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Given the integralg(t) =J e dxUseSimpson s rule with subinterval. n =8 to approximate the value of g(2) Given that the actual value for g(2) is 0.9953, calculate th...

Question

Given the integralg(t) =J e dxUseSimpson s rule with subinterval. n =8 to approximate the value of g(2) Given that the actual value for g(2) is 0.9953, calculate the error for solution obtained in (i).ii.

Given the integral g(t) = J e dx Use Simpson s rule with subinterval. n =8 to approximate the value of g(2) Given that the actual value for g(2) is 0.9953, calculate the error for solution obtained in (i). ii.



Answers

Using Simpson's Rule Approximate the following integrals using Simpson's Rule. Experiment with values of $n$ to ensure that the error is less than $10^{-3}$. $$\int_{0}^{\pi} \ln (2+\cos x) d x=\pi \ln \left(\frac{2+\sqrt{3}}{2}\right)$$

We want to find an error bound on approximating the following integral with the trapezoidal and Citizens Rules. The integral of the integral from 0 to 1 over square one plus X. Where n equals eight. For the trapezoidal rule, remember the rebound is sometimes used as a cube over 12 and square where M is the maximum of the absolute value of double X. On a B S. Double prime is 3/4 times one plus X, divided by one plus actually five half. Since this is constantly decreasing on a B. S. Double prime is maximized. That kind of A or F prime of zero equals three. Force. Thus the error is 347 to minus zero cubed, divided by 12 and eight square or one over 128. For Simpson's rule, the errors and time is running out of the fifth about 180. End of the fourth. This time M is the maximum of the absolute value of four. So F 405 over 16 x was one of nine half again, since the dividing by experts, one of the nine half means F four Xs max at a that's for F 40 it was one of 5/60 years or max. Thus, E is seven over 24,576.

We want to find an error bound for approximating the following integral with a trapezoidal and symptoms rules. The integral of integral from 1231 over X squared dx, where N equals eight number of intervals. So first for the trapezoidal rule, remember that the air around is M times B minus a cube river 12 and square where M is the maximum of F double prime X absolute value for a B. So ethical prime X is six for extra force, which is constantly decreasing on a B. Thus the max is at the time of day or one equal six and the error is 6 17 minutes, one cubed divided by 1200 square or 1/16. Next to Simpson's rule. The urban remember is entitled United State of the fifth, divided by 100 the end of the fourth. This time is a maximum of the absolute value of F four. So for f for 120 over extra six, which is again decreasing on a b, f four A or f four of one equals 1 20 is the max. Thus the error is 1 2010, 3 most, one of the fifth, divided by 180 times to the fourth, or one divided by 192.

We want to estimate the theoretical error. If Simpson's rule with an equal 20 is used to approximate the interval from 1 to 5 of x squared minus four over x squared plus nine. So let's start by defining function F of x equal X Square -4 over X Square Plus nine. And we consider that function to find On the closed interval from 1 to 5. Mhm. In this case the denominator of this function is never equaled Sears or the functions well defined. That is because the expression X squared plus nine is greater than nine for all eggs. Uh And then as if we defend like this, we know that The 4th derivative of F is equal to if we use a calculator, CASS Is equal to -312 times five X. to the 4th minus 90 X square plus 81. Yeah. Over X plus nine X square plus nine. Sorry to the fifth. So this is the 4th derivative. And to Find a bound for the absolute value of this function on the interval 15. Um We use rather geographic or the plot of the function over that interval. Yeah. And we have that over here. Mhm. So we have this graph of the 4th derivative, the absolute value of the 4th derivative. And that is this craft here. And as we can see the maximum values that has a value of the fourth derivative is about 0.1806 which is attained of course at 1.73, 2. More or less. That is The 4th derivative of F. His less than or equal to 0.18 06 for all X. In the Interval 1 5. And that is important to know this bound for the fourth derivative of F. In its value. Because the error in Simpson's rule search that the absolute value of the error, let's call it E N S N. Is the number of seven table as refers to Simpson's rule. And in absolute value, that error is less than or equal B minus A. To the fifth is the lower limit of integration. Visa over limits over 2880 into the fourth times. M. And that M. Which appears here in this bound. Yeah, is an Oprah bound. Yeah. To the absolute value of the 4th derivative of F over the changeable A. B. That is. And is such that the absolute value of the 4th derivative of F at eggs is less than or equal to M. For all eggs India interval A. B. That's the theoretical value bound for the absolute value of the error. Inter upside the rule sovereign. Since this rule, when he is in the hands of intervals. And that Bound depends on the calculation of abound for the absolute value of the 4, 3 vegetables function. That's why we started by doing that, calculated the fourth river there. In fact music uh very system and automatic or computer. The very system to find this formula. And then we plot defunct. Dysfunction in absolute value over 15. As we see here we found that the opera about. And a tight upper bound for the absolute value of the 43 of the T v 0.1807. That is we have this. So we can take here You can take em equal 0.18-6 which is a very good upper bound which is very close to the maximum value of the function. 4th derivative in natural value. So we take em and then The absolute value of the error. And Simpson through with 27 intervals which is the number of end which was given in the statement of the problem is less than or equal to five which is the building the upper limit of integration Minour the lower limit one to the fifth over 2880 22. The 4th Time zero points 1 806. And using a calculator we found this about four points 0133 times 10 to the Native seven. That is the absolute value of the error in Simpson's rule with 27 intervals is less than or equal to 4.0133 times 10 to the -7. And that we For that we found a bound for the absolute value of the 4th derivative of the integral f. Using the graph of the 4th derivative in absolute value and found in a very tight bound upper bound that is about that is slightly greater than the maximum value of the false reality In absolute value over 15. And with that we've made the calculations in the theoretical formula of the absolute value of the error, or the bound for the absolute value of the error in Simpson's rule with and serving tables. And we found this bone over here.

Here we have definite integral from zero to buy of the natural logarithms of two plus co sign of X. And we know that these intervals people to buy times natural, worry them off two plus square to 3/2. To keep the annotation simple. We're gonna call these exact solution. I think I we're going to use that name in our solution here on. We want to approximate this integral use in Simpson rule in such a way that the absolute error we obtain is less than 10 to 93. So we got a fine and numbers of interval, which, which give us on approximation is it seems a rule with absolute error less and 10 to 93. So, first of all, we're going a use. You must say we're gonna use a computer implementation. Oh, since, um, so in particular, we're going to use I metal have implementation of since in through, and using that we're going to proceed as follows. We're going to calculate successively the citizens rule approximations to the Inter girl and calculate for each of those upper summations, we're going to calculate the absolute error and we are going to do that until we find a nearer place in 10 to 93. So remember, debt Seems since rule requires to use a number of serving tables, which is a even number. So we start at s of two. There is Simpson through using to serve intervals. In this case, we calculate the absolute error, which which is as of two minus I and I is the exact solution or the acceptability of the definite into We have them so and we found that these values approximately equal 6.72 times stand to the native to and because these values not less and 10 to 93 we calculate the following approximation within some rule, which is as a four, that is force of intervals. And we calculate the bearer which is in this case ever seemingly equal to 2.6 seven eight times 10 to the end of three. So thistle is not less and 10 to 93 because practice factor here. So we can't relate to following Juan it's of six and the absolute dear in this case, opportunity equal to one point to nine times a tent on and before and this year is lessons into 93. So we can we can take yes of six as a valid answer to your problem. And we have that, um, it's of six machines. The approximation to giving integral is approximately equal to one point 959 888 28 So this is the approximation to given into girl use. It seems some through with a nearer which is less than 10 to native three.


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