# Ibid Ex 4.3 #7

Okay, this is going to be a test run of how we're going to be collecting homework for distance learning. So somebody posted last night what's my question and I believe this is from exercise 4.3 in the new image. And the question asks us to use the first four terms of the Taylor polynomial to approximate the definite, integral of arc tan square root, X between 0 & 1. So in order to answer this question, the first thing we need to know or remember is that the arc tan X has a MacLaurin series expansion. That looks something like X, minus X cubed over 3, plus X to the 5 over 5, minus X to the 7 over 7.

Okay. The general term is X to the 2n plus 1 over 2n, plus 1, minus 1 ^ and would start off negative. If we want to start off positive.

So n, plus 1 will work for us. This formula is on the formula sheet or in the data booklet, but I believe it only has the first three terms. So only this portion here you'd have to figure out what the fourth one is if they work, the first four terms, okay, how is this going to. Help us answer the problem. Well, we can use this to determine the arc tan of the square root of x by simply making a composite function. So this is the square root of x, minus square root of x cubed over 3, plus the squared of X to the fifth over five minus the square root of x to the power of 7 over 7.

Those are the first four terms of the arc time of the square root of x. So we simply just substitute the square root of x in where we see the X. Now we see the questions asking us to evaluate a definite. Integral so we're going to take the definite integral of Arc's, tan root X, which means we're just going to integrate the function that we just determined in order to integrate that it might be easier to write it in exponential form. So it's square root of x is X to the 1/2. And then we have 1/3 times X to the three halves.

And then the next term is X to the five Aspen's. You can put it over five in the last one, which is minus will be to the 7 half over seven. We want to integrate that term by term.

So. The theorem says that if the MacLaurin series is convergent on an interval I, then the integral of the MacLaurin series will also be convergent. And that interval is found by integrating each of the terms of the MacLaurin series term by term. So this will be X to the three halves. And then walked by two thirds because we're dividing by three asses this one will add one to the exponent again in multiply by its reciprocal I'll, just put the one-third underneath X to the five half. And then we have X to the. Seven halves two over seven times five.

And then the last term here is X to the nine halves two over nine, divided by seven. And those are the first four terms that we're going to use to determine the definite integral. And since this is a definite integral, we have to sub in the limits of integration one in the zero, and then take the difference. So this leaves us with two thirds now it's 1 to the three-halves, which will be 1 each one of these X's here here here. And here will receive.

And we substitute. X in for those. And so we know they're all going to be 1 so 2 over 5 times, 1/3 2 over 7 times 1/5 in 2 over 9 times 1/7. Because whenever you put 1 to any power, you get 1, and then we would have to do that we'll go on forever. And then we would have to subtract leave at the value she's up in 0 where you've highlighted all those X's.

Well, of course, that'll just give us 0 minus 0, plus 0 in other words, 0. So in fact, then we can approximate this definite integral by the four fractions we listed above so.2/3, plus 2 over 15, plus 2 over 35 minus 2 over 63 we'll, put a squiggly equal sign there to say, we're, approximating it. Now in order to figure out what that is a decimal we'll just use our calculator off to turn mine on here. Let's see it's been turned off. Okay and let's put that over here on the side of the screen, and we'll clear all that out.

So if T is 2/3 I see made a little error here. This should be minus biggest alternating signs. So minus 2 divided by 15, plus 2, divided by 35 minus 2, divided.

By 63, which gives us an approximate value of about 0.5 5/8 seven-30. So we'll put another approximately equal sign. So two three significant figures. The answer is point, five, five, nine.

Dated : 18-Apr-2022