Problem

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

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Problem 17 Hard Difficulty

Find the exact length of the curve.

$ y = \frac{1}{4} x^2 - \frac{1}{2} \ln x $ , $ 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

it's clear is that when you read here, So we have why is equal to 14 X square minus one have well on offense access between one and two included. And we're trying to find a link. We're gonna first take the derivative and we get 1/2 acts minus one over to X, and we're gonna plug it into our equation. Our art length formula We get square of one plus 1/2 of us minus one over to X where, Deitz. And we see that when we factor this out. This part only you get next square over. Four minus one have plus one over four X square and we're going Thio add one, which comes from here on DS equals one Have plus X squared over four plus one over four. Explain. We could factor out a 14 So 14 next square plus two close one over. X quay. When this becomes this part becomes X plus one over X square. Now we're gonna plug it back in to our link formula. That's a plus one over X square, D x. This is equal to 1/2 from 1 to 2 x plus one over X d x. And when we saw this out, you get three. Fourth was 1/2. Oh, um, have to

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Problem

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

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

Related Courses

Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 8

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Kayleah Tsai

Kristen Karbon

Samuel Hannah

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

Calculus: Early Transcendentals

Grace He

Kayleah Tsai

Kristen Karbon

Samuel Hannah

James StewartCalculus: Early Transcendentals

Grace He

Kayleah Tsai

Kristen Karbon

Samuel Hannah

James Stewart

Calculus: Early Transcendentals