In lectures we have an antiderivative of In(x) is x In(x) – - x. Here we like a formula for the antiderivative F,(=) = / (In(=))"dz where n is a positive whole number. (a) Show that when n = 0, we have Fo(x) = x and when n=1 we have F1(x) = x In(x) - - x. (b) Setting f(x) = (In(x))² and g'(x) = 1, using the “integration by parts" formula | f(2)d'(x)dx = f(@)g(=) – / f'(=)g(=)dx show that we have the relationship F2(x) = #(In(x))? – 2F1 (x). [Hint: by the Chain rule the derivative of f(x) = (In(x))² is f'(x) = 2(ln(x))/x]. (c) Show that an antiderivative of (In(x))² is F2(x) = x((In(x))² – 2 In(x) + 2). %3D
In lectures we have an antiderivative of In(x) is x In(x) – - x. Here we like a formula for the antiderivative F,(=) = / (In(=))"dz where n is a positive whole number. (a) Show that when n = 0, we have Fo(x) = x and when n=1 we have F1(x) = x In(x) - - x. (b) Setting f(x) = (In(x))² and g'(x) = 1, using the “integration by parts" formula | f(2)d'(x)dx = f(@)g(=) – / f'(=)g(=)dx show that we have the relationship F2(x) = #(In(x))? – 2F1 (x). [Hint: by the Chain rule the derivative of f(x) = (In(x))² is f'(x) = 2(ln(x))/x]. (c) Show that an antiderivative of (In(x))² is F2(x) = x((In(x))² – 2 In(x) + 2). %3D
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
In lectures we have an antiderivative of ln(x) is xln(x) - x. Here we like a formula for the antiderivative
Fn(x) = (ln(x))ndx
where n is a positive whole number.
![In lectures we have an antiderivative of In(x) is x ln(x) – x. Here we like a formula for
the antiderivative
F,(1) = / (In(2))"de
where n is a positive whole number.
(a) Show that when n =
0, we have Fo(x) = x and when n = 1 we have F1(x) = x In(x) – x.
(b) Setting f(x) = (In(x))² and gʻ (x) = 1, using the “integration by parts" formula
| f(=)g (x)dx = f(m)g(x) – | f'(x)g(x)dx
show that we have the relationship
F2(1) = x(In(x))? – 2F:(x).
[Hint: by the Chain rule the derivative of f(x) = (In(x))² is f'(x) = 2(ln(x))/x].
%3D
(c) Show that an antiderivative of (In(x))² is
F2(x) = x((In(x))? – 2 In(x) +2).
(d) For n =
3, set f(x) = (In(x))³ and g'(x) = 1, using the "integration by parts" formula above, show that we
have the relationship
F3(x) = x(In(x))³ - 3F2(x),
and using (c), show that an antiderivative of (ln(x))³ is
F;(x) = a((In(x))³ – 3(In(x))² + 6 In(x) – 6).
(e) Use the same idea as in (d), show that an antiderivative of (In(x))ª is
F4(x) = x((In(x))* – 4(In(x))³ + 12(ln(x))² – 24 In(x) + 24)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0e7d93fd-13e6-4ba5-a62f-b19c01acd6ce%2F8ed3aa04-65bd-4cce-b03c-14f680739675%2Fevz2dky_processed.png&w=3840&q=75)
Transcribed Image Text:In lectures we have an antiderivative of In(x) is x ln(x) – x. Here we like a formula for
the antiderivative
F,(1) = / (In(2))"de
where n is a positive whole number.
(a) Show that when n =
0, we have Fo(x) = x and when n = 1 we have F1(x) = x In(x) – x.
(b) Setting f(x) = (In(x))² and gʻ (x) = 1, using the “integration by parts" formula
| f(=)g (x)dx = f(m)g(x) – | f'(x)g(x)dx
show that we have the relationship
F2(1) = x(In(x))? – 2F:(x).
[Hint: by the Chain rule the derivative of f(x) = (In(x))² is f'(x) = 2(ln(x))/x].
%3D
(c) Show that an antiderivative of (In(x))² is
F2(x) = x((In(x))? – 2 In(x) +2).
(d) For n =
3, set f(x) = (In(x))³ and g'(x) = 1, using the "integration by parts" formula above, show that we
have the relationship
F3(x) = x(In(x))³ - 3F2(x),
and using (c), show that an antiderivative of (ln(x))³ is
F;(x) = a((In(x))³ – 3(In(x))² + 6 In(x) – 6).
(e) Use the same idea as in (d), show that an antiderivative of (In(x))ª is
F4(x) = x((In(x))* – 4(In(x))³ + 12(ln(x))² – 24 In(x) + 24)
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 3 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning