Explanation Given: The integral: I = ∫ 4 ∞ d x ( x − 2 ) ( x − 3 ) 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: I = ∫ 4 ∞ d x ( x − 2 ) ( x − 3 ) = ∫ 4 R d x ( x − 2 ) ( x − 3 ) + ∫ R ∞ d x ( x − 2 ) ( x − 3 ) = I 1 + I 2 I 1 = ∫ 4 R d x ( x − 2 ) ( x − 3 ) Assume, 1 ( x − 2 ) ( x − 3 ) = A ( x − 2 ) + B ( x − 3 ) Now, A = lim x → 2 ( x − 2 ) [ 1 ( x − 2 ) ( x − 3 ) ] = lim x → 2 [ 1 ( x − 3 ) ] = [ 1 ( 2 − 3 ) ] = − 1 B = lim x → 3 ( x − 3 ) [ 1 ( x − 2 ) ( x − 3 ) ] = lim x → 3 [ 1 ( x − 2 ) ] = [ 1 ( 3 − 2 ) ] = 1 ∴ 1 ( x − 2 ) ( x − 3 ) = A ( x − 2 ) + B ( x − 3 ) = − 1 ( x − 2 ) + 1 ( x − 3 ) I 1 = To determine (b) To Show: For I = ∫ 4 ∞ d x ( x − 2 ) ( x − 3 ) = ln 2

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 8.7, Problem 41E

To determine

(a)

To Show:

For

$\int_{4}^{R} \frac{d x}{(x - 2) (x - 3)} = \ln | \frac{R - 3}{R - 2} | - \ln (\frac{1}{2})$

To determine

(b)

To Show:

For

$I = \int_{4}^{\infty} \frac{d x}{(x - 2) (x - 3)} = \ln 2$

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Chapter 8.7, Problem 40E

Chapter 8.7, Problem 42E

Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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(a) To Show: For ∫ 4 R d x ( x − 2 ) ( x − 3 ) = ln | R − 3 R − 2 | − ln ( 1 2 )

Solution Summary: The integral: I = a b f (x ) d x = [ F (

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(a)

To Show:

For

$\int_{4}^{R} \frac{d x}{(x - 2) (x - 3)} = \ln | \frac{R - 3}{R - 2} | - \ln (\frac{1}{2})$

(b)

To Show:

For

$I = \int_{4}^{\infty} \frac{d x}{(x - 2) (x - 3)} = \ln 2$

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Chapter 8 Solutions

(a) To Show: For ∫ 4 R d x ( x − 2 ) ( x − 3 ) = ln | R − 3 R − 2 | − ln ( 1 2 )

Solution Summary: The integral: I = a b f (x ) d x = [ F (

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(a) To Show: For ∫4Rdx(x−2)(x−3)=ln|R−3R−2|−ln(12)

(b) To Show: For I=∫4∞dx(x−2)(x−3)=ln2

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Chapter 8 Solutions

(a)

To Show:

For

$\int_{4}^{R} \frac{d x}{(x - 2) (x - 3)} = \ln | \frac{R - 3}{R - 2} | - \ln (\frac{1}{2})$

(b)

To Show:

For

$I = \int_{4}^{\infty} \frac{d x}{(x - 2) (x - 3)} = \ln 2$