Explanation Given: g ( t ) = 1 1 + t 2 − t 1 + t 2 Calculation ∫ 0 1 g ( t ) d t = ( tan − 1 t − 1 2 ln ( t 2 + 1 ) ) | 1 0 To determine To prove: S = 1 − 1 2 − 1 3 − 1 4 + 1 5 − 1 6

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 11.7, Problem 87E

To determine

(a)To prove: $\int_{0}^{1} g (t) d t$ $= \frac{π}{4} - \frac{1}{2} \ln 2$

To determine

To prove:

$S = 1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}$

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Chapter 11.7, Problem 86E

Chapter 11.7, Problem 88E

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Chapter 11 Solutions

Calculus - Standalone book

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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(a)To prove: ∫ 0 1 g ( t ) d t = π 4 − 1 2 ln 2

Solution Summary: The author explains that g ( t ) = 1 1 + 2 1 2 + 1 0 = the Taylor series.

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(a)To prove: $\int_{0}^{1} g (t) d t$ $= \frac{π}{4} - \frac{1}{2} \ln 2$

To prove:

$S = 1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}$

Want to see the full answer?

Chapter 11 Solutions

(a)To prove: ∫ 0 1 g ( t ) d t = π 4 − 1 2 ln 2

Solution Summary: The author explains that g ( t ) = 1 1 + 2 1 2 + 1 0 = the Taylor series.

Concept explainers

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(a)To prove: ∫01g(t)dt =π4−12ln2

To prove: S=1−12−13−14+15−16

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Chapter 11 Solutions

(a)To prove: $\int_{0}^{1} g (t) d t$ $= \frac{π}{4} - \frac{1}{2} \ln 2$

To prove:

$S = 1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}$