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Numerade Educator



Problem 12 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_1^3 e^{\frac{1}{x}}\ dx $ , $ n = 8 $


a. $T_{8}=3.534934$
b. $M_{8}=3.515248$
c. $S_{8} \approx 3.522375$

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Video Transcript

were given an integral in the value of n and were asked in each part to approximate this interval with this value of n using a particular rule. So the integral is the integral from 1 to 3 of E to the one over x. The X and N is equal to eight in part a, whereas to use the trapezoidal rule to approximate this integral with this value then and we're asked around our answer to six decimal places. So for the trapezoidal rule, first we want to find the width of the interval. Delta X, This is standpoint on the right three minus the endpoint on left, one divided by n, which is eight. This is 2/8 or 1/4 which is a decimal is 0.25 Next, let's dry table of function values. So on the left hand side of this table, we have values of X on the right side. We have values of our function. Eat to the one over X. Now for the trapezoidal rule, we want to calculate the values of F at the end points of the intervals. So we're going to start with the left most endpoint, which is one, and then we'll add Delta X 4.25 to get so we get 1.25 and we'll keep adding that. So we get 1.50 1.75 two point 00 2.25 2.5, though 2.75 and then three has a right most endpoint. And to calculate the values in the second column, all we have to do is take the value in the first column X and plug it into a formula. Eats the one over X for example, the first century. This is going to be e to the 1/1, which is E. And since we're really just going to be using decimals here, I'm going to approximate this as 2.71 8 to 8 18 so out to seven decimal places, we're going to want around six eventually. Likewise, the second entry on the right side is two point 2 to 5 540 nine and we can continue in this manner down the table. And so I complete the table and you should get these values at least out to six or seven decimal places. Finally, with our completed people, we have that for the trapezoid rule. The number of terms is equal to and plus one which in this case is nine. So we should have nine terms in our some and our approximation, which I'll call he ate, is given by the formula Delta X or step size over to times you start off with F of X zero is one plus two times f of x, one plus two times ffx too. And we continue in this manner all the way up to two times that those x seven plus ffx eight. So we have nine terms in our some. And if you plug in the values from the table into the some as well as Delta X equals 0.25 the answer is approximately three points. 534 934 to six decimal places now in part B were asked to use the midpoint rule for the midpoint rule. The procedure is similar. We still want to calculate the interval with which is the same as in part a 0.25 Now, though the midpoint rule, we want to calculate functional values at the middle of the interval. So make another table with X and left and our function eat to the one over X on the right. And now, instead of starting with X equals left most endpoint one, we're going to start with the midpoint of the first interval, which which is one point and then half of 0.25 which is 0.1 to 5. And then we add 0.25 to this successively. So we get one point 375 1.6 to 5, and so on We continue this pattern all the way up to 2.875 which is the midpoint of the interval, 2.75 three. Now, in the right column, we plug in the value of X into one of her to eat to one over X, and we approximate this answer to six decimal places or more. For example, the first entry is two point 4324255 The second entry, we plug in 1.375 This gives us 2.69 4 to 9 and we continue this process all the way down the table. And so you should obtain values that are very close to, if not the same as these on the right hand side. Finally, we use our formula for the midpoint rule storm approximation and eight we know it has en or eight terms, and M eight is going to be Delta X, our interval length times half of the first midpoint, which I'll call 60 plus f of these second midpoint, which will call X one all the way up to F of the last midpoint x seven. And if you plug in the values from the table into the some as well as Delta X equals 0.25 M eight is six decimal places approximately 3.515 248 That's our answer for part B. Finally, in part C, we're asked to use Simpson's rule, so Simpson's rule is actually pretty similar to part A. Once again, we have the steps size of 0.25 and once again we want to compute the values of the function at the end points of the intervals, starting with X equals one and ending at X equals three with step size 30.25 So we're going to use the same table of values and the number of terms once again is going to be. And plus one or nine. However the form of are some and our answer therefore, are going to be slightly different. So we still have nine terms, but are some which we'll call s eight s for Simpson's rule. This is now Delta X over three instead of over two times. And then we have f of our first term. So I first turned x zero plus four times at her x one plus two times f of x two plus four times at the Vex three and we continue this pattern up to two times ffx six plus four times ffx seven and then plus one times ffx eight Rexy course three. And if you plug values into this some and calculate the answer is approximately 26 decimal places. 3.52 to red 375 This is our answer for part C.