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Use Simpson's Rule with $ n = 10 $ to estimate the arc length of the curve. Compare your answer with the value of the integral produced by a calculator.

$ y = \sqrt[3]{x} $ , $ 1 \le x \le 6 $

$S_{10}=5.07421$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 1

Arc Length

Applications of Integration

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he has clear. So when Liam Reid here. So we have. Why is equal to the hubris of X and excess between one then six included. We're gonna find the derivative when we get 1/3 x to the negative 2/3 the darkling formula from 1 to 6 square root of one plus won over nine x to the negative 4/3 e x We're gonna find the width of the interval for N is equal to 10 and we get 0.5 for our Simpsons roll you get are with over three f of x sub zero plus four of ex of one plus two of F of except two plus four of f of X up three close to of F Uh, except for plus for times F of X, up five plus two of except six It was for of of ex up seven plus two f of except a plus four f of x m nine plus, uh, of except in. And when we put this all in, this becomes equal to around 5.7 4 to 1. When we calculated on the calculator, we get around 5.7 409 So they're close

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