Problem

Find the exact length of the curve. $ y = \ln …

03:13

Question

Answered step-by-step

Problem 18 Hard Difficulty

Find the exact length of the curve.

$ y = \sqrt{x - x^2} + \sin^{-1} (\sqrt{x}) $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Clarissa Noh

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 8

Further Applications of Integration

Section 1

Arc Length

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 8

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Video Transcript

hands, Claire. So when you married here. So we have. Why this equal to square it of X minus. That's square less arc Sign of square E of X. The domain for Arctic sign is negative 1 to 1. But the square of ex limits that from 0 to 1. So we're gonna have it from 0 to 1. We first take the derivative when we get one, huh? Next minus that square. The negative 1/2 comes one minus. Two acts plus one have X. It's a negative 1/2. Well, we're a square root of one of minus skirt of X square. This becomes equal to one minus two X over two square root Becks and minus X square. So that's one over. His two square affects square root of one minus X. This is equal to one minus two x plus one over two square of X minus that square just equal to to minus two acts over two square root of X minus X square, which is equal to one minus X over square root of X minus fact square. When we multiply by squaring of X minus X square on both sides. When we get square of that's minus X square over X. After we simplify here, we were gonna plug it in to our equation for arc length. When I get zero one square root of one plus square of X minus X square, over at square the X Then this becomes equal to from 0 to 1 next to the negative 1/2 a t. X. So we're gonna use the limit since the integral is improper. So you get the limit as a approaches zero from the positive sign from a one that's they could have 1/2 t x and this becomes equal to limit as a approaches. Ciro Positive two x to the 1/2. We're calculating from a 21 and we get two one minus. So So it's just two times one, which is equal to two.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

Find the exact length of the curve. $ y = \ln …

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace He

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

Calculus 2 / BC Courses

Recommended Videos

Watch More Solved Questions in Chapter 8

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals