Explanation Given information: The integral function is ∫ 0 ∞ d x x ( x + 1 ) . The interval [0, ∞ ) is infinite. The integrand has infinite discontinuity at 0. Hence, the integral function asimproper integral. The integral function is expressed as sum of Type 2 and Type 1 improper integral as follows: ∫ 0 ∞ d x x ( x + 1 ) = ∫ 0 1 d x x ( x + 1 ) + ∫ 1 ∞ d x x ( x + 1 ) Calculation: Show the integral function as follows: ∫ 0 ∞ d x x ( x + 1 ) (1) The integral function is expressed as sum of Type 2 and Type 1 improper integral: ∫ 0 ∞ d x x ( x + 1 ) = ∫ 0 1 d x x ( x + 1 ) + ∫ 1 ∞ d x x ( x + 1 ) (2) Show the definition of an Improper Integral of Type 2 : (a) If f is continuous on [a,b] and is discontinuous at b , then ∫ a b f ( x ) d x = lim t → b − ∫ a t f ( x ) d x Provided this limit exists (as a finite number). (b) If f is continuous on [a,b] and is discontinuous at a , then ∫ a b f ( x ) d x = lim t → − a + ∫ t b f ( x ) d x Provided this limit exists (as a finite number). The improper integral ∫ a b f ( x ) d x is called convergent if the corresponding limit exists. The improper integral ∫ a b f ( x ) d x is called divergent if the corresponding limit does not exist. (c) If f has a discontinuity at c, where a < c < b , and both ∫ a c f ( x ) d x and ∫ c b f ( x ) d x are convergent, then ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x Apply Part (b) of the definition of improper integral in ∫ t 1 d x x ( x + 1 ) part of Equation (2). ∫ 0 ∞ d x x ( x + 1 ) = ∫ 0 1 d x x ( x + 1 ) + ∫ 1 ∞ d x x ( x + 1 ) = lim t → 0 + ∫ t 1 d x x ( x + 1 ) + ∫ 1 ∞ d x x ( x + 1 ) (3) Show the definition of an Improper Integral of Type 1 : (a) If ∫ a t f ( x ) d x exists for every number t ≥ a , then ∫ a ∞ f ( x ) d x = lim t → ∞ ∫ a t f ( x ) d x Provided this limit exists (as a finite number). (b) If ∫ t b f ( x ) d x exists for every number t ≤ b , then ∫ − ∞ b f ( x ) d x = lim t → − ∞ ∫ t b f ( x ) d x Provided this limit exists (as a finite number). The improper integrals ∫ a ∞ f ( x ) d x and ∫ − ∞ b f ( x ) d x are called convergent if the corresponding limit exists. The improper integrals ∫ a ∞ f ( x ) d x and ∫ − ∞ b f ( x ) d x are called divergent if the corresponding limit does not exist. (c) If both the integral function ∫ a ∞ f ( x ) d x and ∫ − ∞ a f ( x ) d x are convergent, then ∫ − ∞ ∞ f ( x ) d x = ∫ − ∞ a f ( x ) d x + ∫ a ∞ f ( x ) d x Here, a is any real number. Apply Part (a) of the definition of improper integral ∫ 1 ∞ d x x ( x + 1 ) part of Equation (3). ∫ 0 ∞ d x x ( x + 1 ) = lim t → 0 + ∫ t 1 d x x ( x + 1 ) + ∫ 1 ∞ d x x ( x + 1 ) = lim t → 0 + ∫ t 1 d x x ( x + 1 ) + lim t → ∞ ∫ 1 t d x x ( x + 1 ) (4) Consider the value of the function u u = x u 2 = x (5) Differentiate Equation (5) with respect to x

Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals, 8th (James Stewart Calculus)

8th Edition

ISBN: 9781305272422

Author: James Stewart, Jeffrey A. Cole, Daniel Drucker, Daniel Anderson

Publisher: Cengage Learning

See similar textbooks

Videos

Question

Chapter 7.8, Problem 55E

To determine

To evaluate: The integral function ∫0∞dxx(x+1) by using the expression as sum of improper integrals of Type 2 and Type 1.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.8, Problem 54E

Chapter 7.8, Problem 56E

Students have asked these similar questions

Use the graph of f to evaluate the following: 6 T 5 4 3 2 f(x) 1 x 1 2 2 3 4 5 6 -1 The average rate of change of f from 1 to 4 = Question Help: Message instructor

Question 9 The plot below represents the function f(x) 8+ 7 6- ·6· -5 4-3-2-1 Evaluate f(4) f(4) = 4 3 2 1 2 2 3 3 4 $ 2 Solve f(x) = 4 I= Question Help: Message instructor

Question 8 Let f(x) = 1 x 25 and and g(x) 6 +5. A Find the following function. Simplify your answer. f(g(x)) = Question Help: Message instructor Question 9

Chapter 7 Solutions

Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals, 8th (James Stewart Calculus)

Ch. 7.1 - Evaluate the integral using integration by parts...Ch. 7.1 - Prob. 2E Ch. 7.1 - Evaluate the integral. 3. xcos5xdx Ch. 7.1 - Evaluate the integral. 4. ye0.2ydy Ch. 7.1 - Evaluate the integral. 5. te3tdt Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Evaluate the integral. 8. t2sintdt Ch. 7.1 - Evaluate the integral. 9. cos1xdx Ch. 7.1 - Prob. 10E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

The integral function ∫ 0 ∞ d x x ( x + 1 ) by using the expression as sum of improper integrals of Type 2 and Type 1.

Solution Summary: The author evaluates the integral function by using the expression as sum of improper integrals of Type 2 and Type 1.

Videos

To evaluate: The integral function ∫0∞dxx(x+1) by using the expression as sum of improper integrals of Type 2 and Type 1.

Want to see the full answer?

Chapter 7 Solutions