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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.8.44
Chapter 8, Problem 8.8.44

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 1 of ((e^(-√x)) / √x dx)

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Identify the integral to be tested for convergence: \(\displaystyle \int_0^1 \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx\).
Examine the behavior of the integrand near the problematic point, which is \(x = 0\), since the denominator \(\sqrt{x}\) approaches zero there, potentially causing an improper integral.
Consider a comparison function that resembles the integrand near \(x=0\). Since \(e^{-\sqrt{x}}\) approaches \(e^0 = 1\) as \(x \to 0\), the integrand behaves like \(\frac{1}{\sqrt{x}}\) near zero.
Use the Direct Comparison Test by comparing \(\frac{e^{-\sqrt{x}}}{\sqrt{x}}\) with \(\frac{1}{\sqrt{x}}\). Recall that \(\int_0^1 \frac{1}{\sqrt{x}} \, dx\) converges or diverges? (Think about this to decide the comparison).
Based on the comparison, conclude whether the original integral converges or diverges. Optionally, you can also apply the Limit Comparison Test by computing \(\lim_{x \to 0^+} \frac{\frac{e^{-\sqrt{x}}}{\sqrt{x}}}{\frac{1}{\sqrt{x}}}\) to confirm the behavior near zero.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Improper Integrals

Improper integrals involve integration over intervals where the function is unbounded or the interval is infinite. In this problem, the integrand has a singularity at x = 0 due to the √x in the denominator, making it an improper integral that requires careful evaluation of limits.
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Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test determines convergence by comparing the given integral to another integral with a known behavior. If the integrand is less than or equal to a convergent function near the problematic point, the integral converges; if it is greater than or equal to a divergent function, it diverges.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test compares the given integrand to a simpler function by taking the limit of their ratio near the problematic point. If the limit is a positive finite number, both integrals share the same convergence behavior, helping to decide if the original integral converges or diverges.
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Limit Comparison Test
Related Practice
Textbook Question

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.

∫ √x e√x dx

Textbook Question

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.

∫₋₂¹ (1 / x⁴) dx

Textbook Question

Evaluate the integrals in Exercises 33–52.

∫ sec⁴(x) tan²(x) dx

Textbook Question

Equations (4) and (5) lead to different formulas for the integral of arctan x:

a. ∫ arctan x dx = x arctan x - ln sec(arctan x) + C [Eq. (4)]

b. ∫ arctan x dx = x arctan x - ln √(1 + x²) + C [Eq. (5)]

Can both integrations be correct? Explain.

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Textbook Question

Use any method to evaluate the integrals in Exercises 55–66.

∫ x / (x + √(x² + 2)) dx

Textbook Question

Use the formula ∫ f⁻¹(x) dx = x f⁻¹(x) - ∫ f(y) dy, y = f⁻¹(x)

To evaluate the integrals in Exercises 77-80. Express your answers in terms of x.

∫ log₂ x dx