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Use Simpson's Rule with $ n = 10 $ to approximate…

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Problem 21 Easy Difficulty

Use Simpson's Rule with $ n = 10 $ to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare your answer with the value of the integral produced by a calculator.

$ y = xe^x $ , $ 0 \le x \le 1 $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 8

Further Applications of Integration

Section 2

Area of a Surface of Revolution

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Video Transcript

in this video will be using Simpson's rule to find some surface area obtained by rotating a curve around the X axis. So we ought to use the surface area Formula 1st which is the following two pi Y. The square root of one plus Y. Prime squared dx. So here's what we need to do. Let's first set up the integral and then we'll talk about Simpson's rule. With that being said. Given a curve, Y equals X times E to the X. We know it's derivative. By using the product rule, it will be equal to the same term with an additional factor of E to the X. And we can substitute all of this into our integral. And I forgot the actual range of integration that we're going to be using. Which would be from in between zero and 1. Uh huh. So our integral for become Will be the integral from 0-1. Will have to pie. Why? Which is X times E to the X times the square root of one plus the derivative squared X times E. To the X plus E. To the X. The whole thing squared dx. So this is our integral. So what I'm going to do is I'm going to let this equal, I'm going to let the entire inta grant be equal to F of X. This will just make our work a little bit easier. Yeah. For Simpson's rule the approximate area under some function will be equal to The change in x divided by three times we would have F of X. Not plus four F of X one plus two. F of X two plus not that. But And so on until the pattern continues. X and minus one plus F of X. In general. Be using this formula. Yeah. And remember that delta X would be equal to b minus a, divided by N. In our case here we have that X is going to be in between zero and one. And were asked to use 10 sub intervals. So we're going to have 1/10 because the last band is just zero, thankfully. All right. At a 0.1. Yeah. So what this means is are integral Mhm will be approximately using this using what we had obtained earlier and substituting values of the specific function we're going to get This is equal to approximately zero point 1/3 times you would have f of zero plus four Times F of 0.1. Go ahead plus two times F of 0.2 plus four times F of 0.3 plus two times F of 0.4 plus four times F of 0.5 plus two times F of 0.6 plus four times F of 0.7 Plus two times f of 0.8 Plus four times f of 0.9 Plus f of one. And you would use again, I said earlier, you would substitute these exercise into the into grand and then determine what the value would be from there. And when you do that, you're going to get that this is approximately going to be 24.1443. So that's the approximate value. But we would we want to compare it to the absolute value or the correct value. And obviously this integral is pretty hard to evaluate by hand. So we're going to use we're gonna use Wolfram Alpha for this. And when we do so When we use Wolfram α to evaluate this integral, we get approximately we get exactly 24.1443. Okay. Actually, the Uh huh. Yeah. When we actually do the approximation, we get about 24.1458. Yeah. Yes, that's how I do this problem.

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