Explanation Given: The integral: I = ∫ 1 ∞ d x x 4 + 1 Formula used: I = ∫ a b f ( x ) d x = [ F ( x ) ] a b = F ( b ) − F ( a ) I = ∫ a b f ( x ) d x = ∫ a c 1 f ( x ) d x + ∫ c 1 c 2 f ( x ) d x + ... + ∫ c n − 1 b f ( x ) d x , Where a < c 1 < c 2 < ... < c n − 1 < b Calculation: The integral we have I = ∫ 1 ∞ d x x 4 + 1 Since, x 4 ≥ x 2 ∀ x ∈ R ⇒ x 4 + 1 > x 2 ⇒ 1 x 4 + 1 < 1 x 2 ⇒ 1 x 4 + 1 < 1 x 2

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Question

Chapter 8.7, Problem 54E

To determine

To Show:

$\frac{1}{\sqrt{x^{4} + 1}} < \frac{1}{x^{2}} \forall x \in R$

And value of the integral

$I = \int_{1}^{\infty} \frac{d x}{\sqrt{x^{4} + 1}}$

is Converges

.

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Chapter 8.7, Problem 53E

Chapter 8.7, Problem 55E

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Chapter 8 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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1 x 4 + 1 < 1 x 2 ∀ x ∈ R And value of the integral I = ∫ 1 ∞ d x x 4 + 1 is Converges .

Solution Summary: The author explains that the integral I = 1 dx x 4 + 1 is convergent by using Leibnitz test.

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To Show:

$\frac{1}{\sqrt{x^{4} + 1}} < \frac{1}{x^{2}} \forall x \in R$

And value of the integral

$I = \int_{1}^{\infty} \frac{d x}{\sqrt{x^{4} + 1}}$

is Converges

.

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Chapter 8 Solutions

1 x 4 + 1 &lt; 1 x 2 ∀ x ∈ R And value of the integral I = ∫ 1 ∞ d x x 4 + 1 is Converges .

Solution Summary: The author explains that the integral I = 1 dx x 4 + 1 is convergent by using Leibnitz test.

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To Show: 1x4+1<1x2 ∀x∈R And value of the integral I=∫1∞dxx4+1 is Converges .

Want to see the full answer?

Chapter 8 Solutions

1 x 4 + 1 < 1 x 2 ∀ x ∈ R And value of the integral I = ∫ 1 ∞ d x x 4 + 1 is Converges .

To Show:

$\frac{1}{\sqrt{x^{4} + 1}} < \frac{1}{x^{2}} \forall x \in R$

And value of the integral

$I = \int_{1}^{\infty} \frac{d x}{\sqrt{x^{4} + 1}}$

is Converges

.