Explanation Given: In the case that f ( x ) is increasing, Figure 20 shows that the part of the error in R N due to the ith rectangle is the sum of the areas A+B+D , and for L N it is |B- E|. On the other hand, A ≥ E Calculation: Suppose f ( x ) is monotonic increasing on the interval [a, b], Δ x = b − a N , { x k } k = 0 N = { a , a + Δ x , a + 2 Δ x , ... , a + ( N − 1 ) Δ x , b } And { x k * } k = 0 N − 1 = { a + ( a + Δ x ) 2 , ( a + Δ x ) + ( a + 2 Δ x ) 2 , ... , ( a + ( N − 1 ) Δ x ) + b 2 } . Note that x i < x i * < x i + 1 implies f ( x i ) < f ( x i * ) < f ( x i + 1 ) for all 0 ≤ i < N because f ( x ) is monotone increasing. Then ( L N = b − a N ∑ k = 0 N − 1 f ( x k ) ) < ( M N = b − a N ∑ k = 0 N − 1 f ( x k * ) ) < ( R N = b − a N ∑ k = 1 N f ( x k ) ) Similarly, if f ( x ) is monotone decreasing, ( L N = b − a N ∑ k = 0 N − 1 f ( x k ) ) > ( M N = b − a N ∑ k = 0 N − 1 f ( x k * ) ) > ( R N = b − a N ∑ k = 1 N f ( x k ) ) Thus, if f ( x ) is monotonic, then M N always lies in between R N and L N ...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 5.1, Problem 95E

To determine

To prove that if $f (x)$ is positive and monotonic, then $M_{N}$ lies between $R_{N}$ and $L_{N}$ and is closer to the actual area under the graph than both $R_{N}$ and $L_{N}$ .

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Chapter 5.1, Problem 94E

Chapter 5.2, Problem 1PQ

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Chapter 5 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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To prove that if f ( x ) is positive and monotonic, then M N lies between R N and L N and is closer to the actual area under the graph than both R N and L N .

Solution Summary: The author explains that if f ( x ) is positive and monotonic, then M N lies between R N and L N.

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To prove that if $f (x)$ is positive and monotonic, then $M_{N}$ lies between $R_{N}$ and $L_{N}$ and is closer to the actual area under the graph than both $R_{N}$ and $L_{N}$ .

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Chapter 5 Solutions

To prove that if f ( x ) is positive and monotonic, then M N lies between R N and L N and is closer to the actual area under the graph than both R N and L N .

Solution Summary: The author explains that if f ( x ) is positive and monotonic, then M N lies between R N and L N.

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To prove that if f(x)is positive and monotonic, then MN lies between RN and LN and is closer to the actual area under the graph than both RN and LN.

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Chapter 5 Solutions

To prove that if $f (x)$ is positive and monotonic, then $M_{N}$ lies between $R_{N}$ and $L_{N}$ and is closer to the actual area under the graph than both $R_{N}$ and $L_{N}$ .