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Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places.

$ x = y^2 - 2y $, $ 0 \le y \le 2 $

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Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 1

Arc Length

Applications of Integration

Missouri State University

University of Nottingham

Boston College

Lectures

01:43

Set up an integral that re…

01:59

01:15

01:09

01:20

02:14

02:47

01:42

01:02

Use either a CAS or a tabl…

00:49

01:39

$3-6$ Set up an integral t…

01:41

you want to find the length of the curve, X equals Y squared minus two. Why on the interval, why is grew greater than or equal to zero, but less than or equal to two? So we'll want to find X prime to plug in. To are, uh, are length formula. So ex prime is simply, too. Why minus two plugging this in zero for a and two for B radical, one plus to Y minus two squared D Y. Um, we can take out the two and square it, so it'll be a little easier to plug this into her calculator, which is what the problem asks for so radical. One plus or times Why minus one squared D y. Plugging this in, you should get approximately 2.9 579

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