Explanation Given: m and M are two constants Formula used: Dominanace Property of Sums: The numbers a k and b k satisfy a k ≤ b k for k = 1 , 2.... , n then ∑ k = 1 n a k ≤ ∑ k = 1 n b k . Riemann sum associated with a function f ( x ) at the sub-interval representative x k ∗ is given by ∫ a b f ( x ) d x = lim n → ∞ ∑ k = 1 n f ( x k ) Δ x Here, Δ x = b − a n and x k = a + k Δ x Proof: Consider the inequality m ≤ f ( x ) ≤ M Replace x by x k . m ≤ f ( x k ) ≤ M Multiply both sides by Δ x . m Δ x ≤ f ( x k ) Δ x ≤ M Δ x Now, using Dominance property of sums, we get ∑ k = 1 n m Δ x ≤ ∑ k = 1 n f ( x k ) Δ x ≤ ∑ k = 1 n M Δ x Now, substitute Δ x = b − a n

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 5.3, Problem 59PS

To determine

To prove:If f(x) integrable throughout the interval [a,b] and m≤f(x)≤M then m(b−a)≤∫abf(x)dx≤M(b−a) .

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Chapter 5.3, Problem 58PS

Chapter 5.3, Problem 60PS

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Chapter 5 Solutions

Calculus

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To prove: If f ( x ) integrable throughout the interval [ a , b ] and m ≤ f ( x ) ≤ M then m ( b − a ) ≤ ∫ a b f ( x ) d x ≤ M ( b − a ) .

Solution Summary: The author explains that m and M are two constants Formula used: Dominanace Property of Sums.

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To prove:If f(x) integrable throughout the interval [a,b] and m≤f(x)≤M then m(b−a)≤∫abf(x)dx≤M(b−a) .

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