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10 Why is J 5 Vx-5 11 =dx an improper integral?

10 Why is J 5 Vx-5 11 =dx an improper integral?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The image contains a mathematical expression that asks: "Why is the integral from 5 to 10 of \(\frac{11}{\sqrt{x-5}} \, dx\) an improper integral?" ### Explanation: An integral is considered improper for one of the following reasons: 1. **Infinite Limits of Integration:** If the interval of integration is infinite. 2. **Discontinuities in the Interval:** If there is a point within the interval where the function becomes undefined or approaches infinity. For this specific integral: - The function \(\frac{11}{\sqrt{x-5}}\) becomes undefined at \(x = 5\) because \(\sqrt{x-5}\) equals zero, leading to a division by zero. This creates a vertical asymptote at \(x = 5\), making the function approach infinity as \(x\) approaches 5 from the right. Thus, the integral is improper due to the discontinuity at the lower limit of integration (\(x = 5\)).
Transcribed Image Text:The image contains a mathematical expression that asks: "Why is the integral from 5 to 10 of \(\frac{11}{\sqrt{x-5}} \, dx\) an improper integral?" ### Explanation: An integral is considered improper for one of the following reasons: 1. **Infinite Limits of Integration:** If the interval of integration is infinite. 2. **Discontinuities in the Interval:** If there is a point within the interval where the function becomes undefined or approaches infinity. For this specific integral: - The function \(\frac{11}{\sqrt{x-5}}\) becomes undefined at \(x = 5\) because \(\sqrt{x-5}\) equals zero, leading to a division by zero. This creates a vertical asymptote at \(x = 5\), making the function approach infinity as \(x\) approaches 5 from the right. Thus, the integral is improper due to the discontinuity at the lower limit of integration (\(x = 5\)).
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