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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 = x \ln x $ , $ 1 \le x \le 2 $

$$\approx 7.248933 \quad \text { (calculator: } \approx 7.24893 )$$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 2

Area of a Surface of Revolution

Applications of Integration

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in this video we're asked to use Simpson's rule in order to approximate the surface area obtained by rotating a certain curve around the X axis. When we do that, we're going to have to we're going to have to set up the internal first and in order to do that we'll just use the green formula that I wrote down. So what we ought to do is to first find the derivative of our function. So if we have Y being equal to X times the log of X where again the log denotes the natural log rhythm. Then we have by, by the use of the product rule we have the derivative as just being one plus log X. Yeah, you can verify this to be true. Now you can substitute all of this into our integration formula so they're integral will become The integral from 1- two. We'll have to pie times Y which is just X times natural log of X Times Square of one plus one. Plus log X. The whole thing squared. Mhm mhm. No. Now what we're going to do is we're going to let all of the integral Aunt, we're going to let it equal ffx And we have to remember Simpson's rule. Now now Simpson's rule states that the definite integral can be approximated by. Okay, A certain multiplying every interval by a certain width divided by three. But the wits are going to be arranged as follows. Plus four. F F X N minus one plus f f X N. Let me make that N -2. subscript a little bit cleaner where we have to find our data. X. Of course. That's just be minus a divided by N. In which case we have two minus 1/10 which is about zero, which is 0.1 right. We're given these bounds earlier. This is going to be a lot. Now. This is going to make our calculations a lot easier. Will approximate then our surface area will be approximately 0.1 over three times F of one. Using our definition of the function we defined earlier Plus four times f of 1.1 Plus two times f of 1.2 Plus four times f of 1.3. Yeah. Plus two times F of 1.4 plus four times F of 1.5 plus two times f of 1.6 plus four times f of 1.7 plus two times f of 1.8 plus four Times f of 1.9 Plus F of two. Mhm. And you can evaluate this in your calculator. And what it gives us is it will give us exact this approximation will give us 7.24893. And if you type this into Wolfram alpha or something, you'll get approximately the same. The same as yes. Wolfram alpha. And that's how I do this question

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