Explanation Given information: ∫ 0 3 x 4 − x d x With n = 5 . where n is number of equal subintervals. Formula used: 1) Midpoint Rule: M = ∫ 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 . 2) Trapezoidal Rule: T = ∫ a b f ( x ) d x ≈ [ f ( a 0 ) + 2 f ( a 1 ) + ... + 2 f ( a n − 1 ) + f ( a n ) ] Δ x 2 Where, a 0 , a 1 , ... , a n are end points of subintervals a ≤ x ≤ b and Δ x = b − a n . 3) Simpson’s Rule: ∫ a b f ( x ) ≈ S = 2 M + T 3 . Where, M the approximate value of integral is calculated by midpoint rule and T is the approximate value of integral calculated by Trapezoidal rule. 4) Integration by parts: ∫ f ( x ) g ( x ) d x = f ( x ) G ( x ) − ∫ f ′ ( x ) G ( x ) d x Where, f ' ( x ) is derivative of f ( x ) and G ( x ) is an antiderivative of g ( x ) . Calculation: For n = 5 Compare ∫ a b f ( x ) d x with ∫ 0 3 x 4 − x d x it becomes a = 0 and b = 3 Therefore, Δ x = b − a n = 3 − 0 5 = 0.6 Now, a 0 = a ⇒ a 0 = 0 x 1 = a 0 + Δ x 2 = 0 + 0.6 2 = 0.3 x 2 = x 1 + Δ x = 0.3 + 0.6 = 0.9 x 3 = x 2 + Δ x = 0.9 + 0.6 = 1.5 x 4 = x 3 + Δ x = 1.5 + 0.6 = 2.1 x 5 = x 4 + Δ x = 2.1 + 0.6 = 2.7 By using mid-point rule, M = ∫ 0 3 x 4 − x d x = [ f ( 0.3 ) + f ( 0.9 ) + f ( 1.5 ) + f ( 2.1 ) + f ( 2.7 ) ] × 0.6 = [ 0.3 4 − 0.3 + 0.9 4 − 0.9 + 1.5 4 − 1.5 + 2.1 4 − 2.1 + 2.7 4 − 2.7 ] × 0.6 ≈ 6.30390 ⇒ M = ∫ 0 3 x 4 − x d x ≈ 6.30390 Also, a 0 = a ⇒ a 0 = 0 a 1 = a 0 + Δ x = 0 + 0.6 = 0.6 a 2 = a 1 + Δ x = 0

Calculus & Its Applications (14th Edition)

14th Edition

ISBN: 9780134437774

Author: Larry J. Goldstein, David C. Lay, David I. Schneider, Nakhle H. Asmar

Publisher: PEARSON

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Chapter 9.4, Problem 18E

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To calculate: The approximate value of integral ∫03x4−xdx by midpoint rule, trapezoidal rule, Simson’s rule with n=5 and find the exact value of integration. Express answers to five decimal places.

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Chapter 9.4, Problem 17E

Chapter 9.4, Problem 19E

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According to Newton's law of universal gravitation, the force F between two bodies of constant mass GmM m and M is given by the formula F = , where G is the gravitational constant and d is the d² distance between the bodies. a. Suppose that G, m, and M are constants. Find the rate of change of force F with respect to distance d. F' (d) 2GmM b. Find the rate of change of force F with gravitational constant G = 6.67 × 10-¹¹ Nm²/kg², on two bodies 5 meters apart, each with a mass of 250 kilograms. Answer in scientific notation, rounding to 2 decimal places. -6.67x10 N/m syntax incomplete.

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Calculus & Its Applications (14th Edition)

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The approximate value of integral ∫ 0 3 x 4 − x d x by midpoint rule, trapezoidal rule, Simson’s rule with n = 5 and find the exact value of integration . Express answers to five decimal places.

Solution Summary: The author calculates the approximate value of integral by midpoint rule, trapezoidal rule and Simson’s rule with n=5.

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To calculate: The approximate value of integral ∫03x4−xdx by midpoint rule, trapezoidal rule, Simson’s rule with n=5 and find the exact value of integration. Express answers to five decimal places.

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Chapter 9 Solutions