Can YOU figure out this INTEGRAL'S STRUCTURE?!

Added: 2026-05-05

[00:00:00]Hi guys. Welcome back to another video of me teaching. And today I have this question on the board for you guys. So, why don't we just get into the question?

[00:00:10]Well, for this question, I'm actually going to show you two ways.

[00:00:13]And the first one will going to be King's rule.

[00:00:17]Now, King's rule is a formula for definite integrals where you can replace the variable with the lower bound plus the upper bound minus the variable.

[00:00:31]So, for example, in our case, we can replace all the x's with a plus b minus x and the integral will stay the same.

[00:00:39]So, because of that, if we use King's rule, this I will still be Now, if you plug in a plus b minus x into this yourself, it will just be actually the same thing.

[00:00:58]And then the top will be a plus b minus x.

[00:01:05]Okay, so since they have the same denominator, why don't we just add them?

[00:01:11]So, 2I will be the integral from a to b of the denominator stays the same.

[00:01:27]But the top will be a plus b because the x cancels out.

[00:01:33]>> [snorts] >> So, now from here, I'm going to take out the a plus b. So, 2I will be a plus b times the integral from a to b of the denominator I will unpack. So, it will be negative x squared and then plus a plus b x and then minus a b.

[00:01:59]And then the top will be 1 d x.

[00:02:03]So, now I'm going to um complete the square for this denominator. And this is already in a classical form of the inverse sign.

[00:02:13]So, we can change this.

[00:02:18]Okay, so the bottom will be Let me first take out a negative.

[00:02:23]So, then it will become x squared minus a plus b times x. Now, what do we need here to create a square?

[00:02:33]Well, you probably know that to make a square, you need to do this half and then you square that. So, you add half of this squared. But since we added this from nowhere, we have to subtract it.

[00:02:54]And then don't forget about this last minus a b. Actually, it should be plus because we are inside of the negative.

[00:03:03]And then 1 d x.

[00:03:06]Okay, so now I leave this as and then the integrand will become So, the bottom will be negative.

[00:03:19]That is of course a completed square.

[00:03:22]So, probably need another one.

[00:03:25]x minus a plus b over 2 squared and then it will be minus and then this if you can do in your or if you do this on your own, it will actually result to something unexpected. It will be b minus a over 2 squared.

[00:03:49]Very surprising.

[00:03:52]Okay, so from here we can unpack.

[00:04:01]So, it becomes square root and then it will be this minus this now instead of this minus this.

[00:04:10]So, b minus a over 2 squared minus this.

[00:04:28]Okay, so now we have this. This is in the in the classic inverse sign form, which is 1 over square root a squared minus x something squared.

[00:04:37]So, we can use the formula now. It will be a plus b bracket and then the inverse sign formula is inverse sign of this x, which is that over the a, which is this.

[00:05:00]And then we have the b and the a.

[00:05:03]So, if I leave this to you, then the solution is actually quite elegant. This indeed will actually just result to pi.

[00:05:14]So, now that we know that 2I is a plus b times pi, then we know that I will be this divided by 2, right? So, a plus b over 2 times pi.

[00:05:27]So, this is how we can solve this integral using the first way, which is King's rule and then a classic inverse sign.

[00:05:35]Now comes the second way.

[00:05:37]The second way comes if you think a little bit deeper here.

[00:05:42]So, let's just see.

[00:05:44]Wait, if you have watched my previous videos, then you know that when there's an elegant solution, it's always because we are observing the integral and we are making some subtle changes.

[00:05:55]Okay, so why don't we just look at this?

[00:05:58]Can you notice any relationship between the numerator and the denominator?

[00:06:02]Well, it's fine if you can't because they look quite different. But once I express this this denominator as y, so if I let y to be the denominator, then I can square both sides.

[00:06:24]So, y squared will equal to this and if we unpack this, it will be negative x squared plus a plus b x minus a b.

[00:06:40]Now, from here, some people might be able to kind of recognize what this is, but I won't I won't reveal it until later.

[00:06:50]So, I'm going to move all the terms to left hand side, so it becomes x squared minus a plus b x And now, from here, I'm just going to complete the square, okay?

[00:07:04]So, we need if you have So, from here, we need if you have watched the first method in solving this, you already know. It will just be a plus b over 2 squared. But then of course, you have to subtract a plus b over 2 squared.

[00:07:22]And then don't forget to add the y squared and then add the a b.

[00:07:28]Okay, and this will be equal to 0.

[00:07:32]So, this is a square.

[00:07:35]So, it will be x minus a plus b over 2 squared. And then I'm just going to add the y squared.

[00:07:45]So, if you have watched the first method, then you already know what this is. It's actually Well, if we move it to the right hand side, then isn't that the thing we just evaluated earlier? b minus a over 2 squared.

[00:08:01]And now, I think everyone should be able to tell what this sort of structure is.

[00:08:06]The relationship between the numerator and the denominator is actually they are both involved in an equation of a circle. How surprising, right?

[00:08:17]Well, from here, we can actually already kind of plot what this circle will look like.

[00:08:24]So, from here, let me just make it a bit bigger. From here, we know that the center will be at a plus b over 2 {comma} 0 and the radius will be b minus a over 2.

[00:08:38]So, if I put this to be a and this to be b, then this is a plus b over 2. So, this is where the center is.

[00:08:48]And the radius is b minus a over 2. So, b minus a which is this thing and then divide by 2 is just this.

[00:08:57]Wow, so nice, right? We actually don't have the full circle. We only have the top half.

[00:09:07]That's better. So, this is more This is more or less what the graph will look like.

[00:09:13]And also, if you thought that this was the only restriction, you're wrong because you also have to draw the open circles at a and b. So, this is actually the correct and valid drawing of this circle.

[00:09:30]Okay, so now we can try and work out what our substitutions will be, right?

[00:09:35]So, we can see if we find any point on this semicircle, x y and then we can figure it out by this, right?

[00:09:46]So, the x value will be Well, obviously, it's just this and then you add this little bit. So, this part is just a plus b over 2.

[00:09:58]And then you add the radius, which is b - a over 2, times the cosine of this angle, which I will call theta.

[00:10:11]Now, you can do the same thing with y, but y will be much simpler. It would just be the radius of this semicircle times sine of theta.

[00:10:23]So, this will actually be our most suitable substitution.

[00:10:28]Now, you see, this is quite cool, because just from this little structure, we can know a very cool substitution that we can use.

[00:10:38]So, we can do the dx, because we're missing that. So, the derivative of this is 0, because it is a constant. And then the derivative of this is negative b - a over 2 sine theta.

[00:10:56]Okay. So, now we can do the bounds.

[00:10:59]Uh let's do it here. When x is a, you can actually figure out that theta has to be pi.

[00:11:08]And when x is b, theta will actually be 0.

[00:11:14]So, from all this, we can actually know that this integral is just the integral from a to b of x over y, right?

[00:11:25]Okay. So, now this I will become integral from a to b of just x over y and then dx.

[00:11:34]And now we can put in our subs, so it will become the integral from pi to 0 of x, which is that, and then over y, which is that.

[00:12:03]And then times dx, which is this.

[00:12:13]Oh, actually I forgot to d theta here.

[00:12:21]Okay. So, now we see that this and this cancel.

[00:12:28]And also we can use this negative to swap the bounds. So, it will be the integral from 0 to pi now of that.

[00:12:46]And then d theta.

[00:12:48]So, from here, this is so simple, right?

[00:12:51]This is one of the easiest integrals that you might have done. So, this will be This is a + b over 2 times theta.

[00:13:02]And then you add uh So, this will be just sine, so b - a over 2 sine theta from pi to 0.

[00:13:20]Okay.

[00:13:21]So, I'm going to put this here. Now, let's just see when the theta is 0, it's quite obvious this will be 0. So, all we have to do is plug in when theta is pi. So, the first part will just be a + b over 2 times pi. And then the second part is sine pi is this is just 0, so this is nothing. And like I said before, when theta is 0, it's also nothing. So, this is how to solve this integral with a very beautiful substitution that requires you to have a deep understanding of how the structure forms.

[00:14:01]So, this is the final answer of this very intriguing and beautiful integral that I found.

[00:14:08]So, thank you so much for watching, and if you enjoyed my video and you want more videos like this, please consider liking and subscribing.

[00:14:16]If you want to master something, teach it.