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Problem

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

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Problem 18 Hard Difficulty

Find the exact length of the curve.

$ y = \sqrt{x - x^2} + \sin^{-1} (\sqrt{x}) $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 8

Further Applications of Integration

Section 1

Arc Length

Related Topics

Applications of Integration

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Watch More Solved Questions in Chapter 8

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Problem 16
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Video Transcript

hands, Claire. So when you married here. So we have. Why this equal to square it of X minus. That's square less arc Sign of square E of X. The domain for Arctic sign is negative 1 to 1. But the square of ex limits that from 0 to 1. So we're gonna have it from 0 to 1. We first take the derivative when we get one, huh? Next minus that square. The negative 1/2 comes one minus. Two acts plus one have X. It's a negative 1/2. Well, we're a square root of one of minus skirt of X square. This becomes equal to one minus two X over two square root Becks and minus X square. So that's one over. His two square affects square root of one minus X. This is equal to one minus two x plus one over two square of X minus that square just equal to to minus two acts over two square root of X minus X square, which is equal to one minus X over square root of X minus fact square. When we multiply by squaring of X minus X square on both sides. When we get square of that's minus X square over X. After we simplify here, we were gonna plug it in to our equation for arc length. When I get zero one square root of one plus square of X minus X square, over at square the X Then this becomes equal to from 0 to 1 next to the negative 1/2 a t. X. So we're gonna use the limit since the integral is improper. So you get the limit as a approaches zero from the positive sign from a one that's they could have 1/2 t x and this becomes equal to limit as a approaches. Ciro Positive two x to the 1/2. We're calculating from a 21 and we get two one minus. So So it's just two times one, which is equal to two.

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