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Approximate the value of each of the given integrals by use of Simpson's rule, using the given values of $n$. Exercises $8-10$ are the same as Exercises $10-12$ of Section 25.5$$\int_{-4}^{5}\left(2 x^{4}+1\right)^{0.1} d x, n=6$$
Calculus 1 / AB
Chapter 25
Integration
Section 6
Simpson's Rule
Integrals
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were asked Approximate Doesn't his role here with the trapezoidal cool. So before we do that, you need to fill out the table on the right side of the table on the left. So are Delta X ISS being minus a over, and so, in this question, are a is negative. One R B is three. In the end, it's four were dealt. Eggs is just one. That means we will start a negative fine. And we go up by one. There are 123 and I was excited. For each of these excise will be one over two three over five, really before one and hurry over to. Do you find these values by just putting your ex I into the back. Such is here. So now, with all these values, we can put them into our formula, for the trap is little. Well, part ay says integral. From 1 to 3 of three over five minus X, the X can be approximated by be minus a over end, which was calculated up here. That's just one times that's half time us through over Size was three over for plus one, and the last term is half of that so half of three over two. And if you plugged in and your calculator, you'll get this is 67 over 20 hate Now Part B. We do the same thing, but this time you seem so as well. So that says into little from negative 123 of free over five minus X yeah, can be approximated by be minus a over Fri end will be Mine is a S 43 times on a 12 and that's times what have plus four times free over five plus two times three over four, four times one last 13 over two. Well, if you put this injury calculator, you'll get it is equal to 33 over in the last part. Pricey. Or ask the justice I await the integral No, that's from negative. 123 five minus x d. X. So the three on top is just a constant, which means they can actually just pull that out. That's three times integral of negative 123 of one over five, minus X d x and the anti derivative of won over five minus X is just why five minus X, but because they're the negative. We have to account for that. Which means they're gonna multiply thy negative one. And that's a relative from negative one three. We'll just move the negative, the front. So that's negative. Three. And that would be one of five minus three. So that's long, too. Line us on a five minus negative one of Lawn six and using our log rule. We know that one of two minus line of sixes just long of 1/3 and using the other law girl that tells us we can put this constant up as a power that's long of won over. Three. The power of Negative three, which is just one of 27. So that's your final answer. Um, if you want, you can put it into your calculator. That is just awfully. Three point 296
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