solve-the-given-problems-by-integration-in-finding-the-average-length-barx-in-mathrmnm-of-a-certai-2

Question

Solve the given problems by integration.
In finding the average length $\bar{x}$ (in $\mathrm{nm}$ ) of a certain type of large molecule, we use the equation $\bar{x}=\lim _{b \rightarrow \infty}\left[0.1 \int_{0}^{b} x^{3} e^{-x^{2} / 8} d x\right]$
Evaluate the integral and then use a calculator to show that $\bar{x} \rightarrow 3.2 \mathrm{nm}$ as $b \rightarrow \infty.$

Julian Wong · Accepted Answer

problem. We&#x27;re asked to use a graphing calculator the approximate this integral. So from 0 to 10 of this function will get 9.9995 from 0 to 50 we get 49.9875 from 0 to 100. We get 99.9500 and from 0 to 1000 we get 9 95.166 in part VII. They&#x27;re asking us to look at these red numbers and determine whether it&#x27;s convergent and divergent. Well, there&#x27;s no obvious signs that it&#x27;s slowing down. So from our answer and 48 we would conclude that this is divergent in part. See, they revealed to us that this integral is actually convergent, so they want us to figure out what it would converge to. Therefore, we have to take the improper and a girl from zero to infinity. So we&#x27;re gonna take the limit as B goes to infinity from zero to be of e to the negative 0.1 x dx. So we have to limit his B goes to infinity. The empty derivative is gonna be negative 100,000. So that&#x27;s five zeros e to the negative 0.1 x and we&#x27;re gonna evaluate this from zero to be, that means limit has bee goes to infinity of negative 100,000 e to the negative 0.1 B minus. When we put zero in there, it&#x27;s minus and negative 100 thousands of 100,000. So as he goes to infinity, this first term will go to zero because you&#x27;re taking E to the negative infinity. So this is zero plus 100,000 which means it&#x27;ll converge to 100,000.

Bobby Barnes · Answer

we want to use a calculator to a fox debate Interval zero to beauty of E to the negative X squared DX four is one ten and twenty. So I went ahead and calculated these beforehand. So on zero to one, we get zero points have been forcing state about on zero to five, we get zero point eight eight six two two about and for these other two, we end up about the same. So depending on what the leader you have, you may end up with just being able to place us indirectly and evaluated. I know when some of the older calculators you&#x27;LL end up with some kind of thing where it looks like of in and and the way you want to plug this in, it&#x27;s going to be your first plugging your function so hee to the negative X squared. Then you plug in your variable that you want to take this in the vault with respect to the X, and then this would be your zero to be in this case. So this is how you build putting this into some of the order calculators, but a lot of the newer ones will actually allow you to put in this definite integral. So the first part is really just a issue of how to plug it into your calculator and the reason why we approximate this is we can&#x27;t really get eh integral for this using methods we know so far there are other things you can do to drive you into this. But with what we know so far, this is pretty much and impossibility for us to do now for part B. It wants to know what we think is going to happen as he goes to infinity. That was going to be coverage in or Virgin and well, from zero to five here with the tenants here with twenty, all are approximately the same value. So I would say, B, I would think this is going to be your car. I would say this is going to converge and I would say is going to converge to something approximately zero point zero point eight six to two. And in this last part says we want to explain why this is the girl should become virgin comparing CT negative X squared with into negative X for X one. And we should know that the actual thing we&#x27;re trying to integrate over zero to infinity is going to convert to the square root of pi over too. So Oh, I should probably actually stop is on page. So we have zero to infinity of e tonegative square, the ex. So they were saying to compare this on one to infinity or at least one ex is larger than one. So let&#x27;s go ahead and break this up into two intervals and girls zero the one of this wass one to infinity need to the negative exports yucks. Now, one thing to note so on one to infinity he a negative expert, will always be smaller than e to the negative sex. Since these air both one over X where e to the X one over. So since it&#x27;s one and larger listen are equal to, I should say, since we&#x27;re including one here, So any number we put in from one to infinity if we square it is going to be larger than just X and thus would have a larger number and our denominator here and so that should be smaller. So we can use comparison to say that this here so this here should be larger than our equal to So the hunt for the integral zero one of either negative effects The X wass Inter goal on this is what we&#x27;re changing one to infinity of each of the negative x the ex And so we changed out the negative expert here for the negative x sense over here where state these are larger now We don&#x27;t know how to integrate this first part, but we know this is going to be a finite about you, finite. So I&#x27;ll just go ahead and call this, eh? So this is equal to so some value which I&#x27;ll just called Capital A and then we could go ahead and integrate this improper interval by first doing no limit as be Kercher&#x27;s in Bennett, you go from one to be he to the negative x, the X and I should probably keep being my London color. There&#x27;s a show going together So those two now we have capital A which is just a constant wass integrating e to the negative x here, So don&#x27;t be a divinity. So that&#x27;s one to be negative. Negative X and then we have one to be and plugging in. Each of these values you end up with this is a close, the limit as being a purchase. Impunity. Oh, so we plug and be so it&#x27;s gonna be negative. Negative into the bean, minus a negative e to the negative one. Okay, well, there&#x27;s not wanted to cooperate right now. It seems so. We have this now the limit as he approaches infinity of e to the negative B is going to get physical back to zero and then the limit of eating negative one. Well, that doesn&#x27;t depend on me. So that&#x27;s just going to go to it&#x27;s so so we can rewrite this on trying screwed all this up. So yes, he was going to be a plus. So these two negatives council Oh, that&#x27;s going a plus one over, he says. It&#x27;s a negative on just being. And so we have that this integral here is going to be less than and made us aside and make this to a little bit more. So over got from this Is that integral from zero to infinity need to negative X squared DX is one to be less than or equal to some constant, eh? Plus one over. And we know that this here is going to be bigger than our equal to zero sense. So you know this here, since each of them that do X squared is always better than or equal to zero. So since this is always equal to zero, this should be bigger than zero. And this is between a plus one over. And actually, maybe I should say a little bit more as to why this right here is actually a finite. The graph of this here looks like this. So this is what the functions graph looks like. So this zero no one is talking about the area from zero two one, and this here is just be a by night value, so we at least know it&#x27;s a finite value. So that&#x27;s why we could say that this here is a finite value and likewise, we know that this was the larger than zero. So the integral of this is something between whatever this value A is or the definite integral zero one is going to be cross one over

Katelyn Chen · Answer

were given only integral on your recipe number things. So, first of all, we are asked to violate this using right William in some with eight and points. So 87 trebles. So, first of all, we need to figure out our what are erecting glue with so to start off with, we know that a rectangle with is equal to being minus a over. And where is the upper limit for mija? Lower limit zero over the amount of soap intervals, which is eight. So we know that our Martina with this 1/2 So we are going to be evaluating our right, leaving some. So we know that this sequel to 1/2 times now we need to come with all our right and points, and so we know that this will be equal to you. Um, so we&#x27;re Axis because we&#x27;re using Brian points in each. One point we know is half. So we will have We have to evaluate our function at 1/2 1 1.5 to 2.5, etcetera. So then, if we put that all inventory calculator, you should come out with negative 5/4 minus two, minus science force minus two. My next five force plus zero plus seven Force Plus for we played that all ensure calculator which get is equal to negative 1.5 That is our right wing and some right over here. This is for part a, right? No, for purple. We were asked a graph office to show it. So if you look here is the graph of our functions in our second graf will be for for part D. So this is for part B. And so we&#x27;re asked to graph and showed us graphically. And so we know that this is based off of eight step intervals. So we know that they sell for right points. We have it will look something like this. So we&#x27;re going based off of our right points. Yeah. One over here and then over here is just like a straight line is right points ago at the X axis. So this is the area under the curve that we have graft. So you, not the area above over here is positive in the area below its negative. All right, So for part c, we&#x27;re asked to evaluate this integral, using the definitions that we have learned. So if we scroll down here, we&#x27;re doing currency. So the first thing that we want to do is we want to evaluate. Um actually Okay, so we know that&#x27;s integral Can be religion as the limit that&#x27;s un approaches infinity of So we have our rectangle with consummation over icicle toe. One of Assad&#x27;s A plus or drink are erecting west, which we know is equal to the limit. It was un approaches infinity and earlier for part a we found our rectangular with we found it to me four over on. I&#x27;m so eyes equal to one of end of f off, you know, a zero plus that should be looked to me with. So then we have our thing with I No, um, I actually can put on the numerator. Okay, so then now that we herbal this, we can simple five down to the limit as an approaches infinity of four over end And then we know that we need to put Is there a post for over and wishes forever and back into its original function. And remember, original function is X squared mice three x So then we can just pocket in So we have for I over end squared plus three times for I over and all right, so now that we have all this, but we can do is we can factor outer concepts. So you get the feeling it was gonna approaches infinity four over on. And then we have. So you factor out a 16 over and squared. So we&#x27;re just left inside with I squared minus 12 on, cause remember, is a constant because it would be any given us are not told in 12 over end times Mission Been icicles, one of I. And so then we know what the summation are. Square is. We know summation. I is. So what we can do is we plug it back in our definitions that we know. So we have four over on. And so then we have 16 over and squared. We know that this entire thing is equal to times 10 times and plus one times 21 plus one over six. We subtract 12 over end times and times on, plus one over. We all went to The next thing we do is we can pull out all of our different constant. So we have the limit was un approaches infinity of 32 or three times one plus one over end times she a plus one over on minus who, me? Four times one plus one over. And we know that one plus one or dilemma as a nonpartisan, many of one over and simply equal to zero. So we&#x27;re left with 64 3rd minus 24 which is equal to they live 8/3. All right, so now the last part of her problem, we are us the graph of what we just found in part C. So if we scroll back of check graph this time, instead of using Riemann some and rectangles, we just found the area and above the curve, so you know that it&#x27;s area underneath the curve. So we have just found the area of this section over here, and then this section down here and this was for a party

Matthew Allcock · Answer

in this video, we&#x27;re gonna go through the answer. It&#x27;s a question of a 70 jumped 8.7. It&#x27;s a restroom. First show that this integral I&#x27;m in here diverges. I never got a few actually things that we&#x27;re gonna show afterwards. First of this diverges let&#x27;s use a a limit comparison test. So first we know that the integral that&#x27;s why the Inter ground is positive. And also it has no singularities on the ranges over to infinity. So that means we can look limit as X goes to infinity. I have the inter around two X next word apples one divided by one of eggs. But this is the limit one plus one of X squared. So as expressed infinity one over X squared goes to zero this limit just because Daddy which is great and zero on finite four since from zero to infinity about one of X, the ex diverges that we can say this integral, uh also diverges so that just then we know that the integral which contains that diverges this integral diverges so that was supposed to show without the limit has to be because you&#x27;re infinity off the interval from minus B two B to US X squared plus one DX converges. Okay, this is very easy to see because we just know that the inter grand is that what function off X is integrated from Ah, minus B two b Therefore, the integral over the negative exes and the school is supposed to have access to counsel each other out is that far This integral converges and covered is 20 So this shows that we can&#x27;t just define this integral was this into

Fuzail Shakir · Answer

Okay, so you have the integral from 0 to 4 of X squared minus three x t x. So using the Ryan Sons with right and Flint&#x27;s, we have our eight just playing everything into formula. You end up with negative 1.5 and you gonna graft this deceit according week, taking the actual integral. We end up with X cubed divided by three minus three over two X squared. We&#x27;re gonna evaluate this from 0 to 4. So this is equal to force the power of three divided by three minus three over two times 16 which is that simplified down fourth you divided by three minus 24 which is equal to negative eight over three.

Problem 42 Easy Difficulty

Answer

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 1 / AB Educators

Catherine R.

Anna Marie V.

Kristen K.

Joseph L.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 28

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Basic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Kristen K.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus