Explanation Given information: The given integral is ∫ 1 2 x 4 + 1 d 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 = ∫ 1 2 x 4 + 1 d x . Compare the given integral with ∫ a b f ( x ) d x . So, the function be f ( x ) = x 4 + 1 . And, a = 1 , b = 2 . Find the value of Δ x . Δ x = b − a N = 2 − 1 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 1 for a in the formula x j = f ( a + j Δ x ) . x 0 = f ( 1 + 0 ( 1 5 ) ) = f ( 1 ) = ( 1 ) 4 + 1 = 2 = 1.414 Substitute 1 for j , 1 5 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 1 = f ( 1 + 1 ( 1 5 ) ) = f ( 1 + 1 5 ) = ( 6 5 ) 4 + 1 = 2.0736 + 1 = 1.753 Substitute 2 for j , 1 5 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 2 = f ( 1 + 2 ( 1 5 ) ) = f ( 1 + 2 5 ) = ( 7 5 ) 4 + 1 = 3.8416 + 1 = 2.2004 Substitute 3 for j , 1 5 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 3 = f ( 1 + 3 ( 1 5 ) ) = f ( 1 + 3 5 ) = ( 8 5 ) 4 + 1 = 6.5536 + 1 = 2.748 Substitute 4 for j , 1 5 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 4 = f ( 1 + 4 ( 1 5 ) ) = f ( 1 + 4 5 ) = ( 9 5 ) 4 + 1 = 10.4976 + 1 = 3.3908 Substitute 5 for j , 1 5 for Δ x and 1 for a in the formula x j = f ( a + j Δ x ) . x 5 = f ( 1 + 5 ( 1 5 ) ) = f ( 1 + 1 ) = ( 2 ) 4 + 1 = 16 + 1 = 4.1231 Now, substitute 1 5 for Δ x , 1.414 for x 0 , 1.753 for x 1 , 2.2004 for x 2 , 2.748 for x 3 , 3.3908 for x 4 and 4.1231 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 ( 1.414 + 2 ( 1.753 ) + 2 ( 2.2004 ) + 2 ( 2.748 ) + 2 ( 3.3908 ) + 4.1231 ) = 1 10 ( 1.414 + 3.506 + 4.4008 + 5.496 + 6.7816 + 4.1231 ) = 1 10 ( 25.7215 ) = 2

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.9, Problem 4E

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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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