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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 = \ln (1 + x^3) $ , $ 0 \le x \le 5 $

$S_{10}=7.09457$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 1

Arc Length

Applications of Integration

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Hey, it's clear. So when you read here, so we have why, as equal to l N of one plus X cube access between zero and five included. First, what we have to do is find the derivative. When we get three x square were one plus execute. We're gonna plug it into our gargling formula from 0 to 5 When we square this we got one plus nine x to the forest over one plus x cute square DX We're gonna find the width of the intervals when N is equal to 10 and we got five minus zero over 10 is equal to 0.5. And when we calculate the values at the interval boundaries, we're gonna use the Simpsons rule. So yet the interval the with over three times f of except zero less for ah of except one plus two f of except two was four f of except three plus to F, except for plus four of except five plus two f of X up six less four f of X and seven Let's to f of X, up eight plus four f of x m nine Klis f of acceptance. We plug these values inside we get around seven point 09457 and then the calculator. We got seven points. 11882 So they're very close. Thanks.

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