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17) Draw the graph of a function K(m), such that S К (т) dm <0, and S, К (т) dm > о

17) Draw the graph of a function K(m), such that S К (т) dm <0, and S, К (т) dm > о

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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**Transcription:** 17) Draw the graph of a function \( K(m) \), such that - \( \int_{-3}^{4} K(m) \, dm < 0 \), and - \( \int_{-2}^{4} K(m) \, dm > 0 \) **Explanation of the Problem:** This problem asks you to sketch a graph of a function \( K(m) \) that satisfies two conditions based on definite integrals: 1. The integral of \( K(m) \) from \(-3\) to \(4\) is negative, meaning the net area under the curve \( K(m) \) between these points is below the x-axis. 2. The integral of \( K(m) \) from \(-2\) to \(4\) is positive, meaning the net area under the curve \( K(m) \) between these points is above the x-axis. To satisfy these conditions, you will need to design the function \( K(m) \) in such a way that the area below the x-axis between \(-3\) and \(-2\) is large and negative enough to make the entire integral over \(-3\) to \(4\) negative, while ensuring the area from \(-2\) to \(4\) is predominantly above the x-axis to result in a positive integral. Consider using piecewise functions or strategically placing zeros or changes in direction to achieve this.
Transcribed Image Text:**Transcription:** 17) Draw the graph of a function \( K(m) \), such that - \( \int_{-3}^{4} K(m) \, dm < 0 \), and - \( \int_{-2}^{4} K(m) \, dm > 0 \) **Explanation of the Problem:** This problem asks you to sketch a graph of a function \( K(m) \) that satisfies two conditions based on definite integrals: 1. The integral of \( K(m) \) from \(-3\) to \(4\) is negative, meaning the net area under the curve \( K(m) \) between these points is below the x-axis. 2. The integral of \( K(m) \) from \(-2\) to \(4\) is positive, meaning the net area under the curve \( K(m) \) between these points is above the x-axis. To satisfy these conditions, you will need to design the function \( K(m) \) in such a way that the area below the x-axis between \(-3\) and \(-2\) is large and negative enough to make the entire integral over \(-3\) to \(4\) negative, while ensuring the area from \(-2\) to \(4\) is predominantly above the x-axis to result in a positive integral. Consider using piecewise functions or strategically placing zeros or changes in direction to achieve this.
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