Explanation It can be stated from the property of modulus that, − | f ( x ) | ≤ f ( x ) ≤ | f ( x ) | Applying the property of inequality, ∫ a b − | f ( x ) | d x ≤ ∫ a b f ( x ) d x ≤ ∫ a b | f ( x ) | d x − ∫ a b | f ( x ) | d x ≤ ∫ a b f ( x ) d x ≤ ∫ a

14th Edition

ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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Chapter 4.6, Problem 79P

To determine

To prove: For all the real numbers a and b with a < b,

|∫abf(x)dx|≤∫ab|f(x)|dx

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Chapter 4.6, Problem 78P

Chapter 4.6, Problem 80P

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Chapter 4 Solutions

Loose-leaf Version for Calculus

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To prove: For all the real numbers a and b with a < b, | ∫ a b f ( x ) d x | ≤ ∫ a b | f ( x ) | d x

Solution Summary: The author explains how the property of inequality can be applied to all the real numbers a and b.

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To prove: For all the real numbers a and b with a < b,

|∫abf(x)dx|≤∫ab|f(x)|dx

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Chapter 4 Solutions

To prove: For all the real numbers a and b with a &lt; b, | ∫ a b f ( x ) d x | ≤ ∫ a b | f ( x ) | d x

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To prove: For all the real numbers a and b with a < b, |∫abf(x)dx|≤∫ab|f(x)|dx

Want to see the full answer?

Chapter 4 Solutions

To prove: For all the real numbers a and b with a < b, | ∫ a b f ( x ) d x | ≤ ∫ a b | f ( x ) | d x

To prove: For all the real numbers a and b with a < b,

|∫abf(x)dx|≤∫ab|f(x)|dx