f(x) < \frac{1}{x^{2\pi}} \). To determine convergence or divergence, consider using the Comparison Test for improper integrals. Compare \( f(x) \) with the integrable function \( \frac{1}{x^{2\pi}} \), which you will evaluate from 1 to infinity. - **Steps to solution:** 1. Evaluate \( \int_{1}^{\infty} \frac{1}{x^{2\pi}} \, dx \). 2. Use the integral comparison test to infer about the convergence of \( \int_{1}^{\infty} f(x) \, dx \). The integral will converge if \( \int_{1}^{\infty} \frac{1}{x^{2\pi}} \, dx \) converges, giving evidence that \( \int_{1}^{\infty} f(x) \, dx \) must also converge, since \( f(x) \) is less than this convergent function. Conversely, if \( \int_{1}^{\infty} \frac{1}{x^{2\pi}} \, dx \) diverges, further evaluation is necessary, or a different comparison function must be used.
Transcribed Image Text:**Problem 7:** Suppose \( f(x) \) is a continuous function and \( 0 < f(x) < \frac{1}{x^{2\pi}} \). Does \( \int_{1}^{\infty} f(x) \, dx \) converge or diverge? Justify your solution with the appropriate calculus and written explanation. --- In this problem, you need to analyze the convergence of the improper integral \( \int_{1}^{\infty} f(x) \, dx \), where \( f(x) \) is a continuous function bounded by \( 0 < f(x) < \frac{1}{x^{2\pi}} \). To determine convergence or divergence, consider using the Comparison Test for improper integrals. Compare \( f(x) \) with the integrable function \( \frac{1}{x^{2\pi}} \), which you will evaluate from 1 to infinity. - **Steps to solution:** 1. Evaluate \( \int_{1}^{\infty} \frac{1}{x^{2\pi}} \, dx \). 2. Use the integral comparison test to infer about the convergence of \( \int_{1}^{\infty} f(x) \, dx \). The integral will converge if \( \int_{1}^{\infty} \frac{1}{x^{2\pi}} \, dx \) converges, giving evidence that \( \int_{1}^{\infty} f(x) \, dx \) must also converge, since \( f(x) \) is less than this convergent function. Conversely, if \( \int_{1}^{\infty} \frac{1}{x^{2\pi}} \, dx \) diverges, further evaluation is necessary, or a different comparison function must be used.
Expert Solution
Step 1: Use theorem

If f(x) is continuous then we can integrate between some limits.

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