To find the arclength of y =3x? + 4 from x = 0 to x = 4, which is the correct integral?OL =V1 + 6x dxOL =VI+ 36xdx4= 10OL =V1 + 36x dx6x to re-write the integral in terms of u.Now that you have the correct integral, use the substitution uWhich is the correct result?24OL =1+ 3u du241OL =24OL =VI+ 6udu241OL =VI+ uduWhat is the value of the integral? Hint: Use a table or technology. You may enter a decimal approximationfor your solution.Question Help: D Video Message instructorSubmit QuestionJump to Answer

Weknow that arc length of the curve is ,L=∫x=ax=b1+[f'(x)]2dxTherefore arclenth of y=3x2+4 from x=0…

Answered: To find the arclength of y 3x2 +4 from…

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...

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Question

To find the arclength of y =
3x? + 4 from x = 0 to x = 4, which is the correct integral?
OL =
V1 + 6x dx
OL =
VI+ 36xdx
4
= 10
OL =
V1 + 36x dx
6x to re-write the integral in terms of u.
Now that you have the correct integral, use the substitution u
Which is the correct result?
24
OL =
1+ 3u du
24
1
OL =
24
OL =
VI+ 6udu
24
1
OL =
VI+ udu
What is the value of the integral? Hint: Use a table or technology. You may enter a decimal approximation
for your solution.
Question Help: D Video Message instructor
Submit Question
Jump to Answer

Expert Solution

Step 1

$W e k n o w t h a t a r c l e n g t h o f t h e c u r v e i s, L = \int_{x = a}^{x = b} \sqrt{1 + {[f' (x)]}^{2}} d x T h e r e f o r e a r c l e n t h o f y = 3 x^{2} + 4 f r o m x = 0 t o x = 4 i s L = \int_{x = 0}^{x = 4} \sqrt{1 + {[6 x]}^{2}} d x (1) T h a t i s L = \int_{x = 0}^{x = 4} \sqrt{1 + {36 x}^{2}} d x N o w p u t u = 6 x T h e n a s x \to 0 t h e n u \to 0 a s x \to 4 t h e n u \to 24 d i f f e e n t i a t e o n b o t h s i d e d u = 6 d x \frac{d u}{6} = d x T h e r e f o r e e q u a t i o n (1) b e c o m e L = \int_{u = 0}^{u = 24} \sqrt{1 + {(u)}^{2}} \frac{d u}{6} T h a t i s L = \frac{1}{6} \int_{u = 0}^{u = 24} \sqrt{1 + {(u)}^{2}} d u$