Problem

Find the exact length of the curve. $ x = \dis…

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Problem 11 Easy Difficulty

Find the exact length of the curve.

$ y = \displaystyle \frac{x^3}{3} + \displaystyle \frac{1}{4x} $ , $ 1 \le x \le 2 $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 8

Further Applications of Integration

Section 1

Arc Length

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Video Transcript

Hey, it's clear. Someone new right here. So we have lying is equal to X cube over three plus one over four X. We're first going to find the derivative. So do you. Why, over DX We got X squared minus one over four X square. Afterwards, we're gonna square. So we get. Do you want? Over. Deac Square is equal to X square minus one over four square square. Um, excuse us X to the fourth minus one, huh? Plus won over 16. Next to the fourth. Now, when we plug it into our arc length equation, yeah, l is equal to from 1 to 2 square root of one plus next to the fourth minus one have plus one over 16 x to the fourth DDX. This becomes X to the fourth plus one Have plus one over 16 x to the fourth. This could be factored into X square plus one over four. Next square square. So when you put it into our equation comes from 1 to 2 square root of X square, plus one over four x square square. The this equals 59 over 24. Yeah,

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Problem

Find the exact length of the curve. $ x = \dis…

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Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 8

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Samuel Hannah

Joseph Lentino

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

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Samuel Hannah

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Heather Zimmers

Samuel Hannah

Joseph Lentino

James Stewart

Calculus: Early Transcendentals