Explanation Given information: The interval, 3 ≤ x ≤ 5 is divides in n = 5 . Explanation: By comparing the interval 3 ≤ x ≤ 5 to a ≤ x ≤ b it becomes, a = 3 and b = 5 . The length of each subinterval, Δ x = b − a n = 5 − 3 5 = 2 5 = 0.4 Consider the interval, a ≤ x ≤ b If the interval is divided in n subintervals, the endpoints of the subinterval are a 0 , a 1 , ... , a n . Where a 0 = a , a 1 = a 0 + Δ x , a 2 = a 1 + Δ x , .... , a n = a n − 1 + Δ x Now, a 0 = a ⇒ a 0 = 3 a 1 = a 0 + Δ x = 3 + 0

Student Solutions Manual for Calculus & Its Applications and Calculus & Its Applications, Brief Version

14th Edition

ISBN: 9780134463230

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

Publisher: PEARSON

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Chapter 9.4, Problem 1E

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The value of Δx and the end points a0,a1,a2...,an of the subintervals where theinterval 3≤x≤5 is divided in n=5 subintervals.

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Chapter 9.4, Problem 5CYU

Chapter 9.4, Problem 2E

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Chapter 9 Solutions

Student Solutions Manual for Calculus & Its Applications and Calculus & Its Applications, Brief Version

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The value of Δ x and the end points a 0 , a 1 , a 2 ... , a n of the subintervals where theinterval 3 ≤ x ≤ 5 is divided in n = 5 subintervals.

Solution Summary: The author explains the value of Delta x and the end points of the subintervals.

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The value of Δx and the end points a0,a1,a2...,an of the subintervals where theinterval 3≤x≤5 is divided in n=5 subintervals.

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Chapter 9 Solutions