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Improper Integral. Explain why 4. .1 dx +0 ,3 -1 * -

Improper Integral. Explain why 4. .1 dx +0 ,3 -1 * -

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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### Improper Integral **Problem 4:** Evaluate the integral and explain why it is not equal to zero: \[ \int_{-1}^{1} \frac{1}{x^3} \, dx \neq 0 \] ### Explanation This integral is considered "improper" due to the presence of a singularity within the integration limits. Specifically, the function \(\frac{1}{x^3}\) has a vertical asymptote at \(x = 0\), where the function is undefined. #### Steps to Analyze 1. **Identify the Singularity:** - At \(x = 0\), the function \(\frac{1}{x^3}\) becomes undefined since division by zero is not possible. This is why the integral from \(-1\) to \(1\) is improper. 2. **Consider a Two-Part Approach:** - Split the integral into two parts to handle the singularity: \[ \int_{-1}^{0} \frac{1}{x^3} \, dx + \int_{0}^{1} \frac{1}{x^3} \, dx \] - Both integrals must be considered separately by approaching the singularity as a limit. 3. **Evaluate as a Limit:** - For instance, evaluate: \[ \lim_{\epsilon \to 0^-} \int_{-1}^{\epsilon} \frac{1}{x^3} \, dx + \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} \frac{1}{x^3} \, dx \] - Typically, both limits will diverge due to the behavior near \(x = 0\), reinforcing why the integral does not equal zero. ### Conclusion This improper integral cannot be computed through standard means due to the asymmetric nature of the function around the point of discontinuity, \(x = 0\). The divergence upon approaching zero from either side leads to an undefined or infinite result, indicating that simply setting the integral equal to zero is incorrect.
Transcribed Image Text:### Improper Integral **Problem 4:** Evaluate the integral and explain why it is not equal to zero: \[ \int_{-1}^{1} \frac{1}{x^3} \, dx \neq 0 \] ### Explanation This integral is considered "improper" due to the presence of a singularity within the integration limits. Specifically, the function \(\frac{1}{x^3}\) has a vertical asymptote at \(x = 0\), where the function is undefined. #### Steps to Analyze 1. **Identify the Singularity:** - At \(x = 0\), the function \(\frac{1}{x^3}\) becomes undefined since division by zero is not possible. This is why the integral from \(-1\) to \(1\) is improper. 2. **Consider a Two-Part Approach:** - Split the integral into two parts to handle the singularity: \[ \int_{-1}^{0} \frac{1}{x^3} \, dx + \int_{0}^{1} \frac{1}{x^3} \, dx \] - Both integrals must be considered separately by approaching the singularity as a limit. 3. **Evaluate as a Limit:** - For instance, evaluate: \[ \lim_{\epsilon \to 0^-} \int_{-1}^{\epsilon} \frac{1}{x^3} \, dx + \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} \frac{1}{x^3} \, dx \] - Typically, both limits will diverge due to the behavior near \(x = 0\), reinforcing why the integral does not equal zero. ### Conclusion This improper integral cannot be computed through standard means due to the asymmetric nature of the function around the point of discontinuity, \(x = 0\). The divergence upon approaching zero from either side leads to an undefined or infinite result, indicating that simply setting the integral equal to zero is incorrect.
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