Explanation Formula used: Left Riemann sum for a given function f ( x ) on an interval [ a , b ] is given by, S L = ∑ i = 1 n m i ( x i − x i − 1 ) Where [ a , b ] is partitioned in n parts like [ a = x 0 < x 1 < x 2 < x 3 .............. < x n − 1 < x n = b ] . m i = inf f ( x ) in ( x i − 1 ≤ x ≤ x i ) and ∫ a b f ( x ) d x < S L . Right Riemann sum for a given function f ( x ) on an interval [ a , b ] is given by, S R = ∑ i = 1 n M i ( x i − x i − 1 ) Where [ a , b ] is partitioned in n parts like [ a = x 0 < x 1 < x 2 < x 3 .............. < x n − 1 < x n = b ] . M i = sup f ( x ) in ( x i − 1 ≤ x ≤ x i ) and ∫ a b f ( x ) d x > S R . Given: The function is f ( t ) = 1 t . The function is ln 2 = ∫ 1 2 d t t . The function is ln 3 = ∫ 1 3 d t t . Calculation: Case 1: Given n = 2 . Calculate left Riemann sum for a given function f ( t ) = 1 t for the interval [ 1 , 2 ] . Then, x 0 = 1 , x 1 = 1.5 , x 2 = 2 . Thus, m 1 = inf f ( x ) in ( x 0 ≤ x ≤ x 1 ) = 1 1 = 1 m 2 = inf f ( x ) in ( x 1 ≤ x ≤ x 2 ) = 1 1.5 = 2 3 Hence, the left Riemann sum for a given function f ( t ) = 1 t is, S L = ∑ i = 1 2 m i ( x i − x i − 1 ) = m 1 ( x 1 − x 0 ) + m 2 ( x 2 − x 1 ) Substitute the value of m 1 , m 2 , x 1 , x 2 in S L , S L = 1 ( 1.5 − 1 ) + 2 3 ( 2 − 1.5 ) = 1 2 + 1 3 = 5 6 < 1 Hence, S L < 1 . Given, ln 2 = ∫ 1 2 d t t and ∫ 1 2 d t t < S L . Thus, ln 2 < 1 . Case 2: Given n = 7 . Calculate the right Riemann sum for a given function f ( t ) = 1 t for the interval [ 1 , 3 ] . Use n = 8 for better approximation. Then, x 0 = 1 , x 1 = 1.25 , x 2 = 1.5 , x 3 = 1.75 , x 4 = 2 , x 5 = 2.25 , x 6 = 2.5 , x 7 = 2.75 , x 8 = 3 . Thus, M 1 = sup f ( x ) in ( x 0 ≤ x ≤ x 1 ) = 1 1.25 = 1 ( 5 4 ) = 4 5 M 2 = inf f ( x ) in ( x 1 ≤ x ≤ x 2 ) = 1 1

Calculus: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134763644

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 7.1, Problem 75E

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To show: The result ln2<1 by approximating and ln2=∫12dtt ln3>1 by approximating ln3=∫13dtt by using Riemann sum.

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Chapter 7.1, Problem 74E

Chapter 7.1, Problem 76E

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Chapter 7 Solutions

Calculus: Early Transcendentals (3rd Edition)

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The result ln 2 < 1 by approximating and ln 2 = ∫ 1 2 d t t ln 3 > 1 by approximating ln 3 = ∫ 1 3 d t t by using Riemann sum.

Solution Summary: The author illustrates Riemann sum by approximating mathrmln21 by using the formula.

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To show: The result ln2<1 by approximating and ln2=∫12dtt ln3>1 by approximating ln3=∫13dtt by using Riemann sum.

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Chapter 7 Solutions

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The result ln 2 &lt; 1 by approximating and ln 2 = ∫ 1 2 d t t ln 3 &gt; 1 by approximating ln 3 = ∫ 1 3 d t t by using Riemann sum.

Solution Summary: The author illustrates Riemann sum by approximating mathrmln21 by using the formula.

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To show: The result ln2<1 by approximating and ln2=∫12dtt ln3>1 by approximating ln3=∫13dtt by using Riemann sum.

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Chapter 7 Solutions

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The result ln 2 < 1 by approximating and ln 2 = ∫ 1 2 d t t ln 3 > 1 by approximating ln 3 = ∫ 1 3 d t t by using Riemann sum.