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Answer: Geometry problems Use a table of integrals to

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 35E Chapter 7.5

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 35E

Geometry problems

Use a table of integrals to solve the following problems.

Find the length of the curve y = ex on the interval [0, ln 2].

Step-by-Step Solution:
Step 1 of 3

Problem 35E

Geometry problems  Use a table  of integrals  to solve the following problems.

 Find the length of the curve  y=  on the interval [0, ln(2)].

Answer;

 Step-1;

                                    If  is a continuous on [a,b] , then the length of the curve y = f(x) , a  is

                                L =  dx.

        If we use Leibniz notation for derivatives , we can write the  arc length formula as follows:

                             L =   dx.

Step-2

            The given curve is   y =  , and the interval is[0,ln(2)].

                         Now , we have to find out the length of the curve  y=  on the interval [0,ln(2)].........(1)

                    If y = f(x) =  , then (y) = f(x) =  )

                                           =  = ……………(2)

                 So , the arc length formula gives;

                                        L =  dx =    dx.

                  Therefore , L =    dx  , since from(1) ,(2).

                                       =   dx  , since =

                                         L =   dx  =  dx  ……….(3)

Step-3;

                           Consider ,

                      For our convenience  take substitution method: put 2 x =t , then differentiation of  2 x =t is ; (2x) = (dt/dx)

                           dx= dt …………(4)

                    Therefore , =  =   dt

                           Substitute  ,  +1 =  =  = p, then the differentiation of   +1 =  is +1) =   

                                       = 2p

                                            dt = dp

                                                =  dp , since  +1 = …………..(5)

            Therefore , the above integral becomes;

                           =   dt  =  (p)dp

                                                                                     =

Step 2 of 3

Chapter 7.5, Problem 35E is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Answer: Geometry problems Use a table of integrals to

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