Explanation Determine the end points and use those to determine the value of M 10 and S 10 for the integral. ∫ 0 π 1 − x 2 d x Given: Sometimes Simpson’s Rule performs poorly – calculate M 10 and S 10 for the integral ∫ 0 π 1 − x 2 d x whose value we know to be π 4 (one quarter of the area of the unit circle). Calculation: Given ∫ 0 π 1 − x 2 d x Let f ( x ) = 1 − x 2 The value of Δ x = 0.1 M 10 = 1 10 [ 1 − 0.05 2 + 1 − 0... To determine (b) How to explain the result of part (a) using the hint that the error bounds are not valid because | f ' ' ( x ) | and | f ( 4 ) ( x ) | tend to ∞ as x → 1 but | f ( 4 ) ( x ) | goes to infinity more quickly.

4th Edition

ISBN: 9781319308858

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 7.8, Problem 66E

To determine

(a)

We usually expect the value of $S_{N}$ to be more accurate than $M_{N}$ . Which of $M_{10}$ and $S_{10}$ is more accurate in this case?

To determine

(b)

How to explain the result of part (a) using the hint that the error bounds are not valid because $| f'' (x) |$ and $| f^{(4)} (x) |$ tend to $\infty$ as $x \to 1$ but $| f^{(4)} (x) |$ goes to infinity more quickly.

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Chapter 7.8, Problem 65E

Chapter 7, Problem 1CRE

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Chapter 7 Solutions

CALCULUS:EARLY TRANS.(LL)-W/ACCESS

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Knowledge Booster

(a) We usually expect the value of S N to be more accurate than M N . Which of M 10 and S 10 is more accurate in this case?

(a) We usually expect the value of SNto be more accurate than MN. Which of M10 and S10 is more accurate in this case?

(b) How to explain the result of part (a) using the hint that the error bounds are not valid because |f''(x)| and |f(4)(x)| tend to ∞ as x→1 but |f(4)(x)| goes to infinity more quickly.

Want to see the full answer?

Chapter 7 Solutions

(a)

We usually expect the value of $S_{N}$ to be more accurate than $M_{N}$ . Which of $M_{10}$ and $S_{10}$ is more accurate in this case?

(b)

How to explain the result of part (a) using the hint that the error bounds are not valid because $| f'' (x) |$ and $| f^{(4)} (x) |$ tend to $\infty$ as $x \to 1$ but $| f^{(4)} (x) |$ goes to infinity more quickly.