Explanation Given: For | x | < 1 , ∫ d x x 4 + 1 = x − x 5 5 + x 9 9 − ... Proof: Let 1 1 + x = ∑ n = 0 ∞ ( − 1 ) n x n = 1 − x + x 2 − x 3 + ... ( 1 ) Now, substitute x = x 4 in the equation (1) and we get: 1 1 + x 4 = 1 − x 4 + x 8 − x 12 + ... ( 2 ) Now, integrate equation (2) on both the side and we get: ∫ 1 1 + x 4 d x = ∫ ( 1 − x 4 + x 8 − x 12 + ...

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 11.6, Problem 52E

To determine

To prove:

For $| x | < 1$ ,

$\int \frac{d x}{x^{4} + 1} = x - \frac{x^{5}}{5} + \frac{x^{9}}{9} - ...$

Use the first two terms to approximate $\int_{0}^{\frac{1}{2}} \frac{d x}{(x^{4} + 1)}$ numerically. Use an alternating series to show that the error in this approximation is at most $0.00022$ .

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Chapter 11.6, Problem 51E

Chapter 11.6, Problem 53E

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Chapter 11 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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To prove: For | x | < 1 , ∫ d x x 4 + 1 = x − x 5 5 + x 9 9 − ... Use the first two terms to approximate ∫ 0 1 2 d x ( x 4 + 1 ) numerically. Use an alternating series to show that the error in this approximation is at most 0.00022 .

Solution Summary: The author explains how the first two terms are used to approximate 0 1 2 d x numerically. The error in this approximation is at most 0.00022.

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To prove:

For $| x | < 1$ ,

$\int \frac{d x}{x^{4} + 1} = x - \frac{x^{5}}{5} + \frac{x^{9}}{9} - ...$

Use the first two terms to approximate $\int_{0}^{\frac{1}{2}} \frac{d x}{(x^{4} + 1)}$ numerically. Use an alternating series to show that the error in this approximation is at most $0.00022$ .

Want to see the full answer?

Chapter 11 Solutions

To prove: For | x | &lt; 1 , ∫ d x x 4 + 1 = x − x 5 5 + x 9 9 − ... Use the first two terms to approximate ∫ 0 1 2 d x ( x 4 + 1 ) numerically. Use an alternating series to show that the error in this approximation is at most 0.00022 .

Solution Summary: The author explains how the first two terms are used to approximate 0 1 2 d x numerically. The error in this approximation is at most 0.00022.

Videos

To prove: For |x|<1, ∫dxx4+1=x−x55+x99−... Use the first two terms to approximate ∫012dx(x4+1) numerically. Use an alternating series to show that the error in this approximation is at most 0.00022.

Want to see the full answer?

Chapter 11 Solutions

To prove: For | x | < 1 , ∫ d x x 4 + 1 = x − x 5 5 + x 9 9 − ... Use the first two terms to approximate ∫ 0 1 2 d x ( x 4 + 1 ) numerically. Use an alternating series to show that the error in this approximation is at most 0.00022 .