which-of-the-integrals-displaystyle-int05_0-cos-x2-dx-displaystyle-int05_0-cos-sqrtx-dx-is-larger-wh

Name: which of the integrals displaystyle int05_0 cos x2 dx displaystyle int05_0 cos sqrtx dx is larger wh
Uploaded: 2017-10-01
Duration: 52 s
Channel: Numerade
Description: Which of the integrals $ \displaystyle \int^{0.5}_0 \cos (x^2) \,dx $, $ \displaystyle \int^{0.5}_0 \cos \sqrt{x} \,dx $ is larger? Why?

Question

Which of the integrals $ \displaystyle \int^{0.5}_0 \cos (x^2) \,dx $, $ \displaystyle \int^{0.5}_0 \cos \sqrt{x} \,dx $ is larger? Why?

Frank Lin · Accepted Answer

So who&#x27;s just Roger on access between zero two Pulling Fife? We know that X Square. It&#x27;s smaller than this. But on the other hand, we know that on this part Cose eyes a decreasing function. When when the argument is is from zero to high way are certainly in that part, which means we have. If we put coz, I hear it&#x27;s going to change the direction of equality. So this one&#x27;s going to be larger than this one. Therefore, we integrate them by the property of the integral. This one&#x27;s going to be quite a report with this one. The left hand side is going to be quick right hand side.

Amrita Bhasin · Answer

we know the art Innovex isn&#x27;t increasing function. Therefore, we know that our content signed X is less than or equal to, which is less than or equal to work. Tin of Axe. Therefore, we know that the integral from wondered too of Arc 10 of axe detox has the largest value because it correlates to this.

Enkhzaya Enkhtaivan · Answer

we have a problem. 24 Chapter 8.3, which is which is into a different intervals are too high off. Square it off, son. Trick function here. No Trig functions when they&#x27;re when they&#x27;re inside a square than that&#x27;s It&#x27;s usually very complicated for something. It&#x27;s It&#x27;s even impossible to obtain close, formless, however, And this one, if you make the correct substitution of things, will be very nice. Well, so when you see your function like oh, assigned to X, the first thing that should be coming to your mind is the no double angle for me. So I I write it here. So co sign off to X. All right, so this is either one cross to curse Sign Square X over too. No, I&#x27;m not. That&#x27;s over too. Actually, just excuse. We have to x here or this is equal to this is native one. I&#x27;m sorry. This is NATO one plus presenters, critics, or this is one minus two times signe square x. No, In our case, we have one minus 2% 2 square x. So if you pick this fifth, pick the second substitution, it seems like we&#x27;re gonna have a nice simplification, they&#x27;re certain. So when I stood when I scored two exes simply to times sine square x so that our our our interim becomes central from zero to pie, square it off two times. Signe sward x. Yes. Now you might be tempted to pull our just sign f ax from this word, which is which is correct. But in Jae in for this problem. But in general, the the only thing you can do is you can pull out the function with the absolute value here. But in our particular problem, we&#x27;re only dealing with the very blacks from there to pie over which the function sign of axis itself positive all the time. What should send non negative? That&#x27;s why you can forget about the absolute value. So Teo Teo really read my point for if if, oh if you&#x27;re dealing with from here dealing with the different interval, for example, if this if we had three over two times pi, then you would have to split in interim into two intervals over which sort of way, whether it depend on depending on whether it&#x27;s sign effects, is positive or negative, and then the absolute value of side effects will become just sign effects itself for the positive part and negative side effects for native part. So anyway, now this is a very simple, integral we know what what they entered here through this, it&#x27;s square of two aunts native co sign X evaluated that zero right, and that is two times screw two.

Robert Daugherty · Answer

Section eight that to number 24. We&#x27;re finding the definite in a row from zero to pie square root of one minus co. Sign of two. Ex First step is I need to be able to integrate this. Find the anti derivatives. So how do I integrate square root of one Money&#x27;s co sign to x d. X? I know that the sine squared of acts is going to be 1/2 one, minus the co sign two X that tells me that, too sine squared of X is equal to one minus co sign of two X. Then, if I were to take the square root of both sides of that equation, I would get the square root of two square bit of science. Squared X is equal to the square root of one minus co signed to X, which gives me the square root of two absolute value. Sign of X is equal to the square root of one minus co signed to X, so this integral turns into the square root of two and then the integral of the absolute value of the sign of X t X. So we&#x27;re asked to integrate this from zero to pie. So this is the square root of two zero to pie. Absolute value Sign of X DX. Now I know that the sign of X is greater than equal to zero when X is in the interval from zero to pie, their quadrants one in two. That means that I can remove those absolute value signs. So this turns into the square root of two absolute value of sine x, d x, and so the integral of sine of excess minus coastline of X. So this is negative square root of two co sign of axe evaluated from zero to pie. And so this just becomes negative square to two co sign and pies. Negative one minus the co sign of zero Um, is one. So, um, and sort out this. There&#x27;s one minus sign that I had missing here. So this is minus squared to the coastline of zeros one. So this is the square root of two minus negative square to two, which is just too square root of two. Okay, so that&#x27;s the value for the definite integral

Ruirui Liu · Answer

so left. Open this. No, it&#x27;s Oh, gross. And turn in Reno skirt with this won&#x27;t be changed. Ex wart. Yes, but the chain, there&#x27;ll be no sign. It&#x27;s worked in this in for a day or two partners, so this can&#x27;t be off before.

Robert Daugherty · Answer

Section eight that to problem number 20 through the definite integral 0 to 2 pi square root of quantity one minus co sign X over too. Okay, so let&#x27;s look at what we know. We know that the sine squared of acts is one minus the co sign of two X over too. That would tell us that the sine squared of X over too is one minus to co sign of X. There were two. So that&#x27;s gonna help us. Because now we can substitute and say that, Well, this is what we need to integrate. So if I&#x27;m looking for the anti derivative here, so this is going to be the anti derivative of the sine squared of X over too T x and square root of all of that. Okay, so again, I just made the substitution. That one minus co sign of X over two is a sine squared. So this just becomes finding the anti derivative of sine squared. OK, Now you know that this is going to be anti driven of the absolute value of the sign of X over too dx. Okay, so that&#x27;s what I would need to integrate. Now, let&#x27;s just go back to our definite integral. So I needed to integrate this over the interval from zero to pry. So that&#x27;s gonna be zero to two pi. Absolute value of sign of X over too. Okay, T X. So what do I know about this particular interval? So I got to figure out with this absolute value, what is the value dysfunction. Can I take the absolute value away, or do I need to get account for being positive or negative? So what I know is when I look at this, the sign of X over too. Sign functions going to be positive and quarters one and quadrant to serve. I know that this is gonna be greater than he go to zero when X is in the interval from zero to two pi because that would make this function living quarters one and two when you&#x27;re looking at the sign of X over to. So that means that this thing is where you see absolute value that is always positive from 0 to 2 pi. So this is the same as the integral from 0 to 2 pi. I can remove those absolute values. Sign of X over too T X When you integrate that you get minus co signs you&#x27;re gonna get minus two co sign X over to evaluated from zero to two pi. And so what this becomes is now the arithmetic problem. So this is, um, I have minus two. The co sign of two pie or two is the co sign of pie, which is a negative one and then minus negative, too. The co sign of zero CO sign of zero is simply one. So this becomes to minus negative to this becomes for being your answer. Okay, So again, what we did was we recognize this one minus co sign X over too. Um, we have an identity for that that we were able to use for the to be the sign square toe. 1/2 facts use the fact that this functions sine squared. One have X is positive on the enter from 0 to 2 pi. Allow me to deal with those absolute values that were Indiana

Frank Lin · Answer

Okay, we know that from 0.5, this court of X has to be greater than or equal to x squared. Therefore, we know that because co sign X is decreasing from 0.5, you know that co sign squared of acts has to be less than or equal to co sign of X squared. Therefore, we know that the integral from 0.5 co sign X squared d X has to be very correlates to this.

Problem

Prove Property 3 of integrals.

Problem 68 Medium Difficulty

Which of the integrals $ \displaystyle \int^{0.5}_0 \cos (x^2) \,dx $, $ \displaystyle \int^{0.5}_0 \cos \sqrt{x} \,dx $ is larger? Why?

Answer

$\int_{0}^{0.5} \cos x^{2} d x$ is bigger

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$\int_{0}^{0.5} \cos x^{2} d x$ is bigger

Calculus

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartCalculus

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

James Stewart

Calculus