Explanation Given information: The given integral is ∫ − 2 − 1 d x x . The value of N is 5. 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 = ∫ − 2 − 1 d x x . Compare the given integral with ∫ a b f ( x ) d x . So, the function be f ( x ) = 1 x . And, a = − 2 , b = − 1 . Find the value of Δ x . Δ x = b − a N = − 1 + 2 5 = 1 5 To find the value of T 5 , first find the values of x 0 , x 1 , x 2 , x 3 , x 4 and x 5 . Substitute 0 for j , 1 5 for Δ x and − 2 for a in the formula x j = f ( a + j Δ x ) . x 0 = f ( − 2 + 0 ( 1 5 ) ) = f ( − 2 ) = 1 − 2 = − 0.5 Substitute 1 for j , 1 5 for Δ x and − 2 for a in the formula x j = f ( a + j Δ x ) . x 1 = f ( − 2 + 1 ( 1 5 ) ) = f ( − 2 + 1 5 ) = f ( − 9 5 ) = − 1 9 5 = − 0.556 Substitute 2 for j , 1 5 for Δ x and − 2 for a in the formula x j = f ( a + j Δ x ) . x 2 = f ( − 2 + 2 ( 1 5 ) ) = f ( − 2 + 2 5 ) = f ( − 8 5 ) = − 1 8 5 = − 0.625 Substitute 3 for j , 1 5 for Δ x and − 2 for a in the formula x j = f ( a + j Δ x ) . x 3 = f ( − 2 + 3 ( 1 5 ) ) = f ( − 2 + 3 5 ) = f ( − 7 5 ) = − 1 7 5 = − 0.714 Substitute 4 for j , 1 5 for Δ x and − 2 for a in the formula x j = f ( a + j Δ x ) . x 4 = f ( − 2 + 4 ( 1 5 ) ) = f ( − 2 + 4 5 ) = f ( − 6 5 ) = − 1 6 5 = − 0.833 Substitute 5 for j , 1 5 for Δ x and − 2 for a in the formula x j = f ( a + j Δ x ) . x 5 = f ( − 2 + 5 ( 1 5 ) ) = f ( − 2 + 1 ) = f ( − 1 ) = − 1 1 = − 1 Now, substitute 1 5 for Δ x , − 0.5 for x 0 , − 0.556 for x 1 , − 0.625 for x 2 , − 0.714 for x 3 , − 0.833 for x 4 and − 1 for x 5 in the expression T N = 1 2 Δ x ( x 0 + 2 x 1 + … + 2 x ( N − 1 ) + x N ) . T 5 = 1 2 ⋅ 1 5 ( − 0.5 + 2 ( − 0.556 ) + 2 ( 0.625 ) + 2 ( − 0.714 ) + 2 ( − 0.833 ) − 1 ) = 1 10 ( − 0.5 − 1.112 − 1.25 − 1.428 − 1.667 − 1 ) = 1 10 ( − 6.957 ) = − 0

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.9, Problem 6E

To determine

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The value of MN and TN

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Chapter 8.9, Problem 5E

Chapter 8.9, Problem 7E

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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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