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00:59

Andy Scott

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

0:00

Virendrasingh Deepaksingh

00:56

Greninjack Dan

02:07

Fangjun Zhu

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Alright. So proving inequalities like this are very useful and analysis. Okay, so you have an integral and you’re just trying to get some kind of lower about. And why would you be interested in a lower bound? Well, if you look at this integral X to the fourth plus one square root, well, it’s going to be very difficult to break that up. Is every month some and I mean you can try it and it’s gonna be difficult to dio and some functions. It’s just hard to find the definite, integral, plain and simple. It’s not always easy. So here, what can we do? Well, we’re going to start with an observation notice that X to the fourth plus one. Okay, so X to the fourth is a positive number. Well, it’s not sorry. It’s a non negative number that doesn’t really matter. And then I’m adding one to it. So if I add one to a number that’s going to be bigger than well, the number If I hadn’t had added 12 so that’s certainly true now because both of these air positive numbers, if I take the square root of both sides, this inequality is still going to be true. And so we still have square root of X the fourth less one. But now the square root of X to the fourth is actually X squared. Okay, cool. So now so I guess I should mention that. Okay. When I take the square root of X to the fourth, I really did absolute value of X squared. So plus or minus X squared, but minus X squared is going to be even smaller than X squared. So definitely this is still true that the square root of X to the fourth most one is going to be greater than or equal to x squared. The positive square root. Of course. I mean, this is a little silly to say that this is greater than or equal to minus X squared. But that’s certainly true, but this is definitely the more interesting statement. Okay, so what we can do then, is we can actually say that the integral of square root of X to the fourth plus one by this kind of domination property. Because we have an inequality between these functions. We have an inequality between their areas. So and a girl from 1 to 3 of x square. So now it’s actually evaluate this definite integral and see what we get. And I bet you that we’re gonna get 26/3. But let’s see why. So we’re gonna partition up the interval from 1 to 3. So we have what? It’s the limit. Well, let’s write down what we’re looking for. So squared, This is the limit is in, goes to infinity of the width. If I break this interval of length to into in pieces so that will be two over n and then some want to end of So I’m starting at one and then adding I times two over in and squaring that because that’s the bunch. Now, this is just going to be some work, right? So this is the limit, as in Goes to infinity two over in, son, let me expand this square. So this is gonna be one plus two times that. So for I over and plus for I squared over in squared. Okay, so let’s come down here. Limit as in goes to infinity of two over in now. I’m just going to use the distributive and associative properties and do these sums really quick. So the first term we’re going to get in, it’s just one plus one plus one in times. Then we have four over in, and then we’re just adding one plus two plus three all the way up to end. So that’s in times in. Plus one over to. And then Plus, we have four over in squared. And then summing, I squared from one to end. So that’s going to give us in times in plus one times, two in plus one all over six. And now. Okay. What are we gonna get? Let’s do this quickly. So this is just gonna be too. So we’re just gonna have two in over in which is to take the limit here. We’re gonna have an eight. Well, the twos, they’re gonna cancel. Okay, so we’re just gonna have this is gonna be in squared over in squared, but we’re gonna have a factor of four. So it looks like plus four. And then here we’ll have a three. So 4/3. But we’ll actually have another factor of two. So it’ll be 8/3 and then in cubed over in cubes, it looks like plus 8/3. What is this? Well, we have 6/3 plus 12/3, plus 8/3, and that is like it. That 26. Every three. And so you see that this inequality is coming from this domination? That X squared is smaller than squared of X to the fourth plus one. So their areas must have this inequality, and then we just compute. Actually, what? This is because it’s not too bad and we get 26/3. So that proves inequality that we want to show. Okay, Okay. Here we have another inequality that we want to show. And where do we start? Well, again, the point is, we don’t know how to evaluate this integral. So we want to relate it to an integral that we do know how to evaluate. And let’s just think for a minute about x times sine x. Well, what do we know about Synnex? What inequalities doing now? Well, we certainly know this one, right, Because, of course, Synnex as being the y coordinate as we go around point, something in a circle will never be bigger than what. And of course, if I multiply both sides by X Now notice that X is really between zero and five or two. So it’s positive. So it’s not going to flip the inequality here. Okay, so we have the X sine X is less than or equal to X. Well, this is looking good, right? Because now he can say that this definite integral that were interested in X times sine x t x is less than or equal to the integral. So you’re the piper to of X t X Now I think we can probably do this one. Let’s see what happens. So I’m guessing that we’re going to get pi squared of rate, but let’s see, so x t x, That should be the limit his in goes to infinity. Okay, so the width if I break this up into in equal parts, that’s gonna be pi over two into pi or two over n, which is the same thing. Is Piper to in and then we add up all the heights of the rectangles. So that’s I equals one to end. Our function is just X okay. And then our points air just I times pi over to it. Okay, so those air kind of our partition points and we are evaluating our function and these partition points. But of course, you know, the function is just X, so we’re just that’s just the identity function. Okay, So take the limit and goes to infinity weaken. Factor out this factor of pie or two in to get pi squared over four in squared. And if I add from one to end just the values I 123 all the way up to end I’ll get in times in plus one over to. But at this point, I hope you’ve done so many of these limits that they’re really easy. The numerator is in squared, the denominator is in squared. I mean, what I mean is that Ah assim tonic, lease in squared or in other words, the leading term is in squared And then the denominators in squared. So I just take the ratio of the coefficient. So I have pi squared on top. I have eight on bottom, so I get pi squared over a and there we see here’s our inequality. So the inequality first comes from this inter girl being smaller than this in a girl. By this simple inequality about sine X. And then we just evaluate what this inequality is. What? Sorry, what this integral is. And then we get pi squared over eight. So that gives us the upper bounds on this definite integral.

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