Consider the curve segments: 1 $1: y = x² from x =to x = 3 and 3 1 52: y = Vx from x =to x = 9. Set up integrals that give the arc lengths of the curve segments by integrating with respect to x. Demonstrate a substitution that verifies that these two integrals are equal. Substitution u = 3x² made in the integral L2 1 + -dx verifies that the length of the second segment is equal to 4x /- the length of the first segment: L1 = 4x² + ldx. Substitution u = vx made in the integral L2 = 1 + 4x -dx verifies that the length of the second segment is equal to the length of the first segment: L1 = :/ V4x² + 1dx. Substitution u = Vĩ made in the integral L2 = 1+dx verifies that the length of the second segment is equal to 2x 3 the length of the first segment: L1 = /2x + 1dx.

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...
icon
Related questions
Question
1
Substitution u =r made in the integral L2
1 +
4x
-dx verifies that the length of the second segment is equal to
the length of the first segment: L1 =
/ V4x? + 1dx.
1
Substitution u = x² made in the integral L2
1+dx verifies that the length of the second segment is equal to
4х
3
the length of the first segment: L1
/ V4x? + 1dx.
Transcribed Image Text:1 Substitution u =r made in the integral L2 1 + 4x -dx verifies that the length of the second segment is equal to the length of the first segment: L1 = / V4x? + 1dx. 1 Substitution u = x² made in the integral L2 1+dx verifies that the length of the second segment is equal to 4х 3 the length of the first segment: L1 / V4x? + 1dx.
Consider the curve segments:
1
to x = 3 and
3
1
$1: y = x from x =
S2: y = Vx from x = to x = 9.
Set up integrals that give the arc lengths of the curve segments by integrating with respect to x. Demonstrate a substitution that
verifies that these two integrals are equal.
9
Substitution u = 3x made in the integral L2 =
1+dx verifies that the length of the second segment is equal to
4x
3
the length of the first segment: L1 = / V4x? + 1dx.
9
Vx made in the integral L2 =
1+dx verifies that the length of the second segment is equal to
4x
Substitution u =
3
the length of the first segment: L¡ =
V4x + 1dx.
Substitution u =
Vx made in the integral L2 =
1 +
-dx verifies that the length of the second segment is equal to
2x
the length of the first segment: L =
I V2r + 1dx.
Transcribed Image Text:Consider the curve segments: 1 to x = 3 and 3 1 $1: y = x from x = S2: y = Vx from x = to x = 9. Set up integrals that give the arc lengths of the curve segments by integrating with respect to x. Demonstrate a substitution that verifies that these two integrals are equal. 9 Substitution u = 3x made in the integral L2 = 1+dx verifies that the length of the second segment is equal to 4x 3 the length of the first segment: L1 = / V4x? + 1dx. 9 Vx made in the integral L2 = 1+dx verifies that the length of the second segment is equal to 4x Substitution u = 3 the length of the first segment: L¡ = V4x + 1dx. Substitution u = Vx made in the integral L2 = 1 + -dx verifies that the length of the second segment is equal to 2x the length of the first segment: L = I V2r + 1dx.
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
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: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning