Explanation Given information: ∫ 1 5 ( x − 1 ) 2 d x with n = 2 , 4 where n is the number of equal subintervals. Formula used: Midpoint Rule: ∫ a b f ( x ) d x ≈ [ f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n ) ] Δ x Where x 1 , x 2 , ... , x n are mid-points of the subintervals of the intervals a ≤ x ≤ b and Δ x = b − a n Calculation: For n = 2 Compare ∫ a b f ( x ) d x to ∫ 1 5 ( x − 1 ) 2 d x ; it becomes a = 1 and b = 5 Therefore, Δ x = b − a n = 5 − 1 2 = 2 . Now, a 0 = a 1 = 0 x 1 = a 0 + Δ x 2 = 1 + 2 2 = 2 x 2 = x 1 + Δ x = 2 + 2 = 4 By using the mid-point rule, ∫ 1 5 ( x − 1 ) 2 d x = [ f ( 2 ) + f ( 4 ) ] × 2 Here, f ( x ) = ( x − 1 ) 2 At x = 2 ⇒ f ( 2 ) = ( 2 − 1 ) 2 = 1 At x = 4 ⇒ f ( 4 ) = ( 4 − 1 ) 2 = 9 Substituting all the values in the formula, it becomes ∫ 1 5 ( x − 1 ) 2 d x = 2 [ 1 + 9 ] = 20 For n = 4 Compare ∫ a b f ( x ) d x to ∫ 1 5 ( x − 1 ) 2 d x ; it becomes a = 1 and b = 5 Therefore, Δ x = b − a n = 5 − 1 4 = 1 . Now, a 0 = a 1 = 0 x 1 = a 0 + Δ x 2 = 1 + 1 2 = 1.5 x 2 = x 1 + Δ x = 1.5 + 1 = 2.5 x 3 = x 2 + Δ x = 2.5 + 1 = 3.5 x 4 = x 3 + Δ x = 3.5 + 1 = 4.5 By using the mid-point rule, ∫ 1 5 ( x − 1 ) 2 d x = [ f ( 1

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Chapter 9.4, Problem 8E

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To calculate: The approximate value of the integral ∫15(x−1)2dx by midpoint rule with n=2 , 4 and exact value of the integral ∫15(x−1)2dx. Express the answer to five decimal places.

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Chapter 9.4, Problem 7E

Chapter 9.4, Problem 9E

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The approximate value of the integral ∫ 1 5 ( x − 1 ) 2 d x by midpoint rule with n = 2 , 4 and exact value of the integral ∫ 1 5 ( x − 1 ) 2 d x . Express the answer to five decimal places.

Solution Summary: The author calculates the approximate value of the integral by midpoint rule with n=2, 4 and the exact value by expressing the answer to five decimal places.

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To calculate: The approximate value of the integral ∫15(x−1)2dx by midpoint rule with n=2 , 4 and exact value of the integral ∫15(x−1)2dx. Express the answer to five decimal places.

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Chapter 9 Solutions