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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.$y = x \sin x$ , $0 \le x \le 2\pi$

## ${L \approx S_{10}=\frac{\pi / 5}{3}\left[f(0)+4 f\left(\frac{\pi}{5}\right)+2 f\left(\frac{2 \pi}{5}\right)+4 f\left(\frac{3 \pi}{5}\right)+2 f\left(\frac{4 \pi}{5}\right)+4 f\left(\frac{5 \pi}{5}\right)+2 f\left(\frac{6 \pi}{5}\right)\right.}{\left.+4 f\left(\frac{7 \pi}{5}\right)+2 f\left(\frac{8 \pi}{5}\right)+4 f\left(\frac{9 \pi}{5}\right)+f(2 \pi)\right]}$$\approx 15.498085$

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Applications of Integration

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### Video Transcript

Hey, it's clear. So when you're in here so we have the link of a curve from zero to two pi square root of one plus the of a sign over D s square P s. This is equal to from 0 to 2 pi square root of one plus ex co sign plus sign square the ex just equal to from 0 to 2 high square root One plus square co sign Square Signed Square plus two X Sign Co Sign the ex missus equal to 10 to 2 pi square root of X square co sign square. Let's sign Square plus X I kn to X plus one the ex So L is equal to from 0 to 2 pi and want to make this up of X and we'll make f of X equal to X square co sign Square plus sign Square was ex signed two x plus one. So using the Simpsons rule going to calculate the change in X So to pine Linus zero over 10 which is equal to pi over five. Someone calculated Well, yet over three of zero was for of pie. If it's cause too, uh, two pi over five plus for, uh, no three pi over five plus two ah, four pi over five plus for a five pi over five plus two. Uh, no. Six ply over five plus four f A seven pilot, five plus two of a pie over five plus four, nine pi over five close f of 10 pi over five and we get this to be equal to 15 0.498 085 And when we used a calculator where this we get this to be around 15.37 or 568 and they're very close to one another.

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Applications of Integration

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