Explanation Given information: The given integral is ∫ 1 4 d x x . The value of N is 6. Formula Used: The integral is in the form of ∫ a b f ( x ) d x . The N th trapezoidal approximation is: T N = 1 2 Δ x ( x 0 + 2 x 1 + … + 2 x ( N − 1 ) + x N ) Where, Δ x = b − a N and x j = f ( a + j Δ x ) . The formula for the midpoint rule M N is: M N = Δ x ( f ( c 1 ) + f ( c 2 ) + … + f ( c N ) ) Where, Δ x = b − a N and c j = a + ( j − 1 2 ) Δ x . Calculation: The integral is I = ∫ 1 4 d x x . Compare the given integral with ∫ a b f ( x ) d x . So, the function be f ( x ) = 1 x . And, a = 1 , b = 4 . Find the value of Δ x . Δ x = b − a N = 4 − 1 6 = 3 6 = 1 2 To find the value of T 6 , first find the values of x 0 , x 1 , x 2 , x 3 , x 4 , x 5 and x 6 . Substitute 0 for j , 1 2 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 0 = f ( 1 + 0 ( 1 2 ) ) = f ( 1 ) = 1 1 = 1 Substitute 1 for j , 1 2 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 1 = f ( 1 + 1 ( 1 2 ) ) = f ( 3 2 ) = 1 ( 3 2 ) = 2 3 Substitute 2 for j , 1 2 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 2 = f ( 1 + 2 ( 1 2 ) ) = f ( 2 ) = 1 2 Substitute 3 for j , 1 2 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 3 = f ( 1 + 3 ( 1 2 ) ) = f ( 5 2 ) = 1 5 2 = 2 5 Substitute 4 for j , 1 2 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 4 = f ( 1 + 4 ( 1 2 ) ) = f ( 3 ) = 1 3 Substitute 5 for j , 1 2 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 5 = f ( 1 + 5 ( 1 2 ) ) = f ( 7 2 ) = 1 7 2 = 2 7 Substitute 6 for j , 1 2 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 6 = f ( 1 + 6 ( 1 2 ) ) = f ( 4 ) = 1 4 Now, substitute 1 2 for Δ x , 1 for x 0 , 2 3 for x 1 , 1 2 for x 2 , 2 5 for x 3 , 1 3 for x 4 , 2 7 for x 5 , and 1 4 for x 6 in the expression T N = 1 2 Δ x ( x 0 + 2 x 1 + … + 2 x ( N − 1 ) + x N ) . T 6 = 1 2 ⋅ 1 2 ( 1 + 2 ( 2 3 ) + 2 ( 1 2 ) + 2 ( 2 5 ) + 2 ( 1 3 ) + 2 ( 2 7 ) + 1 4 ) = 1 4 ( 1 + 4 3 + 1 + 4 5 + 2 3 + 4 7 + 1 4 ) = 1 4 ( 4 + 4 5 + 4 7 + 1 4 ) = 1 4 ( 787 140 ) = 1.405 Thus, the value of T 6 is 1.405 . Now, find the value of M N . To find the value of M 6 , first find the values of c 1 , c 2 , c 3 , c 4 , c 5 and c 6 . Substitute 1 for j , 1 2 for Δ x and 1 for a in the formula c j = a + ( j − 1 2 ) Δ x

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.9, Problem 5E

To determine

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The value of MN and TN

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Chapter 8.9, Problem 4E

Chapter 8.9, Problem 6E

Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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