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

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Problem 9 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_0^2 \frac{e^x}{1 + x^2}\ dx $ , $ n = 10 $


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Yuou Sun

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 7

Approximate Integration

Related Topics

Integration Techniques

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Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

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27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Video Transcript

were given an integral and a value of n and in each part, whereas to use a particular rule, approximate this integral with this value of end and to round our answer to six decimal places. The integral is the integral from 0 to 2 of e to the X over one plus X squared DX in the value of n is 10. Now in part A were asked to use the trapezoid rule to use the trapezoid rule. Virtual wants to calculate the values of dysfunction at the boundaries of the intervals. Before we do this, we need to figure out what the intervals are. So the interval wit Delta X is going to be the end point to minus the other end 0.0.0 over end, which is 10 or 1/5 which as a decimal, this is point to and now we set up a table so the first row will be the values of X, and the second one will be the corresponding values of the function F. Now we're going to start with X equals 0.0 and move all the way to X equals 2.0. In fact, maybe I shouldn't do rose. I might want to do columns here because it's going to be rather long. So we have X and F of X extort cells as 0.0. If you plug in 0.0 to the function, eat the X over one plus X squared. You get one. Since we want to round to six decimal places, I'm going to calculate the function out to six decimal places. You could get a more accurate answer if you calculate more decimal places. Next we have the X equals 0.2. So we add the step size plugging in 0.2 F of X is approximately 1.1744 to 6, and so on. Next we add two. We get four, for which X is approximately 0.4 and FX is about 1.286056 We continue in this manner, increasing the value of X by 0.2 at each step and finding the value of ffx to six decimal places. I'll let you do this on your own. So completely calculations. I get this table of values now you know, for the travels away rule that there are and plus one which is 11 terms in the some and the some, which will call the trapezoid Rule t 10. This is approximately are integral and exactly is besides a bar interval. Delta X, over to times air function evaluated at the 1st 0.0 plus two times function directed at the second point. X one plus two times the function evaluated at the third point, x two and so on, all the way up to two times a function of values It the 10th point x nine plus the function evaluated at 11 point x 10. And then, if you plug in the values from above to Delta X is too, and we have f of X zero throughout the next 10. On the right hand side here, you'll find the some is approximately 2.66 0833 Now, in part B here asked to use the midpoint rule, so this is going to be similar to part A. Our interval with is still point to, and now, instead of calculating values of the function at the end points of the intervals, you want to calculate them at the mid points of the Internet. Hence the midpoint rule. I'm going to draw another table here. So our first calm will be X and the second column will be the value f of X. And so we start off with instead of exit 0.0. We start with the midpoint of 0.0 and 0.2, which is 0.1 then will increase by 0.2 all the way until we get to the midpoint of 1.9 and 2.0. I'm sorry. 1.8 and 2.0, which is 1.9 Sorry. This should be three 1.5 0.7 and so on. And just as in part a. Well, simply plug in these values of X into our function ffx, which is eat the X over one plus X squared. So, for example, if you plug in X equals 0.1 F of X is approximately 1.94229 and we can continue in this manner. And so we obtain this cable of values, starting with 1.9 4229 and ending with 1.450302 At this point, I'm going to erase some of the results from part A to make room for those for part B. So, um, because we're using the midpoint rule, we know that the number of terms instead of being end plus one, is going to be only and minus two or a total of eight terms. I'm sorry, should be a total of end turning. So we have a total of 10 terms here and our approximation, which will call em 10. This is equal to formula is Delta X Times and then we simply have the some of our function values of these midpoint so f of x zero plus f of x one all the way up to effort X nine for these 10 turns and plugging in the values from the table into this, some in calculating using the fact that Delta X's point to this song is approximately 2.664377 So that's your answer for part B. Finally, in part C, we're asked to use Simpson's rule to approximate so using Simpson's rule is similar to the previous parts. I'm going to override part a here, so Part C begins in the same way that part eh began the step. Size is going to be the same 0.2, and for Simpson's rule, we're going to use the same table from part. So we calculate the value at the interval boundaries. So we start off with X equals 0.0. We go all the way an instrument 0.22 x equals 2.0, and we see the F of X ranges starting at 1.0 and then our table will end with 1.477811 We simply plug in the value of X and to eat the X over one plus X square to get the right side of the table. And with these function values, recall that Simpson's rule as yes and plus one or a total of 10 plus one or 11 terms, just like party had 11 terms. And here I'm going to erase the results from Part B to make room for the results from Part C. Now our approximation, which we'll call s 10. It's Simpson's rule in the n equals 10. Well, this is in the formula. The steps sides Delta X, which we know is point to over three times and there's some is f of the 1st 0.0 waas four times, half of the second point x one plus two times f of the third point, x two plus four times f of the fourth point x three. When we continue this pattern all the way until we get up to because two times ffx eight plus four times f of x nine and then plus just one times f of x 10. So are some does contain 11 terms And if you plug this result into your calculator and you use the results from our table, this son is approximately 26 decimal places. 2663 244

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Calculus 2 / BC Courses

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Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

Join Course
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