Problem

Use (a) the Trapezoidal Rule, (b) the Midpoint Ru…

00:30

Problem 17 Easy Difficulty

Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson's Rule to approximate the given integral with the specified value of $ n $. (Round your answers to six decimal places.)

$ \displaystyle \int_0^4 \ln (1 + e^x)\ dx $ , $ n = 8 $

Answer

a. $T_{8}=8.814278$
b. $M_{8}=8.799212$
c. $S_{8}=8.804229$

Claire B.

October 19, 2021

Simpson's rule with n=50

Claire B.

October 19, 2021

Sorry, I need to elaborate: how would you use the TI-84 to perform Simpson's Rule for something as high as n=50?

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Video Transcript

were given an integral in the value of n and in each part where asked to use a certain rule toe. Approximate this interval with this value event. Okay, the integral is the integral from 0 to 4 of the natural log of one plus e to the X. I mean where n is equal to eight in part a were asked to use the trapezoidal rule. So first for this rule you want to determine the interval lengths Delta X These are be right most endpoint four minus the left most endpoint zero over n, which is eight. This is 4/8 or in a decimal notation 0.5. The next step is to find required function values. So in trapezoidal rule, we want to find the values of the function on the end points of these sub intervals. So I'll make a table on the left side, we have the end points X and on the right side we have the value of our function, which is natural log of one plus e to the X mhm. Let's start off are left endpoint zero and will increase by Delta X or 0.5 until we reach four. So we have zero 0.51 point 1.5 and so on. Mhm. Now, for the right side, all we have to do is simply plug in attention X from the left side. So, for example, plugging in X equals zero. You have the natural log of to for our first entry in decimal notation, this is to eight decimal places 0.693 14718 Likewise, we plug in X equal 2.5. We get approximately 0.974076 98 We continue in this fashion all the way to X equals three. And so your table of values should look something like this paying and how accurate want to be. You may have more or less decimal places on the right. Now that we have these values, we'll use the formula for the trapezoidal rule for approximation, which will call T eight for trapezoidal. We know this has n plus one or a total of nine terms, and it's some and that t eight is equal to our formula is the in the interval length felt x over two times f of the first point x zero So this is zero plus two times ffx, one plus two times ffx, too. And we continue in this fashion all the way up to two times ffx seven and then plus just one times ffx eight, which is three. And then we simply plug in the values for Delta X, which is 0.5, and all of our values from the table into the some in the answer is approximately 8.804227 to six decimal places. That's the answer for Party now in Part B were asked to use the midpoint rule. This is similar to party, except for in the mid point rule. We have the same steps size Delta X equals five, but now we want to find the values of the function natural log of one plus e. D. X at the midpoint of these intervals. So we're going to start drawing another table. First column X second column is the natural lot of one plus e to the X. Yeah, and instead of starting with X equals zero, the left endpoint we're going to start at the midpoint of the first sub interval, 0.5, which is 0.25 Then we're going to add five until we reached the midpoint of 3.540 which is 3.75 So we have 25 plus 250.5 point 75 plus five is. Mm hmm. 1.25 plus 0.5 is 1.75 And so on, then, just as in part A, we're going to plug in the values and left to get the values on the right. So, for example, plugging an X equals 0.25 and two natural log of one plus e to the X. This is approximately 0.825939 42 However many decimal places you want, I would suggest a least six. Likewise, we plug in 0.75 We get 1.136 871 and I'll continue this all the way down the table. Table of values should look something like this. And finally, we're going to use the formula. So we know that the midpoint rule our approximation will call M eight. We know that this is going toe, have the total of n or eight terms in its some and the formula is the N eight is the length of a sub interval Delta X, which is 0.5 times the values of our function at the midpoint of each of these intervals. So we have from the table above f of x zero plus f of x one plus all the way up to F FX seven. So we have a total of eight terms in our son. And if you plug this into your calculator, this is approximately 8.799 212 to six decimal places under part C, whereas to use Simpson's rule approximately interval integral I mean now Simpsons World actually pretty similar to the trapezoidal rules. I'm going to overwrite heart a here, so this now becomes part C just is in part a and part B. The interval length is 25 and also just is in part a. The table is going to consist of on the left all the end points of the sub intervals on the right, the value the functions at those end points. So this table is the same as in part egg. But now our formula is going to be a little bit different, so below party here, I'll finish part C. So we know that s a r approximation giving substance rule as n plus one for nine terms just like part A and nine terms. But we know that these form of S eight is going to be the some interval with Delta X over three now instead of two times f of x zero plus four times value of after the next and point X one plus two times the value that for the next 10 point at thanks to plus four times that I have effort the next endpoint x three. We continue in this fashion, alternating between four and two until we get all the way up to four times half of x eight. Sorry, it should be four times f of x seven plus two times epic Lexi. Well, it's not really four times after that. Seven since sevens on. And then just one time different X eight puts the last turn. It's a total of nine terms in ourselves. And so if you plug this and do a calculator using the values from our first table up here, this is approximately 8.804 229 out to six decimal places. Okay,

Chris T.

Ohio State University

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