Applied Calculus for Engineers - Part 3: Single & Multivariable Integrals, Intro to Diff. Equations | Mark Misin | Skillshare

Applied Calculus for Engineers - Part 3: Single & Multivariable Integrals, Intro to Diff. Equations

Mark Misin, Aerospace & Robotics Engineer

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29 Lessons (6h 30m) View My Notes
    • 1. Promo Video

      1:13
    • 2. discrete VS continuous functions

      10:44
    • 3. integrals intuition 1

      10:30
    • 4. integrals intuition 2

      14:36
    • 5. integrals intuition 3

      11:56
    • 6. integrals intuition 4

      13:10
    • 7. integrals intuition 5

      13:42
    • 8. integrals intuition 6

      17:47
    • 9. integrals intuition 7 indefiniteVSdefinite integrals

      15:02
    • 10. DefiniteIntegrals 1

      22:04
    • 11. DefiniteIntegrals 2

      12:40
    • 12. Improper Integrals

      20:42
    • 13. integrals formal definition

      20:41
    • 14. CircleArea1 initialConditions

      13:29
    • 15. CircleArea2

      13:58
    • 16. Integrals VolumeForSolid

      15:58
    • 17. integration by parts

      14:03
    • 18. arcLength 1

      16:19
    • 19. arcLength 2

      10:26
    • 20. arcLength in space parametricEquations

      12:31
    • 21. multidimensional Integrals intuition 1

      16:55
    • 22. multidimensional Integrals intuition 2

      17:31
    • 23. multidimensional Integrals intuition 3

      9:54
    • 24. tripple integrals intuition

      7:50
    • 25. quadruple integrals intuition

      6:26
    • 26. tripple integrals application

      16:00
    • 27. differential equations intuition 1

      16:09
    • 28. differential equations intuition 2

      12:42
    • 29. 156 time frequency domain

      5:28

About This Class

Dear Students,

Welcome to the course! I am very happy to transfer my knowledge to you. This is Part 1 of the course (out of 3 parts + EXTRA).

WHY should you take this course? Fair question!

In Science & Engineering, literally EVERYTHING is based on Calculus. From Mechanical & Aerospace to Electrical & Computer to ARTIFICIAL INTELLIGENCE & MACHINE LEARNING and many more.

If you REALLY understand Calculus and you are trained to APPLY it to REAL LIFE problems, learning and understanding more advanced material will be much EASIER. Your life will be much EASIER!

In my course, I WILL DELIVER just that. I WILL make you understand Calculus INTUITIVELY & in terms of APPLICATION. Just give it a shot!

These are the topics that the course covers:

  1. Single variable integrals

  2. Multivariable integrals

  3. Introduction to Differential Equations

  4. Extra Wisdom - waves in time and frequency domain

I wish you good luck in learning!

Mark

Transcripts

1. Promo Video: Welcome to the applied calculus course. My name is Mark, and I want to help you master calculus. After this course, you will walk away with strong intuition and understanding of calculus. And you will be trained to apply it in real life to the level not seen in other courses. You're going to learn about single, double, triple, and even quadruple integrals and also introduction to differential equations. And in addition to that, I'm going to create this very important bridge between theory and real-world applications. This is really powerful stuff. My promise to you is this. If you give me a chance to teach Calculus, I will make it second nature to you. It will be a great investment. Take a look at some of my free videos. And if you like what you see enrolling in the course, and let's get started. And looking forward to seeing you there. 2. discrete VS continuous functions: Welcome back in this video, we will start making our way into the world of inter girls. My explanation will be slightly different from the standard explanation to properly explain to you what they are. I first need to show you the difference between discreet and continues functions that will make the explanation of integral a lot more intuitive. So let's get started. So I have a typical function here, a problem y equals x squared. And here this and this kind of function is called a continuous function. And you can know it because if you look at the line, it's a continuous line. You don't have any discrete steps. Say that I want to discuss ties dysfunction now. Well, I could do it, for example, with Resolution one. And I could do it, for example, like this, so I can take the middle point 00 and then I can go 0.5 to the right and 0.5 to the left. All right. So I would have 0.5 here and minor 0.5 here. So the total distance of this red line here is one so 0.5 on one side and my 0.5 to the other side. So I could just say that instead off having discontinues black line here, I can say that. Okay, since at 00 my function, my y is zero. Then when I considered the interval from minus 0.5 to 0.5 than a love AL use from minus 0.5 to 0.5 will be zero. Now, the next interval that I will consider is from 0.5 to 1.5. All right, so I have 1.5 here and then I'm gonna take the middle point of it, and I'm going to see what the values and the continues function from the middle point, which now is one. So it's here. So if my ex is one, then my continues function tells me that wise one. So I'm going to say that from 0.5 till 1.5 my value my why will be one like this, you see, So it's another continues line anymore. It's a step. It's like a stare. So from 0.5 to 1.5 my values one and the same thing from minus 0.52 minus 1.5. So the same thing here, miners one is the middle point. And that will give me why equals one. So I can say that from minus 0.5 to minus 1.5 for a while. These access my wife will have only one value, which is one. And then if we go from 1.5 to 2.5 and then I can take the middle value with just to the mill axe. And then why would before? So I can say that from 1.5 still 2.5, I will have four old values For these access from 1.5 to 2.5 will be four and same thing on the other side from minors 1.52 minus 2.5. My milk point from minus two will give me four. And from minors 1.5 till my stupid five. When x equals from minus 1.5. Still mine. Stupid. 50 My wife will be four. So you see, now you have this kind of steps here in red. You don't have a continuous line. You have some kind of steps and some kind of discontinues changes from my 0.5 to 0.5 year zero. And then all of a sudden you jump onto the next level and then again, you jump onto the next level and this is called at this Crete function. There you go. In red, you have now discreet function. No, continues line, just certain levels. And then they change dis continuously. So it's discreet function in black. It's a continuous function, but in red is discreet function. And since my resolution here was one, for example, from minus 0.5 to 0.5. So that distance here was one and again from 0.5 to 1.5. It was one. I can say that one of these intervals, I could call them Delta X. All right, So I have an equal interval. I chop my axe into equal intervals and I can call it Delta X. And in this example, my Delta X would be one. However, what if I decide to increase my resolution? What if I want to chop my axe into smaller pieces? What if I want my Delta X to be 0.5 Well, if my Delta X equals 0.5, then you can see that I will have more levels here. So again, the black line is the continues function. But since my delta X is smaller than I chopped my axe into into a bigger amount of intervals and the for example, here, this interval would span from mine is your 0.25 up until 0.25. All right, so the total resolution is 0.5 eso that told distance. The told that the exes is your 0.5 and then if I go from 0.5 tells Europe on 75 I would have this level from 0.75 Still 1.25. My level would be here and etcetera, etcetera. And you can see that now, compared to this function here, I will have a lot more levels. So here have 12345 And here are 123456789 So you can see that if I decrease my Delta X, then I increase my precision and notice One interesting thing. The smaller this Delta X becomes, the more precise I become and remember, the the most precise scenario is a continuous function continues. Function gives you the most precision, right? With discreet functions, you lose a little bit of precision because in reality when you're is Europe on five here, if your exes you open fire, then you're why would be 0.25? But then, if you describe ties, it than your zero point five would be zero. So you see you with TheStreet functions, you lose some kind of precision. However, if you describe ties your functioning to smaller and smaller pieces, then you will have more and more off those pieces, and your function becomes more and more precise. And so the smaller your Delta X is, the more precise your function becomes. And if your Delta X approaches zero right, so the closer it gets to zero, the more percent your function becomes. So if you're dealt, thanks approaches. Zero, then you're discreet function. I'm gonna call DF. You're a disgrace. Function approaches. You're continues function. Does that make sense? Yes, it does. So the smaller your death tax is the mawr intervals you will have on the X axis. The more levels you will have the more precision you will have, the more closely you approximate discontinues function because in the end this was functions, the approximate continues functions and then depending on the resolution, then you can choose your precision off how you approximate your continues function. So the higher your resolution, the more closely you approximate your you're continues function. And so if you're dealt, thanks approaches. Zero then you're discreet. Function approaches your continues function. In other words, if you're del tax approaches zero, then something like this something which is discontinues like this set of stairs would approach this continues line. 3. integrals intuition 1: welcome back. And now we have a problem. We have a building here that looks like a parabola, and our problem is to find the area off the war on that building. So it's some kind off hangar, right? It kind of looks like a hanger. If you think about it. If you look at it in three dimensions, So some kind of hanger, maybe some kind of military hangar or hangers like that you can find them in airports and then you have airplanes inside, right? And this side of the hangar has this parable of shape. OK, so of course, hanger this hangar in three dimensions would look something like this. But then we're interested in this side here, okay? And we want to calculate the area of this side and we can mull this side exactly with a probably function because it was built using the parabolic function. That's one thing that this hangar, these ears, your point is not here. 00 point is here. How would this function look like if it was symmetric? How about those ears? Your point? Well, the problem would be something like this, right, But now it's shifted its It's here. So everything starts from here. Here you have the 00 point. So you have tow shift the function in this way, right? In other words, this point needs to come here. So remember how we shifted functions. If this would be original function and you can see that why here? The highest point of the hangar is 16 meters, right? And then the other dimension do with of the hangar is eight meters. Then if we started counting our hanger from here, if this was our 00 than our function would be, why equals 16 minus X squared, right? That would be this. But now we need to shift the function. Right. So this parable here, we have to shift it in this direction and that we would do that is we would take the X like we have learned before, we would take the axe on Lee, the X and then we would subtract for from the X So we would have Why equals 60 minus and then x minus four and then all that squared. And now you would have Why equals 16 minus X minus four squared. And I can see that if I now take my eight right. Then you can see that my wine here is zero. And I could check it like this. Why equals 16 minus eight mines four squared. I will get 16 miners. Four squared 16 minus 16 equals zero. So it works. And if I take for example for I need to get 16 so I can say why equal 16 minus. And then for minus four squared, I would get 16 minor zero and I will get 16 and I would get 16 here, so everything works. But now, in order to calculate the area off the wall off the hanger, one approach that I could take is to describe ties. This proble like this with levels. Right? So I have these different levels here. And then once I have done that, I can calculate the area underneath those levels so I can calculate this area, and I can calculate this area in this area so you can see that I have rectangles here. So if I calculate the areas of these rectangles, then I will approximate the area off this hanger side so I can approximate the area of this war, okay? And I can also number. These rec time was one to three. And 45 six, seven and eight. Okay. And the way I would, uh, approximate them would be like this. Remember this interval here is Delta X. Okay? And this is equal everywhere. I'm considering equal intervals right now. So this would also be Delta X, and this would be Delta X. All right. This would also be Delta X. Now, to calculate the area of the wall. My first will need to calculate the area off each rectangle. And then I have to some those rectangles up. So the way I would do it is like this. I could do it like this area. And this is now approximated area. Okay. Approximated area. This approximated area would equal delta X times. Why one so for the rectangle won the level is why one for the rectangle to the level is Why too. For the rectangle three, the level is why three. And then why four here? I would have won five. Why six. Why seven. And why eight. And then the area would approximately be the thanks times. Why one plus delta x times? Why to write? Because if you think about it. You take death tax this interval and you multiplied by the height off the rectangle by white to if you do that, if you multiply Delta X by y to you will get the area of direct angle. Second, continue plus Delta X times. Why three and I can just say that plus three dots plus Delta X times. Why eight? So that's how you would approximately calculate your area. Now, of course, I can also write it like this. I can simply factor out my Delta X right. And then I can have they'll thanks times and in the parenthesis I will have. Why one plus Why, too? Plus why three plus three dots up until why? Eight. All right, I can also write it down like this, or there is even a more compact version of this. I could say that my area approximately equals to Delta X times. And now I'm gonna put this sign here, which is a summation sign. This is a summation sign here, and then I'm gonna just right here. Why I and I'm going to say that this I goes from one up until N equals eight, So Okay, Death X here is the same like this. And this entire thing, right? Is just this entire thing in the parenthesis. All right. Says why one plus why, two plus, why three up until I was eight. So this is a compact way to write this. But remember, this equation here on Lee gives me an approximated area. It's not the exact area. I have errors here. You see, I have errors. These are errors here. So this is I'm not taking into account this part and this part and this part and this port , and there is no reason to think that. Okay, this cancels out this or this cancels out this. So it's these air just errors. And no matter what I do here in this formulation, I don't get my exact area. I get an approximated area, so that means that I have some kind of error. Now, what should I do to decrease my errors and to get an area that is more and more precise 4. integrals intuition 2: Well, what I can do, I can simply steak a smaller interval. A smaller del tax I can take their thanks equals 0.5. Right. So if I take Dell tax equals 0.5, then I will chop this X here into smaller pieces. So the way it would look like would be something like dish soap is gonna do it intuitively that Okay, now, my Delta X would be this all right. And this and these were the same lengths. So my del tax would be this and this. So you see, essentially, I take my del thanks, and I make to delta access out of it. If I do that, then what will happen here with my rectangles? Well, first of all, for each red rectangle now I would have to black rectangles so I would have a rectangle here. And then I would also have a rectangle here. So I just got right now like this, you see, because what I do now, I take this milk point from 7.5 and eight. I take this middle point, which is seven point 75 and then I go to that continuous curve and I see that. Ah ha. This is my value here and then From 7.5 to 8, I have this constant value. And then I make a rectangle out of it. And then I do the same for all other red rectangles. And then I do the same for all the other cases. And then you can see intuitively already that for each red rectangle, I will have to black rectangles. And now my errors are smaller and I will get a more precise area. So in this case, I'm gonna do I'm not gonna do it for everything, but I'm gonna do it for one more. So I'm gonna take this part here, and I'm gonna go to this black line and then, Okay, this would be one of my levels. And then here from six still 6.5, I take the middle 0.6 point 25. I go to this curve here to this black curve and then from six still 6.5, I will hold this value constant. And now I will have two rectangles to black rectangles instead of one red rectangle, you see? So now I will have mawr rectangles and I can do it for all red rectangles. So in total, I will have 16 rectangles. Now, if I want to model it mathematically, what does mean? It means that this term here de thanks becomes smaller. Right? But then the amount off wise will increase. So I will have an area that I can approximate. And in this case now, it would be now Delta X. I'm just gonna put it small, smaller? And then here I would have y one plus. Why too, Plus why three plus up until plus why 16? Okay, Or the summation sign would be. I were Delta X small. I equals from one till end equal 16. And then why I So you see on one hand this Del tax becomes smaller here. My Delta X was one. And now here it's 0.5. So my dad takes decreases, but the amount off wise crease And now I have a better approximation off my area Now what happens if I decrease my day attacks even further? Well, I will simply have more rectangles. My death actual be smaller. But I will have more wise. And if I continue doing that and I say that my delta X approaches zero. Then the amount off wise will approach infinity, right? If my Delta X approaches zero, then the amount of these terms will approach infinity. And in the end, I will get my exact area. Because remember, right now, with rectangles, I'm in my discreet world, right? So right now, as I approximated my area, I'm getting a discreet area. But let's say the the most precise area would be a continuous area. So a continues area I get if I work with a continuous function and this probably here it's a continues Arambula. So in the end, I'm interested in knowing the exact area under this black per bullet curve. So I need to get this continuous area. But I have been able to approximate that continues area with with rectangles or with the discrete area. But now if I say that my interval, my delta X approaches zero, then consequently the amount of wise who approach infinity and then this entire multiplication this entire product in which one term approaches zero and the other term approaches infinity. This entire product will ultimately resemble into my exact area. In other words, my discreet area will approach my continues area. Okay, And here I want to note that when I say that my del tanks approaches zero, it's not the same. Like saying built the X equals zero. It's not the same these air to different concepts. All right, when my delta X equals zero that obviously my term here would be zero and then I shouldn't have any area at all because everything that I multiplied by zero would be zero. But in this case, if my Delta X were to be zero, then my then the amount of wise here would be infinity. So there some would be infinity, so I would have some kind off multiplication off zero times infinity. Right, So this will give me infinity and this will give me zero. And that kind of thing is just undefined. Mathematically, it's impossible toe. Define it. So it's undefined. Meaning you don't know what you will get here. Could be eight 50 two million. You don't know. You just can't determine what this product would be if you were to multiply zero times infinity and thats why this equality sign? All right, this belongs to algebra and algebra is too weak to solve. Ah, problem like this. And that's why calculus was created. And calculus came up with a new concept. Instead of equality, I'm going to say that something approaches zero. So when deal thanks, approaches. Zero. Then I can see what my area will approach. If Bill thanks approaches zero, then. Well, then, of course, this turn will approach infinity. And then I can use the limits that we have covered in the past limits to find what? This total equation, this entire product, what it will approach and remember it will approach something. It will approach your exact area. It will not equal. It will approach. You're exact area, so your exact area would be like a theoretical limit. In other words, you can think of it like this for simply as Del thanks gets closer and closer to zero. All right, then you're these great area gets closer and closer to continues. Area You can always get closer to zero. You can always have 0.1 and then you can have 0.1 and 0.1 You can always get closer to zero , and the way you should think of it is like this. The closer you get to zero, the closer you're discreet. Area approaches your continues area. In other words, the errors that you get here they get smaller and smaller and smaller. So as you're del tax approaches zero the error. The errors also approach zero and you're discreet. Area approaches continues area. And now when that happens, then you're you're some because essentially, it's a son, right? The other name for this is some because you multiply something, then you multiply something else and you multiply something else, and then you have a some. And that's why you have this summation sign here and now this summation sign is a discrete summation sign. Discreet summation sign. Okay. And when you're dealt, thanks approaches. Zero, then you're discreet. Some approaches continue some, and for the continue some, we have a different sign. If this is the discrete summation sign, then when you're discreet, some approaches your continues some. Then this sign here, which is for discreet summation, will become this sign. And this sign is a continue summation. So you use this sign when you're dealt, thanks approaches zero. And when you're the amount off wise will approach infinity. And when you go from your discreet world to your continues world, then you have this continues summation sign which is called into girls. And there you have it. One way to look at Integral is that it's a continues summation. All right, this sign here is discreet summation Your something some kind of blocks. But as the width of those blocks approach zero, then as you go from your discreet world to your continues world, then decimation becomes an integral and therefore, if you know, apply the concept of integral to this function. Here you will calculate the exact area. That technique will allow you to calculate the exact area off this wall. And in the next video, I will show you the integral from a different point of view. And I will also dive into the math off integral so that you could actually mathematically calculate the area over this wall, the area under this black curve. So that will be the objective for the next video. Now, you know that if I take my function, which is this one y equals 16 miners and then Francis X minus four and squirt, if I take the integral off it. All right, then I will get the exact area of the wall. And in the next videos, I will show you how you can do that, mathematically. So see you in the next video. 5. integrals intuition 3: welcome back in this video we will look at integral a bit in a different way and then we will see that it is really the same thing. So let's get started. So I have a car here and this car drives along the X axis and I have three different cases . In the first case, the car travels with this function X equals time. In the second case, the court travels X equals time squared and in third case, X equals time. Cute. And if I take the derivative for all the cases, then here they are ecstatic was one meters per second extract equals to T meters per second and ex dot equals three d squared meters per second. Now, if I were to graph ALS these cases, then I would have something like this. There you go. So when I had take my original function than in the first case, it would be a straight line function. X equals t. In the second case, it would be X equals t squared. In third case, it would be X equals t cute. And now if I take the derivative here will get one. All right. I will have one here. When I take this operator, you see the era and D over DT. That's my operator. And I apply this operator to dysfunction and I get my derivative, which is one. And then if I do the same thing here, I will get a function which would be ex dot equals to t. And then if I do it here, then I will get a operable, uh, which would be ex dot equals three d squared. So that's how you would graft them. So now let's consider this graph here. All right, this first derivative and let's consider this graph and calculate the area under the curve or in this case, under the horizontal line. And we're only going to consider the first three seconds, all right, and it will be a discreet case. So what I mean by that is that we're not going to be calculating the area continuously. But what we're gonna do, we're gonna chop this function into three pieces, so I would have three pieces here, and then I would calculate the area in three parts, so I'm gonna have area one here. I'm gonna have area do here, and I'm gonna have area three here. So that this is my first part. This is my second part, and this is my third part, and we're gonna call this interval here. Between them, we're gonna call it Delta T and Delta T here, obviously is one second. So what I'm going to do here? I'm going to calculate the accumulated area under this horizontal line. Now, when I say accumulated area, what do I mean by that? It means that Let's imagine that I'm gonna put some kind of scanner here. All right? So just some kind of scanner that I'm just gonna draw in red. So have some kind of scanner that I started T equals zero seconds. So imagine my Ben. It's like a scanner, and then I just start scanning the area under the line. Okay, so I just I have to spend here and I start moving to the right, and as a move to the right, I scan the area under the horizontal line. And now, since it's a discrete system, we record the area covered everyone second, starting from the beginning. So that means that as I start scanning in this direction, once I have covered one second. I'm gonna see. And I'm gonna record how much area I have covered in one second, starting from the beginning. All right, so let's say he financed are scanning. And then once I reach one second, I'm going to see how much area I have covered. And I can see that I have covered well, death, the tea times one area. So atomic was one. I can say that I have covered a one. Right, So I can put here a one. And then I can just, uh I could just graph it like this. All right? A one. So you have nothing Here is just here at time because one second I have area. Warn? No. When I continue scanning the area, once I reach T equals two seconds, I'm going to see how much area I have covered. Starting from the beginning. So still starting from T equals zero seconds. I'm going to see how much area have covered. Once I have reached d equals two seconds and then I have covered area one and area too. Right? So antique was two seconds. I have covered area one and then area two is equal to area one And so I have also covered area to right, which means here you will have a one plus a to right. So now, now I have to consider the previous area and the new area the area that that covered from 0 to 1 seconds and then the new area that I covered from 1 to 2 seconds. So you see, I have already accumulated my area. Not only I take into account the new area, but also take into account the old area. And then when I go from two seconds to three seconds, then I will still consider the area I have cover starting from the beginning, starting when Deco zero seconds. So when d equals three seconds, then I have covered a one a to and a three. So here I would have a one plus a two plus a three. Right. That means they have a one a to and a three. Right. And that's the accumulated area or the accumulated some that as your scan as your skin of goes along, this function here at time because one second, it gives you the area calculus, the area at time. Two seconds. It calcalist the previous area plus the new area and the timing was three seconds. It calculates the previous area, plus the previous area plus the new area. And of course here. Since areas are the same, we can say that area one equals area to equals area three, right, and then area one equals delta T times one. All right, so you have delta t here, but then you multiplied by one. So there's a T times one and then you get your area notice one thing that here, in this case, our area is not meter square. Okay, It's not meter squared, because here our function is not geometric. You don't have distance on this number line. And on the vertical number line here, you don't have meter squared here. What you have on one axis, you have time. All right. You have seconds. And then on the other axis, you have x dot which is velocity, right? So you have meters per second. So if you calculate the area under the line, that means that you multiply time by velocity or seconds by meters per second. So when you calculate the area under this horizontal line, what you're doing, you're multiplying time, times velocity. And when you do that, you get a displacement vector, right? Let's remember velocity is a vector and if you multiplied by time, you'll get a displacement Victor. Okay, But even though you don't get meter squared but you get the meters because you multiply seconds times meters over seconds, it's still area under the line. So if you have ah, a grabber, you have velocity here and you have time here then here the area has a different meaning. It's not a geometric meaning, but this area still gives you something and that something is displacement. And sometimes I confuses with distance for what I really mean. I mean displacement because we lost. He's a vector of the last times. Time equals displacement. If I had speed here, then this would be distance. So I could also right times, times speed, which is a scale of quantity, would equal distance. So this is the difference. So now I would be working with the scale of quantity issue here. In this example, it doesn't really matter here. We're trying to understand the concept of Integral OK, but now let's take our delta T. And let's make a smaller Let's make our delta T equal zero point five seconds 6. integrals intuition 4: So now, since my delta t equals Europe on five seconds. Now my scanner. As I start scanning the area under this curve, my scanner records the covered area every 0.5 seconds. So you can see here I have a 0.5 seconds. 11.5 to 2.5 and three. And on this graph, I'm going to record the accumulated area and the recording happens every 0.5 seconds. So I start scanning this graph. I start scanning the area under the line under the horizontal line. And every time I cover 0.5 seconds, I record the covered area. Remember from the beginning, because it's an accumulated area. I always record the covered area from the beginning from Tommy Koh. Zero seconds. So here, as I start scanning the area at time because you're a 0.5 seconds, I will have area warm. Right? I will have area one. No, as they reached equals one seconds. I will have area one and two. So we'll have here Area one, and then I will also have area to. And then when I reach 1.5, I have covered area one area to and area three. Once they reach two seconds, I have covered area one area to area three and area four, 2.5 seconds. It's area one to three, four and five. And when I reached three, I have covered a of 12 three for five. And, well, six seats. Area six. Okay, so here you would have a one here, you would have a one plus a two here, a one plus a two plus a three. And let's say I'm too lazy to write at all those things here. So what I can do? I can use this discreet summation symbol, right? I can have the summation. I equals one from my cause. One to end equals four, and then I can have a I hear it would be like a one plays a two per se three se four. I can have the same thing here. Hi equals one to and equals five a i and then here as well. Well, I'm just going to write it down here. I from Mike was one toe end equals six. I can have I a And that will be then this. Okay, So there you go. This is our accumulated area. But now what's gonna happen when my delta t when I decide to approach zero when my delta t approaches zero seconds? Well, the closer and closer I get to zero the more of these lines, these vertical lines I will have. Right? So I will have a lot of lines here. And then if I connect all the tips of the lines, right, all these tips, then I will get a line. You see, in the end all those tips, they get closer and closer to each other. So would have something here. I would have something here. I would have something here, etcetera. And then when I connect all these tips Okay, I'm connecting old this tips. When I connect all these tips, I get a straight line and this straight line will give me Let's say, if I take time equals one. All right, then it will give me a one plus a two. But what is a one plus a two. What was a one? A one? Was that the tee times one. Right. So we're now looking at this first case area one. Was that the tee times one. So that 30 here in the first case was one second. That means that here it was the area one was one meter because remember, we multiplied time, times philosophy. And then we get displacement that has the unit off meters. So here, in this case, area one is one meter right now. In this case, Area one was 0.5 times one. So here my area one would be zero point five meters. Right, Because again, it's meters over seconds, times seconds. And you get meters. No, If I'm a time equals one, then I will have 0.5 plus 0.5 right? Which means I get one if I'm at a time. Because two seconds I have 0.50 point 50.5% 0.5. So I will have two meters. So notice I get exactly the same graph like here X equals d, right. So this red line here that I got right this red line here it's X equals D Does it make sense? Right? If I take time, ACOs three, then my ex should be three. Since each of this area is Europe on five then and I have six of them. All right. Six areas and each area. 0.5. Then I have 0.5 times six equals three meters. Right. So, look, I get this graph here. X equals T is the same thing. So when my delta t approaches zero seconds, then I'll have this straight line and the straight line is exactly like my original X equals T graph. Now it means that when my delta t approaches zero seconds Well, then first of all, my delta t would then become DT, right? A differential. And then the other observation is that my accumulated area goes from the discrete world. It goes two continues world. So my accumulated area now which at the beginning is in the discreet world. No, it becomes continues. So you will have. You will have this continues line. So you will have the continues accumulated area that is represented by this line. So what it means is that our scanner as we scan through our function here, right, it means that recording happens continuously. It happens all the time without any discreet stops. It means that our scanner, as it scans through this function ist goche and scans through this function as it records the area. The recording happens all the time. Our scanner records the area since the beginning. Since time equals zero seconds continuously, it does not stop recording the area, not even for Nana second. So it just as it goes through the function, it just continuously records it. And that's why you can imagine like that as our scanner scans through the function, it just keeps recording without any stops there no stops in time at all, said the recording off the area starting from the beginning. Starting from here, it happens all the time. And that's why instead of these tips here, right off the accumulated area, instead of these tips, all these tips merge. They become closer and closer to each other. And then, in the end, those tips here they will form a straight line. And when that happens when you enter this continues world, then it means that your summation becomes Anita girl. So you now, First of all, you had this sign, and now you have this integral and this cold and integral operator, just like we had d over DT. That was a differentiation operator. This is an integral operator. And then the other thing that we can conclude now is that if you take the integral off d x over d t which is this graph here, if you take the integral off dx DT, you will get a continues accumulated area and this continues accumulated area is the same, like X equals t here, which is our original function. Okay, now that means that okay, if I have some kind off original function x as a function of t right, and then I apply the differentiation operator D over DT, I will get the X d t. And that's called a derivative, right? The river tiff. But then if I have my d x over de t as a function of time and now if I apply the integral operator, right, So now this is the integral operator. Then I will get X as a function of time and that is taking the integral All right, integral. Yeah, in other words, into girl has another name. Now, as we know, it's continues some okay? Or we can now also call it that the integral is nothing else but anti derivative. Okay, so if you go from extra the X tt you take the derivative, but going from DX DT two X, you take the anti derivative or integral or continues some. So these are the conclusions so you can see the into grow. Essentially, it's the same thing, but the you can think of it from different points of view. And remember, this integral sign is just an operator just like D over. DT is an operator. Integral is also an operator. You apply this operator to your derivative and then you get your original function. 7. integrals intuition 5: But note this. If I have a function like this, X equals t plus five. Right? And then I applied the differentiation operator, which is D over DT. Then I will still get d X over d t equals one because this constant here, this five, it becomes zero. Right? So now if I want to go back and I want to go from DX back to axe, if I want to go from DX DT back to acts right then, okay, if my DX DT equals one and apply the integral operator because that's what I need to do I need to apply the integral operator to go from the X DT to axe. So I will still get X equals t right? So even though now I had a constant five here and I took the derivative if I got one now, when I went back, then I got my X equals T. But that's not the same like X equals t plus five, right? So why is that? Well, that's because when you go from X to the derivative effects, then you lose information because remember derivatives the only capture information about how your functions change so so maybe a good way to explain it would be like this. So imagine that I have some kind off so kind of function here, right? That looks like this. And then I have the same function. But then I add some kind of number two. It's so so I have the same function like this. It has the same shape, right? That's exactly the same shape. The only difference is that I added a constant see to it. So let's save. This is my Why equals X squared then this could be my y equals X squared plus C and the sea can be any number. Now if I want to take the derivative at, let's say this is X. This is why, If I want to take the derivative at X equals to right, then remember the derivative. What does it do? It takes the information about the functions changed. So it shows you how the function changes with respect to the input. In other words, it gives you the slope of the tangent line, right? That's what the derivative does. It gives you the slope off detention line, But the slope off the tangent line here is the same like it is here, right? So these two slopes there the same. So these two lines here, they're parallel to each other. So whether you take the derivative of this function or this function, it doesn't matter because the sea become zero and you only get the slope of it. So when you take the derivative of a function, you lose information about all the non changing components and the sea, right? The sea is an unchanging component. It doesn't change is just the constant. So you lose information about that, and then when you go back when you know, take the integral of it, then you cannot retrieve that information. Okay, Once you take the derivative, you lose information about non changing components. And when you take the into grow off your DX DT, you can't retrieve that information anymore, which means that when you go from the X et to back toe X, you have to add some kind off arbitrary constants, some kind of see Okay, you have to do that. Otherwise, your answer is not complete because you're your initial original function. It can have for whatever constant possible can have 1233.5. Whatever. So if you really want to generalize it and you don't know your initial condition and knowing your initial condition means that you know your initial constant right, like here. You knew your initial condition. Your initial condition. Waas five. So at Tom Nichols zero, that's your initial condition when your time equals zero seconds. You knew that your ex equalled five when your time equals zero seconds. But when you don't know your initial condition, then you have to assume that your function is x equal steep plus C because you can have any possible See, So you need to assume that you have many graphs here. Okay, so they all have the same slope. But since you don't know your initial condition, you have to put an arbitrary see here. And that means that you really haven't infinite amount of possibilities here. However, if you're not, take the differentiation operator. Then you will get only one answer. You will get here. You would have X adopt or dx DT. You will only get one possible answer and that is one. All right. So here x dot equals one, because if you take the derivative with respect to t you get one and the sea becomes then zero. So all these functions that you have an infinite amount off, they all become one. And how would it mathematically look like when you take an integral off our function? So you start from this. This is your derivative, right? D x, with respect to DT how your exchanges with respected time and we know that this equals one . So what shouldn't need to do now? If you treat this thing as a ratio, then you can write it down like this. D X equals one times D t. Okay, now think off DT, as your scanner. A super super thin scanner. Okay, it's your super thin scanner. Okay, so it's extremely thin thinner than a nano meter. And as this DT now, So it's a super thin scanner and of course it's It's an interval, right? It's Ah, it's DT's. Uh, it's your very, very thin interval. It's when you're Delta t approaches zero, then you can think of it as a differential. So it's like a very thin interval. But then as this interval, as it scans through the function as it goes through the function like this As it scans through the function, it continuously records your function and gets the accumulated some right So d x you can think of DX as a super small increments off your area and there you go. Your DX is a super small increment off your area. And what do I mean by that? What I mean by that is like this that as you scan through your function, right, you calculate your accumulated area and your accumulated area is X so you can write it down like this. So you have external This is your ex or your accumulated area at time equals T plus D T right equals your accumulated area. Xto a time equals t plus de x. So you see that time, Christine, let's say t equals five seconds. Your accumulated area is this. But then this d X is your super small increments off your area. So, to that area, time equals, let's say five seconds. You add this small increments off your area and now at time equals five seconds plus a little bit more. Your total area now is what you had here, plus a small increment and then that would be your new toll area the old total area, the old accumulated area plus a small increment. And to initiate the scan right to initiate scanning this function. All right. To initiate this process, you apply the integral So you on both sides. You put an integral sign here and you put it here and now. One of the properties of the integral operator is that if you have a constant here than he can go out out of the integral so you can also write it down like this. D X equals one times integral DT, but only when you have a constant. And now if you apply this integral than here, you will have X. And here you will have one times and that becomes t so essentially what you're doing, you have these small lu intervals, right? And remember, integral is a summation in the continues world. So as you sum all these small DTs together as you scan it through the function along along the variable t right, So as you sum them all up like this, so you have so you start scanning and then you have DT and then you add DT DT DT. And then when you reach an arbitrary point, let's say here, t then if you some all those DTs up, then you have a t So this DT becomes a T. And remember, Since you don't know the initial condition, you have to put the constant C here and there you go. So once again, once you apply this integral sign, you initiate the scan and then by initiating the scan, you continuously record your accumulated area, which is at any given time. You record your area covered starting from the beginning, and you can model that accumulated area with this function. So this DT becomes tea. And since you don't know the initial condition off your function, you have to put this constant here plus c. And now in the next video, we're gonna look at the other functions that we had remember we had a parabola, right? We had tea here and then we had some kind of parable a x equals d squared. Then we had also t we had we had a function X equals d cubed. And then we took the derivative of those functions. So we're gonna see in the next video how this entire process that we did in this video and in the previous videos how this process would look like when we work with these functions here. So see you in the next video. 8. integrals intuition 6: Welcome back. Let's take a look at this function of X equals T squared. So this is the graph, Fred. And if I take the derivative it with respect to time, then I have eggs dot equals two times t and this is my graph here. This is my line two times t and let's do the same thing. Now, let's take, uh, areas, But now, discrete areas under the line. Okay. And again, we, first of all, choose our delta t to be one second, which means that we're gonna initiate our scanner, and we're gonna start scanning from ah zero up until three seconds. And at every second, we're gonna record the area that we have covered since from the beginning. So when we're at equals three seconds, then we want to know the entire area off this triangle. So starting from the beginning, and we already know that if we have this derivative graph where you have eggs dot here and you have time here and this ex dot it's ah Well, if we think off vectors, it's the velocity fear we think off scaler, zits, the speed. And we know that in this case, if we calculate the area under the line. Then we're gonna get either displacement when this is velocity or distance when this is speed. So that's the meaning off the area in this case. So let's start doing it. So imagine we started our scanning. We start scanning our area underneath this line. So once we reach time equals one second, then, ah, we're gonna record this covered area here, so we're gonna have it here. First of all, we need to know what eggs dot is a time because one second a time, because two seconds and three seconds so we can write it down very simply like this. So x dot at time equals one. Well, if it's one than one times two, it's gonna be two meters per second, right? No ex dot at time, because to equals two times two is four and x dot at Tomic was three, three times two equals six meters per second. Remember, this is meters per second. Here. These are the units to make it a little bit more practical and less abstract. No, knowing that we can start calculating the areas. Right. So when we reached time equals one second than our area of one would be Delta T, which is the interval here. Delta T Times ex dot At T equals one And now, since in the previous video we just had the rectangle But now we have a triangle. So to calculate the area of the triangle we have to divide this thing by to and that would give us then ah, one does a tease one times This one is too divided by two and we then get the one meter. Remember, it's meter here because here the area has a different meaning. It's not geometric. It's velocity times time, which is displacement that has the units of meters. So area to now would be now. Since we measured the accumulated area, we always measure it from the beginning. In this case, from time equals zero seconds. And by the way, Onley in this example we start measuring from time equals zero seconds. In many other examples, you can start measuring a your time from, for example, miners one or minus four or minus three. But that's not the case here. Here we measure it, sounding from time equals zero seconds. So this is something that you can choose for yourself. So in other words, you can choose your start. You can choose your beginning, but from the moment you have chosen your starting point, either it's time and zero seconds or miners one seconds. Or maybe you want to start calculating your accumulated area from Tom because one. But then you always have to calculate your accumulated area from your starting point. Say, in this case, we start from Tommy Koh zero seconds. Now, in fact, we're not interested in area two were interested in area two plus area one, right? We want this entire area here starting from Tommy Koh zero seconds. So to calculate that we have to multiply. Well, first of all, we need to multiply Delta T by two because now we have to Delta tease here Delta T one and Delta T two two times Delta t times x dot at time because two seconds divided by two. And that will give us that will give us two times one times ex Don two was four divided by two. And that will give us four right? In this case, four meters. No, that means that if we record the first area, then it would be one that time. It was one. Right now, time equals two. It's four now. So our area here would be this. And now I'm just going to write it down here If I want to get my A one plus a two plus a three. That would equal Delta T. Now, how many adult that these do I have 12 and three. 30 times, three times x dot At time equals three, divided by two. That will give me one times three times six, divided by two So I can cancel them out to this becomes three. And I will get nine. Okay, in this case, nine meters, because here the area has a meaning off displacement. So now here I have nine. Okay. And nine would be, I guess, somewhere here, out of the scope of this graph and just gonna break the rules and put a nine here. That time was three. The accumulated area would give me nine, You see? Nine. So it would be here. So I have one point here. Another point here and now the point here and notice. Now you don't have a straight line because now you're area one does not equal to area to an area two does not equal to area three. So Area one does not equal to area two and then does not equal to area three. So there different now. But of course, if you think about it now and you think that Okay, What if instead of Delta T equals one second? What if I choose it to be 0.5 seconds? Get what's gonna happen then? Well, then, obviously you're gonna have more areas, right? So I'm just gonna do it in red. So you would. Your interval would be smaller, right? It would be half off your previous interval, so therefore you would have more areas. So if now we use red for the new case when Deltek was 0.5 seconds, then you'll see that you have now more areas you have here. You would have area one, and then here you would have area to area three area four, area five, an area six. Okay. And then So I'm just gonna write it down like this. Here. Your three area, four area five, an area six. And so if you do the same procedure like we did here, but with more areas. Then you can imagine that the you would your accumulated area graph would be something like this. If you performed the calculations, you're accumulated. Area graph would be something like this. And in fact, if you let your delta t approach zero, then remember, you go from your discreet world, you're continues world. And if you if, you know, connect all the tips off off these levels, Right? So you so I'm just writing this, uh, line just for visual purposes. This vertical line, the really important things here are Are these points these tips? Right? So as your delta T approaches zero this tips, they become closer and closer to each other, and you will see that the it will resemble into a problem. Right? So if you if you think about it, if you look at this function and your time, because three seconds right there what? What will you get? X equals three squared, which would be nine. Okay, but then here You also had nine, right? Remember here at a time. Because three seconds You also had nine here. So as your deputy approaches zero, you approach If you connect all these dots. Then in the end, you will get a problem. And there you go. Your proble here would equal to your parabola here. So what it means is that when you're Delta t approaches zero than this scanner that you have here once you start scanning this function the scan it starts recording your accumulated area continuously every time without stopping. And now if you record the area off a triangle continues because now you have this triangle . You don't have a square anymore. You have a triangle. And now, if you scan the area of your triangle and you record your area continuously, then your answer will result in a proble okay, in a Preval. So let's initiate the scan mathematically. Now, how would it look like? Well, we can do it like this. D x over DT equals to a t right. That means that we converted like this. D X equals to t DT. All right, this DT goes here to initiate the scan. We apply the integral operator on both sides one here and another one here. But now you already know that the constant can go before the integral, so you will have DX equals two integral t the t No. How would you take the integral off this? Well, this side will be the same, right? It will become X right X, or if you want to be very precise, x as a function of time. And now when you have this four teeth times DT, then, uh, what will happen is that you will have 1/2 here than your tea becomes squared. And then you have to add the constant now because remember, even though that was your initial function. But when you go and take the derivative, then you only capture information about change of the function. So when you go back, then you have to add the constant because going from the derivative back to the probably you're not sure whether this was your initial function. Or maybe it was this. Or maybe it was this. You are not sure. And that's why you should put this constant here. And if you take the derivative of dysfunction again, then this constant will become zero anyway, so it won't matter. And by the way, notice that what I did now not only I made this t squared, but I also divided this thing. This t's court. I divided it by two. And why did I do that? Well, first of all, there are rules, and we're gonna cover those integral rules later. And I'm also going to show you the formal definition. Often integral, however, just intuitively understand. It is because it's always good to have an intuitive understanding to intuitively understand why I divided my t scored by two. You have to look at this denominator here. Right? So when I calculated my every years, right, my area one than my area one plus area to in my area one plus area to place area three. Now, notice now, I was calculating the areas off my triangle. So, for example, when I calculated the area of this entire triangle, then what I did, I took three times. Still the tea. All right, so that's this three times delta t times My ex dot at time equals three. All right, X, that atomic was three. Is this part here? All right, So when I multiplied them together, the new fact when I got I got a square, right, So I got a square, so I got the area off the entire square. But I was only interested in the area of my triangle and therefore I had to divide this square by two two on Lee get my triangle. And that's why that's why I had to divide by two. And that too. If you know, let your death fatigue go to zero, then you won't get rid of the fact that you're at some point. Even when you're delta T is very small, you will still when you calculate your areas of your triangles a cumulatively, you will still first calculate the area off a rectangle and then then you always need to divide your rectangle by to to get your triangle. And that's why you have to hear. And then, of course, you can canceled them out and you will have t squared plus c. And then again, if you want to go from x two x dot When if you take the derivative of it, then this will become to D and see you will become zero, right? So that will be this 9. integrals intuition 7 indefiniteVSdefinite integrals: welcome back. And we also had one more function. So I thought that I was gonna squeeze it onto the same page because then it's easier for me to explain certain concepts by referring to the material that we had in the previous video . So this is another function that we had X equals t cute. So now this is a cubic function. And of course, if you take the derivative for cubic function, then you will have ah squared function. So x dot equals three times t squared. And, uh, the logic is exactly the same. We take this function and we first will start taking the accumulated areas dysfunction discreetly when Delta t equals one second. So you would have area one here, you would have area to here, and you would have area three here and again, they're not equal to each other. And if you take this areas, those accumulated areas so we can say that area one here would be delta t times ex dot at time equals one. And now notice one thing that I want to mention now we don't have a triangle. We had a pribyl. So what happens is that when you have a parabola, all right. And let's say you have one here and to hear, and you want to take the integral of it and you want to take the area of it underneath the curve. Then you see when you first, uh, take, uh let's say if you first take this distance, which would be two times delta t right, and then you would take this distance, which is EC start at time because to if you multiply them together right to delta t times x dot at time because two. Then again, you will have a rectangle. Okay, you will get the area or for rectangle, but you don't want a rectangle, right? You just want the area under this parabola, and in this case, to get it, you would have to divide this thing by three. So this area under the probably it's 1/3 of the area of the correct angle. And that's why you have to put three here and then area one plus area to. In fact, we have already done it. Area one plus area to would be this and then area one plus area two plus area three would be three times Still the tee times x dot at time because three divided by three. And of course, then you can take a smaller delta T, and you would have, for example, is your 0.5 seconds. Then you can get a smaller delta T, and you would have 0.1 seconds and etcetera. And as your delta T approaches zero, you would get from the discrete world. Do the continues world, and then you would get this X equals t cute again. But remember, when you go now from here back to here, then in order to be correct, you holes have to add the constant because you don't know whether your constant at the beginning was zero. Or maybe you have something else, so you should not forget. Plus, see, when you go from your ex, stop to your X and then mathematically taking Dita girls would look something like this. So you you have d x over d t would be three t square. Okay. And now what you do You have the X equals three t squirt DT. To initiate scanning the area under the curve, you have to apply the integral operator on both sides here you would have X and then here three goes before the integral sign. Then you would have t squared d t here X equals. And now because of this denominator here, right, Because of this denominator here, you have to take your T. Now what's gonna happen with t t itself? The power increases, right? It's gonna go from squirt function to a cubic function so your tea will become cubed. But because of this denominator here because he instead of the area of the squared, you need the area under the parabola. You have two divided by three and then plus see okay. And then, of course, you can cancel out the threes and then you would simply have t cute toe a C. And in fact, there is a general rule that if you have some kind off function, let's say DX equals eight times t and and then you have this DT here, so d X equals eight times t to the power of end. Then you have DT here. If you have the integral if you apply the integral, then you will have X equals a times. And now, since you have the power of n here in the denominator, you always put n plus one and then your tea now will also have and plus one. And then you have to add the constant. So that's a general who and that rule is derived from a formal mathematical derivation for integral. I'm gonna briefly cover that formal definition as well in later videos. But that's one of the rules off taking the into gross. So if you need to take the integral than one of the rules that you can use is this rule and now you know intuitively why you have this n plus one in the denominator. Okay, so so far, what we have been doing, we have bean learning about how to go from ex dot two x. In other words, we were taking the anti derivative or the integral right and what we got our result was a function right. It wasn't some kind of number. It was a function. For example, we had t cubed plus C. But it's not a number. It's a function. You see, you have an input here, you have a variable. You have tea, right? So what we have been doing so far is called Indefinite into girls. All right, so I'm just gonna write it here, indefinite into girls. So in the next video, I'm gonna teach you about something called Definite Integral. And I'm gonna talk about it more than but for now, I just want to tell you that indefinite into girls as a result gave you a function. Okay, You had a function. This is your function with a variable with an input and an output, people would be t on output would be X, and a definite integral will not give you a function. As a result, a definite integral will give you a number as a result. So when you calculate a definite integral, you will get a number. And that is because the definite integral simply means that you take your function that you have here, right, that will give you the accumulated area. You choose your starting point like, for example, zero. And then you choose your ending point for now in indefinite. Integral is your ending point with some kind of abstract variable t a variable that you could vary but then indefinite Integral, you're gonna choose your tea. You're gonna choose a specific number for your tea. You're gonna fix your variable, right? Let's say you're going to choose that T equals three seconds. So that's your final T. Okay. You also choose your initial t, which in this case would be zero. Right? And if you do that, then a definite integral will calculate the area from ti ko zero seconds till teak was three seconds. And then you will get the area of it. And here it would look like this. So if your initial t was here, then your accumulated area at T equals zero is zero, right? But then if your final teas three, then your final accumulated area is here, which would be three to the power of three, which would be 27 from not mistaken. Three times Street was nine times streaks. 27 yes, 27 right. And then you subtract your initial accumulated area from your final accumulated area. So you would have 27 minus zero, and then you would have 27. And since right now we're dealing with displacements and velocities. It would be 27 meters. In other words, from here you could see that. Okay, If this is my velocity function than if I consider my time from zero till three seconds. Then I have covered 27 meters. Okay, so here on this function, I could represent it as an area. But then if I go one level higher, then since this is the graph of my accumulated area, then my final accumulated area is 27 at my final time TF miners. My initial accumulated area, which was zero at the initial time, equals zero second. And then if I subtract them, I get my tool area covered here. Or I can represented here simply as a graph. It's game. My total area courtiers 27 and then here it zero. I don't have to take my initial time. Zero seconds. I can take my initial time. One second. All right. And then I can take my final time. Two seconds. Which means I will only consider this part here. Okay. From one till two. And in this case, what I would have I would have here on this graph. I would have for one times what one cubed. We give me one right, But in two cubed would give me eight right to Cuba. I would give me eight. And then if I now subtract eight minus one, I will get seven. And in this context, this seven would mean meters. And that means that I will Onley calculate the this area this part under the curve. So I forget about this part and I forget about this part. I only consider this part this area and you see now instead of function, I will get a concrete number. I will get an area. And, uh, this area in this case would be seven meters. And that means that if I go from time equals one second to time because two seconds than my car has, uh, covered seven meters. Okay. And so this is what the definite integral gives me. And I've already given more explanation than I was planning to, but then I will elaborate on that in the next video as well. In the next video, I will give you a specific example for indefinite integral. And I show you how you can formulate it mathematically. So see you in the next video 10. DefiniteIntegrals 1: welcome back in the previous video, I explained the logic off the definite integral that instead of a function, it gives your number. Because when you compute the area under the curve, as you scan through your function, you define concrete initial and final inputs, and then you compute the exact area between them that will give you a concrete number. So I define my initial input and my final input. And then I compute the area in between off X, I and X f and I will get the exact number, which is the area under the curve. All right, it's not a function, it's a concrete number. Now we're in a position to compute the area of the wall far hanger. Remember, from from the previous VDs we had a small hangar, all right? And we wanted to We wanted to compute the area of this side. So now we're in a position to do that. So we defined this this wall, ride the side of the hangar. We defined this wall with this kind of function. Why equals 16 minus ex minds four and then squared, and so graphically it looked like this that we had our y axis here, and we had our X axis here. And then our parabola went like this. Okay, so I'm gonna extend my X axis. And so here, Why equalled 16 and here X was eight. And here X was zero. And then you can see that it's a proble, and, uh, this wall starts from here to here. So we Onley Consider X from zero 28 All right, so we only consider the X from zero 28 That's the only thing that we do. And then we take them to grove it and we are in a position to calculate the area off our wall to compute the area off our wall. We do the following We take this function dysfunction here and we say that why equals d a over DX. In other words, the area differentiated with respect to X. Okay. And then if we say that if we say this statement, why cause d a DX? Then we can write that d a over de X equals 16 minus sh x minus four squared. Or we can write it down like this d a equals And now de actual go to the other side so we can very simply write it down like this. 16 minus X minus four. Squared d X now D X. It's like we can treat It is a constant so we can distribute it along this function So we can also write it down like this d a equals so I can write it down like this 16 D x minus X months four squared dx Okay, And now, in order to calculate the area under the curve from 0 to 8 Okay, from zero to it, we only consider the excess from 0 to 8 and we want the concrete number not a function, but a concrete number. In order to do that, we have to initialize our scan from zero and we and then we finish at eight. And the scan then would be with continues area recording. So as you scan you record your area and then you and then, of course, you grafica cumulatively. But that was indefinite integral here. We're interested in a concrete number, the area and by the way, here the area uh here the area would be meters square because now we have two number lines and both are meters axes meters and wise meters. So the area under the curve here has the geometrical meaning and it's really and squared before, In the previous videos, we we multiplied time times, uh, velocity. So it was seconds times meter per second. So we got meters and so the area under the curve there the meaning of that area Waas displacement and, uh Or if we had speed instead of lost it and it would have been the area under the curve, the meaning of it would have bean distance. However, here the area under the curve, the meaning of it is in fact the distance squared or meters squared. And now, in order to calculate the area under the curve from 0 to 8, we initialize the scan and we do it by applying the integral operator. However, since this is the definite integral in the definitely to go, we have to define our initial point in our final point. And we do it like this. Here we write, X initial equals zero. And here, above this, I need to go sign. We're gonna right X final equals eight. Okay. And here is well, ex initial equals zero and X Final equals eight, and then the next step is that is like this. We can write it down like this. Hey equals. And now we can distribute this interval operator by those two terms and we can write it down like this 16 because Dean Ferguson can also be in front of the constant 16 the integral, the X. And by the way, you don't have to write those x I and except for the time you can just right zero and eight and then you can put minus here and then another integral sign. 08 and then X minds four squared D X. So now you have two terms you have term one and then you have turned to and then you can compute them separately and then you can subtract term to from term one. So later on, you can subtract term to from term one. So let's start with the term warn. Let's see what term one will give us. Turn one would be 16 into grown from 0 to 8 d x, and then what I have is 16 than into those brackets. I get the result here, which is dxy to go. I would get X from the previous videos. You know that you would get X, but then you have the limits. You have zero and eight and what you do now, you take 16 and then you take your ex and you put your final eggs inside it. Eight inside X and then you subtract and you take your ex and you put zero inside acts. Your initial X goes into acts, so you have zero here and then you'll end up with eight miners. Zero is eight, so you'll end up with 16 times eight equals 1 to 8. OK, so that's our first term. Our second term would be the following. So the integral from 0 to 8 and then you would have X minus four squared and D x. No. One way to do it is like this so you can take your X minus four squared and then you can expand it right? You can write it down like this X squared minus two times X times for so minus eight X and then plus 16. So four times for 16. And then you can put this thing here and you can calculate being to grow. However, there is another way to do it. There is an easier way, and that's called the substitution method. And I'm gonna show you the substation method instead because it's a very useful thing. All right, so what you can do instead, instead off having this kind of form where you have 08 and then you can put this form inside here and X squared minus eight X plus 16 D x, and then you would essentially distribute it. But DX here and here and here and then Dita girls as well. You would distribute it on this terms. That's one way. But then you can also just do it like this from 08 Then you have your ex minds four squared DX and what you can do. You can take your X minus four all right, and you can see that it equals you. You just take another variable and you say that X minus four inside here, inside this parenthesis this equals you. In this case, you would have 08 right? This is your ex initial X final, and then you would have you squared dx and then that would be your for now. This form is not enough Because what you need to have also you need to take your DX and from the X you have tow, find a way to go to d you. And also you need to take your ex initial and you need to go to you initial. And then you need to take your ex final and you need to go to your final. Why? Well, that's because when you go from X to you, right, then you have a different function. Now, you you have a function where your input now instead off and X is a you right? And here this is your function f as a function of you. That means when you take the integral of it all right, do you have some kind of function? When you take the integral of it here, you need do you and not the X. And then also you need your initial you and your final You. So you need to go from D extra de you. And then from X I x f Do you i u f and the way you go from DX to do you is very simple. If you know that you're you equals X minus four. Then if you differentiate, this is also function right, so u equals X minus four. So if you were too graphic, then it would look something like this. So you have your you here, which is your output, and then and then your ex isn't input and then you have minus four here and then you have a positive slope. So you see, this is also a function. So you have some kind of mathematical relationship between you and X, which means you can also take the derivative off you with respect to X. So I can say that d'you d X equals and I get one right. And now I can express d'you as a function of d X so I can say the d u equals d X. You know, the worst. I can write it down like this. If I have my d x here and I have my do you hear, then I have a straight line, right? Do you equals d X? Of course. Ah ah I can have a different kind of situations as well. For example, if my you equalled, let's say three X then if I work too, take d you with respect to the X, Then I would get three. And in this case, of course, my D X right. The X goes here would equal to 1/3. Do you correct? Because I could write it down like this. D you equals three D X and then my DX would be 1/3 d you? That's not the case here. Okay, In this case, we have d u equals d X. That means that instead of d x, you would put just a d you, in this case, if we had this case instead of DX, you would put 1/3 do you? But in our case, instead of the X, you can simply put and the u okay, And now to get Are you initial? Well, it's very simple. You take your you initial equals ex initial minus four, right u equals X minus four. So you initial equals X initial minus four, which is zero minus four equals minus four. And then are you final Would be X final minus four, which would be eight minus four, which would be four. Okay. And there you have it. We can now form Are you went to grow and it would look like this. So from you, initial equals minus four to you Final equals four we put Are you here and squared. And since our do you equals D X. Then instead of DX, I simply put, do you? And now I'm just gonna follow this previous integral rule that I torture in the last video where if you have this kind of situation, you have this exponents here this too. Then it simply becomes one over two plus one. And then here you would also have two plus one. And then you put it inside the brackets and you go from minus four, 24 right. So you would go from ah, miners for to, let's say four wherever it is. OK, But you was not our initial variable our initial variable waas X. So now what we're gonna do, we're gonna go back from you were gonna go back from X okay, so we can write it down like this 1/2 plus one. That's 1/3. All right. But instead of you and here's U to the power of Cuba now to pour three. You cute, But our you was X minus four so we can write it down like this. Ex miners four and our two plus one. That was cute. All right. And now, since we went back to X, we need our ex initial and, uh, and X Finals are ex initial waas zero in our X final. I was eight, and now we're just gonna fold the the regular procedure where we're gonna have this kind of equation 1/3 times. And then, first of all, you take X munch for cute and you put X final inside of X. So you see, Now you have to It's very important that you look at this entire function as one as you put eight inside, and then as you put zero inside So first of all, you put eight inside and you have eight minus four. Cute, right? And then you subtract And then again, you take the entire function and you put zero inside. So you have zero minus for cute. Okay, It's very important that you do that. You have to be careful with that. And so you will end up with 1/3 then eight mines Force four. So it's for cute. Minus zero. Minus four is minus forces minus minus four. Cute. And then what you will get in the end is think it can squeeze it here. So all this equals 1/3. And then here you would have four cute would be 64 minus minus four. Cute would be minus 64. Right. So you will end up with 1/3 times 64 miners. Miners would give plus plus 64. So you will get 1 to 8 over three. And that's your second term. All right, that's Onley Your second term, by the way. Now our first term, Waas 1 to 8. And now we have to subtract our second term from our first term. 11. DefiniteIntegrals 2: So there you go. Your first term was 128 in your second term was 128 over three. And now what you have to do. You had to performed this operation one minus two. In other words, you take your 1 to 8 and you subtract my 12 8/3 and what you will get you will get approximately 85 0.3 square meters. That means that the area off your wall So you have X equals eight here and X equals zero here. And then this is why. And this is why equal 16. So the height here is 16. That means that the area here remember, this is just a function. This is a parabolic function, right? And we only consider we only consider this part here. And the area of this part or the area of the wall is approximately 85.3 square meters. Now I want to also show you that we could have done it in an easier way. I could have just shifted this function and I could have a function like this. So I have X here I have Why here and then have minus four. Here have four here. And then my problem would be like this. I would have 16 here, and then I could have just have a function like why equals 16 minus X squared. Okay, I could have that kind of function, and then I could have written it like this. D a over DX. No problem. And then, of course, I could have written like this D a. Equals 16 minus x squared d x, I could have taken the integral. And now, from minus 4 to 4, my ex initial would have been minus four mikes. File would happen. Four the same thing here from minus four, 24 Right. And I could have just written down like this. I could have said that. Okay, my area equals and I could have written it like this. 16. Integral from minus 4 to 4. The x minus, integral from minus four, 24 x squared the X right. Then here I would have had 16 and again X from minus four, 24 And then here I could have had, uh, minus. And now one cert right, two plus 1 1/3 x cute. And then that would have gone from minus four. Still four. So I could have written it like this. 16 now, four minus minus four, four minus minus four and then mine ish. 1/3. And then here I could have had ex cute. So have had four cute minus minus four. Cute. Like this and again here. 16. And here I would have four minus minus four, so that would be eight. So this thing would give me eight so would have 16 times eight. And here I would had, uh, miners 1/3. And then here I would have had 64 miners miners for Cuba, 64. So, you see, look, I would have had exactly the same answer 16 times eight. It's want to eight minus want to 8/3. And then that would have been approximately 85 0.3 squirt meters. So I could have done it like this. However, I decided to do it in another way because I wanted to teach you this institution method is the same answer. But I really wanted to shoot to go through the substitution method because in many cases it's very important that you know it and it can simplify your life greatly. So now you know the substitution method as well. So to in war, I like that very quickly. I want to cover another case. So look at this function here. This is a straight line function. Why equals X and I want to calculate the area under the curve will hear it would be under the curve here. It would be in the negative zone. It would be above the curve. Right. So I want to calculate the areas here, So I want to consider the inputs from minus two to to So let's see what I will get So I can write. Why equals d a over DX, right? So I can have d a equals x dx. All right. And then I can go from minus two till two. And here is well, from minus two till two. So my area will be then, uh, 1/3. My mistake. My area would be since here, your exponent is one. You would have one plus one here so you would have 1/2. And then here this exponent becomes too. You would have x squared and then you would have from minus two you would go to to so your answer would be 1/2. So you would put here to two squared minus minus two square. You know the words, you would have 1/2 and that would be four miners and then minus two squared would be also for positive. Four right, minus two squared would give you positive for right so you would have four minus four and then you would have 1/2 times zero and you would have zero. So look, your answer here would be zero. That means that if your function is in the positive region, if your range so this why now this This is your output, your range. So if your range is above zero, if it's if you're in the positive region, are here in this region right here. Your line is in the positive region, so if you calculate the area here, it will give you positive area. And if you're in the negative region here, it will give you the negative area and then if those areas are equal, well, even if they're not equal than your total area, would be the difference of your positive area and your negative area so you would have something like this, plus and minus and then the negative area. So if this big plus if this is, let's say 16 right? If this is 16 square meters and then this is 15 square meters, right, then your total area would be 16. Minus 15 would be one. But if those areas air equal, if you're positive area and your negative area are equal, then they will cancel each other out. And your toll area will be zero in this case, zero square meters. But really remember that here in calculus, the area very rarely it means m scored here. The area has different meaning. In other words, here the area under the curve has many different meanings. It all depends on the context. All right. It all depends on what you model. It all depends what your input is and what your output is. All right, Thank you for attending this lecture. And in the next lecture, we're gonna take a look at improper internals and just to give you a quick introduction, an improper integral would be that if you have some kind off initial point in your function and some kind of final point in your function, right? And then you have your function, and then you have your d X. Then you're you have an improper integral When, Okay, now, when you take the interval off, your if is a function of thanks, then we're going to say that that it equals big F as a function of X. So the capital, the capital F So this is like the anti derivative. It's like the result. It's like here you got from you went from D X or even better here. You had your X square, so that was like your f affects. All right, that was your X squared. And then you all of sudden you had 1/3 times x cubed. So that's your capital f of X. And now you put it in the brackets and then you would go from a to B. In other words, you would put your being side all right. You would put your be here minus you would put your a here. Okay, so this is your That's the X final. And this is your ex initial and improper internals happen when either a goes to minus infinity or your V goes to plus infinity or both. In other words, if you have some kind of function here, right, some kind of function, then you're a would go two miners infinity and your be would go toe plus infinity. So that's that's what an improper integral is called either when a goes to minus infinity, biggest plus infinity or both. So we're gonna talk about it in the next video. See you there. 12. Improper Integrals: welcome back. So let's continue with our improper integral. So in this video, we're gonna talk about improper, integral and the proper integral is when one or both boundaries approach infinity. So what do I mean by that? If I have a function de y dx and I say this is just X All right, then your graph would be the following, so your ex would be here and then Ah, your D y d x would be here. This would be your range. And then if it equals acts, then you just have a straight line, all right? And now let's say that the you want to calculate the area under the line starting from X equals zero up until when x approaches infinity. And remember, here X does not equal infinity. It approaches infinity. All right, so it's a different concept. It approaches infinity. It cannot equal infinity, but it can approach it. It can get larger and larger X approaching infinity essentially means that your ex gets larger and larger. And then if you take the integral of dysfunction, then you get the area under the curve and intuitively thinking what would be the answer? Well, it's probably be inventing, right, Because if you think about it, then as your ex approaches infinity, then we'll hear your area gets bigger and bigger and also your accumulated area. This continues recording starting from the beginning gets bigger and bigger. So obviously, if exports infinity then, uh, in this area here who will get bigger and bigger. And it will also approach infinity. No. How would we formulate that mathematically though So mathematically we would formulated like this we would have. Do you? Why equals x d x? Then we would put our Diggle science here. We would put zeros and zeros here and then here we would put acts and ex says just a variable. And then we would put the limit here limit as X approaches infinity and here is well, limit as X approaches and fanti And then you would simply calculate being to grow. So it would be the following. It would be why equals limit as X approaches infinity. And now you already know that the answer would be 1/2. Then here you would have x squared and then you would go from zero to X and X approaches infinity. So your why would be then 1/2 and then here you have X squared minus zero. Then you should not forget that you have to put limit here as X approaches infinity. And if you do that, then here this X gets bigger and bigger. So your why in the end will approach infinity. Okay, It's logical, right? It's logical because essentially, if you think about it again that integral are our area under the curve. That's one way of looking at it. But then the other way of looking at it is that you're continuously measuring the area. In this case, under the line, you're continue sleep, measuring the area from the start and then integral means that you take this area that you're that you're continuously measuring. All right, You still have the same input, X. But now you have different output. Why? And your output? Why is continuously measuring your area? So this area that you have here under the line, you now make a range out of it, right? So you first you have this kind of function, your axes, your input and then your F one. That's a function effects. And then you get do you y and D X Then if you take the integral off it, then you still have the same input X right. You still have the same input. But now you have a different function if two as a function effects. So if one as a function of X here was just X But then if two was a function of X now here would be 1/2 x squared. Right? So this would be your function here and then your output here would be different. It would be why and this why it's your area under the line. And as you go along, X, then you always measure how much area you have already covered starting from the beginning and for us. Now the beginning is zero. So you know, take this area and you say that Okay. My area is now represented on this number line on this output number line and then you get your probable here and a parable here Will you get this function and dysfunction? Is it parabola like this? And it's logical that if you're X goes to infinity, then your why or your accumulated area Your area under the line also goes to infinity right . So that's why you have your infinity here. But let's consider another function. Now, let's consider that your function now looks like this. So your input is still X, then here you're still have the wide the X. All right. But then you're function Looks like this. And for example, if you have one here, then you also have one here. But then, if you have to hear, then you have 0.25 here. That means that here in your do you why d x here it would equal one over x squared. So now this is your function. And your goal now is to find the area under the curve. Starting from X equals one up until when x approaches. Infinity. Okay, starting from here. You want to know what's the area under the curve as X approaches infinity. Well, my first intuition tells me that it should be infinity. Because if x never ends, if this ex never ends, then the area should never end either. Right? So the area should also approach infinity. But is it true, though? Let's find out. So let's calculate this. I need to go and see what we will get d y equals one over X squared d X. You can also write it down like this x to the part of minus two DX. So we very simply put the integral is here and here and here. And then we go from zero to X. Yeah, my mistake from 12 x Here we go from 12 x want to acts. But then we should not forget that we have to put the limit here as X approaches infinity and here is well, limit as X approaches infinity and limit as X approaches infinity. So to calculate this integral well, let's first will take the centre girl like this so x to the power of minus two times DX. Well, here in the denominator, we put minus two plus one and then here we would have minus to plus one, which means that we would have minus one times X to the part of minors one or minus one over X. All right. And now we can say that why equals limit as X approaches infinity. So we would have miners one times one over X and we're gonna put it in the brackets. I should also initially put this miners one inside the practice, but then you can factory doubt. So let me show you what I mean. So instead of here, I'm gonna put mine. Is one here? Right? I'm gonna put miners one here and then I'm gonna go from one two x No, that would mean that I will have limit as X approaches infinity. So first of all, I would have minus one times one over X and then I have minus. And then again, the new function would be minus one divided by and now I'm gonna put this one inside here instead of banks. Now, if you see, then this constant can be now factored out. That's where when you have a constant, then you can just factory it out, put it here and then you simply have one over X and then minus one like this. Okay, if our X approaches infinity and this becomes zero, right? And that means that your entire answer is one. So there you go. If you go from one up until in fut than your entire area under the curve is one. So how can it be? It's not intuitive at all. Well, I can give you the intuition in two ways. The first way is if we look at this equation here that we get All right. So what do we have here? We have. Why equals And then this entire thing. If we calculate like this minus one and then you have one over X miners one, you can also write it down like this. If you multiply everything by miners one, the needs would be one minus one over X. Okay, so that would be the function that you would get if you take this, uh, indefinite, integral, minus the constant. Of course. Because remember, with indefinite integral see also need to put the constant. But here we here, we're not dealing with the indefinite integral. But that would be one part of the indefinite struggles that we would get. All right. And then we would simply put the initial and final values inside X that How would it look like graphically? Well, you look like very simply like this. You have. Why here? And remember now our area under the curve. Right now it's measured on this. Why number line and your input states the same. So if my input is one. And I'm also gonna put to here. And I'm also gonna put four here. So if I look at this function, then if I have one, if my ex is one, what will I get? Why equals one minus 1/1? Well, I get zero. All right, So I'm here now, if I have to, then why equals one minus 1/2? I get 1/2 or 0.5, right? So I'm gonna say that this is your 0.5. There you go. Now we 54 If my Xs 41 minus one fourth, this is 0.25. So I will get your 0.75. Right? So I'm gonna get it here. 0.75. So here's my point. And in fact, if my ex approaches infinity, then this is one here, right? And as my ex approaches infinity than this term here becomes smaller and smaller. And if x approaches invented in this term disappears, it becomes zero. So therefore, my function would be something like this. It goes through this points but then it will never It will never, ever grow bigger than one. So it will always approach one. You see, this is your theoretical limit here. And as your ex approaches infinity, you get always closer to one. But you never quite reach it. You can Onley get closer to it as your ex approaches infinite. As your ex gets bigger and bigger, you get closer to warm. And that is because this is your theoretical limit. And this term here becomes smaller and smaller as your ex becomes bigger and bigger. And of course, dysfunction would continue also below one when your ex is smaller than one than your function would be here. But we only consider from one until when X approaches infinity. Right? So we start from here and then we simply approach one like that all the time and another way to look at it. It's like this. Let's consider a rectangle, all right? And let's say that this rectangle or the square, let's consider that the area of the square is one. Okay, we're we don't care about the units right now. Well, we can say that. Okay, this exceeds meters And why's meters? In that case, it would be one meter squared, OK, But you see, this square is empty and this square is empty when X equals one. You see, when X equals one, your area is empty. You're areas zero. You see your Y zero, but you're why equals your area. So when X equals one, your areas zero No, When you're X equals two, then my area is your 0.5. So I cover half of this square. Okay, so that is when my ex equals two. Now, when my X equals four, then my area is 0.75 which means I take the remaining empty now rectangle and I cut it in half. So now this is when my X equals four. You see, when my X equals four, I have covered this area here and also this area here. So in total, until if it if the total area off Mirer off my square is one. Then once I have covered four access starting from one gay where we start from one. Once I've gone from X equals warm two x equals four. I have covered 0.75 meters squared or 75% off my off my area in the square. And of course, if my ex now equals eight okay, When my X equals eight, then I will have covered 1/8 off the area. And then if my X equal 16 then it's like this 32. It's like this and I can go on and on and on, and you can see that I can always take the remaining empty area and I can cut it in half. And if I do that I never quite reach area equals one meters squared because I can always take the remaining empty area and cut it in half, right, So you can always cut it in half. So I will only be able to cover the entire rectangle when my X would be in finished. But it can be infinity. So that means I will never be able to cover my entire rectangle or, in this case, my entire square. I will never be able to color it fully. I can Onley approach the square equals one meter squared. I can only approach the state as my ex gets bigger and bigger and bigger. And so, as my ex approaches infinity, then then I'm able to color in my square more and more. But I'm never able to quite reach area equals one meter squared. All right, so this team to issue behind it. Okay, well, thank you very much for attending this lecture. And in the next lecture, I'm gonna show you how integral Czar formulated mathematically in a former way. See them? 13. integrals formal definition: Welcome back in this video, I want to show you a formal mathematical definition for definite integral. And to do that, I'm just gonna draw a random function here. And this is my input here, X, and this is my output. Why? And let's say I have some kind of random function and I want to calculate the area under the curve, starting from a and then I want to finish at B. So my a is X initial. So I'm gonna call it Ex initial or X with an index off zero, and that will be my A All right now, we already know how to calculate theory, end of the curve, or at least the logic of it. And so what we do, we divide this area into small rectangles and and then we just calculate these rectangles, the areas off these rectangles. And now the goal of this video is to first of all, show you how to mathematically represent Ah, the calculation off off these small areas here and then how you would mathematically represent the addition of these areas. And then how you're going to arrive to integral is when you do that. So let's continue. If this is X zero right, A 60 then we're going to say that the interval, right? The interrupt between these or the width of the small rectangles is that the X all right and the with is the same for all of them. So if this is eggs Eero, then I'm going to say that this point here is X one. Then this point here is X two and then I'm going to say that this point here, after whatever amount of rectangles than this point here's X and mine is one and then be is X with an index off. And okay, so I have x zero x one x do Then after some rectangles, X in minors won the penultimate point. And then I have X with index of n here, which is be now my goal here now is to find some kind of discrete function. Alright, Right now we're in the taste great world right now I want to find at this crease function Just gonna put a d f. This will function right off. Which output is this discreet area under the curve, right? So in this case it was look, something like this and then something like this, Right? And here is, well, something like this. So there will be a lot of rectangles here, and then if we send them up, we will get our discreet area. So our goal is to find this discreet area, and our input here would be n or the amount off intervals here or the amount off these rectangles. All right, so I need to find a function, a relationship that takes the amount of wreck things that I have, And then I have my discreet area. So I have to find this function here. So how would I do it? Well, first of all, how would we get this discreet area in the first place knowing, for example, the interval or the with off one rectangle? Well, our discreet area could be written like this that if I take, ah, my function why? And then I put some kind of point inside Xie. It can be x one or x two or exited minds one. It doesn't matter if I do that. Then what will this term give me? It will give me the height. At that point, let's say if I put if I put here inside the why and then I put X two inside the parenthesis . Then what I will get. I will get this height here. Right? So why at X two gives me this height now? If I now multiply that height by Delta X, then I will get this little area off this rectangle, right? I'm gonna get the area of this rectangle. So if I put Delta X here, then whatever x I put inside here inside the parenthesis, then I would get the area off that rectangle. If I put X in, mine is one here. Then I will get this rectangle here if I put be inside here or X. And then I will get this rectangle here because why at Exxon will give me this height? If I multiply by Delta X, I will get ah, the area. So in order to do now, some all those rectangles up, right? This formulation gives me the area off one rectangle, depending on X sub I so he found went to some of them all up. I just need to put the summation side and then I will have from I close one up until and okay, so So I started going from X one. X one gives me this height and then times they'll takes. Then I will get this area here and then, Ah, the next one would be x two. So I can even right down here to make it more clear. So first, why X sub warn times Delta X plus, Why? At X two times Delta X plus, it's gonna continue here. Plus, then you would have why xn minds one times Delta X and plus why at X and or be times built that X right. And this formulation gives me the total discreet area under the curve. So, for example, this this term here, well, give me this area here. All right, X, and mine is one times Delta X. It goes to the left. And then that's how I will get my discreet area. But right now, right now, my discreet area is not as a function of end. It's a function off Delta X and then X I right. So right now I have this kind of situation where I have built X and then I have some kind of X I and then I have dysfunction here, which is, uh, why at excite Times, Delta X, and then I get my discreet area. Okay, but my goal was to find it as a function of end as a function of the amount off wreck things that I have, which means that I have to take my to impose that I have here, and I have to write them in terms of an in terms of the amount of rectangles. Okay, so let's do that. And if we do that, then, uh, I can show you how you can get into girls out of it. So first of all, let's think about death tax. How could we write Delta X as a function off of the amount of rectangles? So now we are in this kind of situation where we have an input, which is Delta X Right now, we'll need to find some kind of relationship so that we would have Ah, in fact, now what we want to do, we want to have an input as the amount of rectangles, right, And then we will get dealt acts as an out. So now we want to find this kind of relationship. And how would we do it. Well, if you look at it, then if you look at B in a All right, what will you get if you subtract a from B? If you have this kind of expression, what will you get? B minus A. Well, you will get this distance. Right. So this is at point A. I'm just gonna right down here, and this is be You will get the distance from a to B. Right? You will get how far? Be ease from a You will get this distance here. Okay, That's what you will get if, uh, you have B minus a. And now what will happen if you divide this distance by the amount of rectangle so by n so , if you're understand, if you divided by 10. So if I divide this by the amount of rectangles, what will I get? Well, I take this distance right from a to B, and if I divided by n, I will chop it into small little pieces. And then the distance of that piece is Dell tax. Right? And there you go. That's how you can write the Delta X as a function of end. Okay, so you just calculate the distance between A and B, and then you divided by the amount of rectangles. And then you get your deal. Thanks. And now so dysfunction goes here and now you can write You're dealt X as a function of n which is what we want it because now we can put this expression here. Now let's go to the next one X abi. What would we do there? So again we want and and then and these are input. And then let's say now we want another function. Let's call it H. And here we want X up. I So we want end as an input on some kind of impression should give us X I Well, one way to do it is let's just pick around the point. Let's just pick X two, right? So how would we get two x two? Well, I could write down an expression like this. I could say that for example, X two all right equals a plus. Two times Delta X doesn't make sense. Yes, you're at point a and then your ex too Well. In order to get two x two from a, you need to add two plus Del tax. So you have to deal taxes here, the tax and then another the attacks. So you see, you can get it like this also. If, for example, if you want X one then it would be a plus. One time still tax write a plus one time still tax and your addicts one eg zero a plus zero times the tax. Right. So this term will be zero. You're at a ex n miners one a plus. And then here you would have. And mine is one times Delta X And you're at this point accent equals a plus and times Delta X and you're here. Okay, No, this Delta X we already had the expression for Dell tax. So it was this expression. So in fact, this term here, this is like counter right, You count whether it's two x one x two in minus one. This is the sub index, right? So we can say that this up in Vicks, we can say that it's I right, it's a counter. And then we can rewrite the attacks like this so we can say that X I equals So this is a sub index I a plus i times. And instead of Dell tax, we put b minus a divided by an You see, So this becomes I hear these these terms they become I there this up in assists and then instead they'll tax you put this you put this expression and no, you very simply take this function for our discreet area that have re written here. All right. And then you simply substitute Delta X with the with this expression that we got earlier with this expression and you also substitute X I with the this expression that we got earlier this one here. And so this expression here is our disagreed area. Okay, are discreet area. So we get a lot these small rectangles of went and when we discreetly sum them up, then we get an approximate area under the curve. But it's an approximate area with errors because we're doing it discretely. So now we have to go from this creek area two continues area and in order to do that, we have to take that end that we have here right? And the end has to approach infinity. What will happen when n approaches infinity? What will happen that means that the amount of rectangles increases more and more and more . So you will have more and more rectangles here, right? But at the same time, Delta X becomes smaller and smaller and smaller. So if you look at this term, this is the Delta X term, right? This is this. This is the Delta X term. This term becomes smaller and smaller because, and becomes bigger and bigger. The amount of rectangles increases. And so your interval be miners A. This distance stays the same. But if you have more rectangles, that the amount of rectangles is higher than your death tax will be smaller. So this part here this Delta X will go down because smaller but at the same time, at the same time, you will have ah, more, let's say X eyes, right? So if you look at this part here, then the amount off excise Well, I just say that it's a number off excites the amount of X eyes increases right, So the logic is that the width of the rectangles goes down, but the amount of rectangles goes up and then you will get closer and closer to the continues world. So you will approximate the area under the curve better and better, right? And then when that happens than your discreet summation becomes, you're continues summation, which is an integral right. And so this is the forum mathematical definition off definite, integral. The formal definition off a definite integral is obtained like this. So you take this this create area here, the formula for the described area and like we discussed before If if n approaches infinity and now you take this entire formula and you take limit of it as n approaches infinity. So what will happen this term becomes smaller. Also, this term here becomes smaller. But then the amount of eyes increases because your end gets bigger and bigger and bigger here. So now I goes from one up until, ah, higher number. So before it was 100 now it's maybe 1000. So now you will have 1000 summation terms, so this eye becomes bigger. Okay, this becomes smaller here, Del tax becomes smaller, but I becomes bigger and then a remains the same and be as well. And there you go. You take this entire formula that we have derived and as an approaches infinity. Then you go from this good area to continues area. And this is your definite integral from a to b your function, which is this and your DX, which is this. And then if you take the integral of it, you get your exact area under the curve. And that's how a definite integral is formally defined. So now you know the formal definition of it. All right. Thank you. And see you in the next lecture. 14. CircleArea1 initialConditions: Welcome back in this video we're going to apply. Integral is to compute the area off a circle, and we're gonna do it in two ways. So let's get started. So I'm just gonna draw a circle here. All right, So I have a circle. We're going to imagine that the radius here is Ah, five meters radius equals five meters. Now, just to give an idea of how to calculate the area of the circle, we should think of it like that. So let's suppose that I'm gonna take a random radius here. A smaller radius, Let's say our apostrophe, right? And then I'm gonna rotate it 360 degrees. So what will I get? I will get us a conference, right. Well, if I take another radius, let's say this radius here, then I'm going to get another circumference. And in fact, if you want to calculate the area off of the circle, then what you need to do you need to sum up all the conferences off the circle, starting from radius when radius equals zero meters up until radios equals five meters. In our case. So that's the general logic. If you want to calculate the area of the circle. Then as you go from radios equals zero up until radius equals five meters. You need to sum up all the possible, sir conferences. And then you will get your area. So how would we do that? Well, first of all, our radius all right, goes from zero up until five meters. And what was our formula for circumference? Same conference was two pi times are right. We covered it in the past. And if if you want a graft dysfunction, then he would look something like this. So you have radios here, then you have soon conference here, and then you have a straight line here. So straight line function and C equals two pi are. And now if we take this radius apostrophe here. All right, let's say radius here is five. So this is five meters. So this is our full radius. It was a circle, right? And in this case, than if radius equals five, then the same conference would be too pie. Times five would be 10 pie meters. Okay, sir. Conferences in meters and radios as well as in meters. And now we also had this radios apostrophe which was a small radius. So let's say somewhere here and then the same conference, this in conference here would be two pi times radius, apostrophe, Right. So you can see that as you start from here, when your radios equals zero meters and then you move up until radius equals five meters. In order to calculate the area of a circle. Like we said before, you have to sum up, allow the possible suit conferences that you have here. But by looking at this graph, then you represents their conferences as a vertical line. Right. This is your circumference access. So in order to sum up all the possibles or conferences for a circle, you need to sum up all these vertical lines here. In other words, you have to calculate the area of this triangle, or you can use integral to do that. And we're gonna use Dean two girls to do that. So was do this. So you had area of the circle equals. And now our function Is this a conference functions two pi oh, are all right. And now, if we want to calculate the area off the circle, we have to calculate the area under the the line here, so we have to take the integral from Ah, well, right now, let's just take, uh, an indefinite integral. Let's just assume that, uh, it's just an arbitrary constant are that? Let's just forget about five for a second. We'll just say that is just are okay, because we want the general formula for the area of our circle. So, in order to cover the area, if our circle this is our sin conference formula here, and then you take the integral of it sometimes D r. So the are would be the small differential here D r. So it would be like a super thin area off the rectangle. And so if you calculate the integral, then what will you get? You'll get two pi are d r. Two pi. And now, since you have one here, you're gonna have to here and then you're gonna have r squared here. And since we're now calculating the indefinite integral, remember, you had a constant. You have to have a constant. But now we have to look at our initial condition. Now, what is an initial condition? When I say initial condition, I mean that Okay, when my radius equals zero. Then what's my area off the circle? Well, if my radius equals zero than the area of the circle equals zero square meters as well, right? Therefore, if from this integral I see that the area off the circle is and by the way, I can cancel those two things. Those twos here if the area of the circles pi r squared plus C and I know that when radius equals zero, then the area of the circle equals zero meters squared as well. And that means that I'm gonna make this are here zero. And I'm gonna make this a C here is your as well. So I'm gonna have zero equals pi times zero squared plus c and that means that my C equals zero. So my initial condition means that if I take mine very evoke my input and I make it equal to in this case zero. Then what will be my area of the circle when my rate is equal zero? My area of the circle is zero and therefore this constant see also equal zero. Therefore, I can conclude that the area of the circle equals pi r squared and There you go. We have now derived our area of a circle. And now if we say that our radios equals five meters, then we very simply just calculated, like this pie times five squared equals 25 pi meters squared. But I won't go back to this initiative condition case And I want to emphasize that that's not always the case that when you're in Portico zero, then your output equals zero as well. For instance, let's say that if you have a car, all right, you have a car in the car drives in this direction and then we know that displacement on the car was going to say D, which is a vector. Displacement of that car equals 10 meters when time equals zero seconds. Okay. And we also know that, uh, the velocity off the car is the following. So this is my velocity here, and this is my time. And I know that the function off my velocity is the equals five t plus five. Right. So this part here is five okay, meters per second. And now you want to calculate the integral of dysfunction because if you can call it the integral of dysfunction, you will get displacement, right? Because if on one axis you had velocity and on the other hacks is you had time, then here, the area under the line, the meaning off this area here would be displacement. So let's do that. Let's calculate the integral off this function and it's gonna be the indefinite, Integral. So five T plus five DT would be five integral t DT The Loss five Integral DT equals Now here. I would have 5/2 t Square plus five t. Okay. And now, since it's an indefinite integral, I have to put constant here. Okay, so this function is my displacement. This function is for my displacement vector. But now we know that atomic 00 seconds, my displacement equals 10 meters. So I'm gonna make my time. Zero right, 5/2 times zero squared, plus five times zero plus c. But now my displacement, atomic all zero seconds. My displacement equal 10 meters. Okay, now this term will be zero, and this term will be zero. So that means that my C equals 10. So that means that in this case, only in this specific case the function for my displacement would be 5/2 t squared. Bless five T plus 10. So now your constant is 10 and you get it from the initial condition. And remember, the initial condition means that Okay, if you have your input t and then you have your function and then you have your output, which is displacement. So the initial condition Is that okay? If my t equals zero seconds, what is my displacement? If my displacement was, let's say 20 meters than my constant would be 20 meters here. Then I would have 20 meters here. Okay. All right. Thank you for attending this lecture. And in the next lecture, we're going to see how to calculate the area of a circle in another way to see you there. 15. CircleArea2: welcome back. So in this video, I'm going to show you another way how to calculate the area off a circle. So again, I'm gonna draw a circle here. And this time we're gonna do it like this, that I'm gonna say that I have a small area here, A differential area. Of course. In reality, the differential area would be a lot smaller, but I'm just going to say that this is my the A Okay, so I have some kind of radius here, and I wrote they re just a little bit okay, Just a little bit up until here. And I have covered a small area here. Now, how would I represent this area? So I have radius here, and then I also have my arc length here. Since this is a differential area than this arc length here, the small arc length would be D l. Okay. And in order to calculate my area, I would have to do it like this. D a equals D l times are divided by two. Does that formula make sense? Well, let's see if I have some kind off radius here. All right, so this is my radius and then let's say I have rotated a little bit. So I was here and I'm here, so I have my d l here. So if I have my deal here and I multiply that deal, but are then what I will get I will get this area here, so I'm gonna get twice the area of what I need. So I'm going to get, like, this entire area, you see, because my are is here, and my deal is here. That means that if I multiply our by D l then I'm gonna have the same deal here. So this would also be my deal. And this would also be my radius. So it would be like a rectangle, but here deals, they would be arcs. Okay, but then you you only need this area. Therefore, you have to divide this thing by two. Now, how would you Rippers and D l you could represent deal in the following way. Deal equals yeah, times. And now we're gonna say that since I have rotated just a little bit, I have rotated by D feta. Okay, so this is my differential area, and I have also rotated my radio's by a differential angle, so by defeat. So this is my defeat, that here and as you know, in order to calculate the the ark in general. So if you have some kind off random circle and you want to calculate the ark, then you just need to take your radius and this is your angle, and then you take your radius and you multiplied by your angle. And of course, your angle has to be in radiance. And then you will get your our clinks. Okay, So since I have rotated by defeat A here, then my deal can be written like this are times d feta. That means that I can write my d A. Like this. Ah, Times defeat. Er times are divided by two. Or, in other words, I can write my d A. My differential area. 1/2 r squared the theater. Okay. And now, in order to calculate the entire area of the circle, what I need to do, I need to take a definite integral from since now, my differential is defeated and it's in radiance. I have to take it from zero radiance up until two pi radiance. Okay, From zero up until two pi. Why two pi or because two pi would give me three on 60 degrees so I would rotate one full circle if I go from feet equals zero up until feet. Eichel's two pi. Okay. Remember these Air seat us here. The lower limit, the upper limit. Okay, so what will we get when we do that? And remember, now this is very important. Now our differential is defeated. So in fact, our input is Sita not are in fact, are now is constant. Okay, When one axis, we have feta and then on another axis, we have d a over de cita. All right, so here on this axis, we have the change of our area with the respect to feta. And here we have feta. So we have some kind off function here, and, uh, will. This function, you can see is a straight line function, but it's completely horizontal. And the height off this part here is R squared over to So you're radius is now given your dino your radius. Okay, You don't move your radius, You move your Sita, okay? You're Sita is your variable and your radius is already constant. You already know your radius. You don't do anything with your radios. It's a constant is given. So you're very Sita. And, uh, well, if you look at this function, all right, so since you have this d a equals this or let me just write it down here again D a equals 1/2 radius squared de cita. Okay, so you can take this defeat. Uh, and you can put it here. Okay. And now you have d A over defeat equals 1/2 r squared. So there you go, de a d feta. It's 1/2 times r squared, and your radios here is constant. So since ah, you don't have any feet that variable here in this formula. Since you don't have this Tita variable, then, uh, your function here is a straight horizontal line. And to calculate the area off the circle, you just need to go from zero up until two pi and then you. Once you calculate the area underneath this horizontal line, you will get the area off your circle. So that means that we can write this formula here like this 1/2 times R squared de cita from 0 to 2 pi and, uh, that would be then 1/2 R squared feta from 0 to 2 pi and that would be 1/2 R squared two pi minus zero. So you would have to pie r squared over to you can cancel out twos and look again. You're left with pot r squared since the same formula. You used another approach to calculate the area of the circle, but it's the same formula. Okay, now, one thing that I want to note is that, and I just want to make sure that we understand it. I want to go back to this part here when we have our normal function. Right, we have some kind of function, and in order to calculate the area underneath the curve, we chop the area into small rectangles. Right? That's what we did. And now these things here these were my errors. And as my so let's say that this is my Delta X. This is my accent. This is my why, and as my Delta X approaches zero, these errors would also approach zero so they would become smaller and smaller because I would have more and more rectangles that would be thinner and thinner. So I would have, like rectangles here, and and then I would have rectangles here so my errors would become smaller and smaller. Okay, Now, in this case, in this case here, when I had my radius and then I had my our clanks my d l And then when Ah, I multiplied my art times d l. Then I got this kind of shape here like a rectangle. But then I had this instead of straight lines. I had this arc, so I needed this area. But in fact, when I multiplied our times, d l I got the entire area off this, okay? And then we had to multiplied by two to get this area. Now, I just want to point out that this is not an error, okay? Not error. And the reason why it's not an error is because here, if my defeat, er here, if my well, let's say that it's not defeat that let's say that it's Delta Theater. It's not differential. It's It's a bigger change. Let's say Delta feta equals one radiance. Okay, But now when my Delta theater now approaches zero radiance right when it approaches zero radiance, then this part here won't become smaller. Okay, so in this part here, these errors, they would disappear as my Delta X approach zero. But here, as my daughter theater approaches zero. Then this part here, it wouldn't disappear, so the proportionality would still be the same. So I would simply have something like this, so I would have a smaller delta theta. All right, so I would have a small delta feta here, but then I would have exactly the same thing here. So I would also have a Delta theta here, and then I would have my deal here and deal here. But you see, this part wouldn't disappear. So you would still need to multiply these two guys here by two. So as your delta feet approaches zero, the proportionality here won't change. So this part wouldn't disappear. So therefore, it's not an error, so I just wanted to point that out. All right. Thank you for attending this lecture. In the next lecture, I'm gonna show you how you can use integral to calculate the volume or for solid. So see you there 16. Integrals VolumeForSolid: welcome back. So here in this video, I'm gonna show you how to calculate the volume off a solid that looks like this. So let's make a graph here. So I have my Z axis here. So this is my Z axis. And this here is my Y axis. And I'm gonna have four here, and then I'm gonna have to here. So my function will be well, you can already guess on a function will be Z equals square root of why, Okay, so imagine if I now calculate the area under the curve here, and then I just take this entire plate that I have this entire area and I rotated 360 degrees about the Y axis. So let me depict it in a three dimensional space. So let's say you have your ex here. Your ex now is popping out of the plane. Positive. So if I now try to draw it in Ah, three dimensions. Okay. You would have. Why here? You would have x here, and then you would have Z here. Okay. And now you would have four here. You would have to here and now. Okay, so this this is the same graph like this. Okay, But now, if you rotate it by, let's say by 360 degrees about the y axis. Then you would have something like that. Let's say a few, first of all, rotated by 90 degrees. Then give. This is two. Then you would also have to here and well, then you would have something like this. And then same thing here you would have to. So this is now. Mine is easy access here. So I guess you would have something like this here one point, and then this is minus X axis. You would have well minus two here, and this would also be minds to. In other words, if you imagine it, then you would have some kind off solid that would look like this. Okay, so you would have some kind of solid that would look like this. So just imagine that you take this plane and then you rotate it three and 60 degrees, and then you would have this solid here. Okay? And you want to calculate the volume of the solid. So how would we do it? Well, the logic is very simple. As you go from why equals zero meters up to four meters. So everything is in meters now Here meter Meter actually is also in meters. So as you go from, why cause zero meters up into four meters? Every time you move just a little bit You take this small height in the Z direction and then you rotated three and six degrees about why access? So let's say if this is my why star here? I'm just going to say that this is my wife star and that will give me Z star here. Okay, so my Z star here would be Z start here. So this is my wife star, my Z star. And now if I rotate this Z star here, then I will have like, uh let's say like, small disk. You see, I will have a small disk and this Z star here is like a radius of that disk. Now if it take another why let's say why a posture feet Then I will have Zia posture over here. So if my if this my wild apostrophe in this my Z apostrophe here right then this would z apostrophe here at when Ah, I have wild apostrophe here. So here the Z apostrophe would be like a radius off this disc that I will get here. See? So in other words you take, you're your value for Z at each y. And remember, you're Z would be different for each wine because, well, because this is your function, right? So for each, why you would have a different Z and then as you go from, why call zero up until four. You just take each Izzy that you have, and then you've rotated about why access and then you get a disk and then you calculate the area of that disk. And if you go from, why cause 0.24 you will have an infinite amount of disks. Okay, so you would have a lot of discs here, and then you would just sum up along these areas off these disks, and you will get your volume. So how would you calculate that? Mathematically, Let's see. So the area off some kind of disk let's say, if this was my y apostrophe and this was my Z apostrophe than the area of this particular disk, right, this particular disk would be a apostrophe, so this a apostrophe would be pie times now the area of the circle, remember? Was pi r squared. But here instead of r squared, I have Z squared. So I have ze apostrophe squared. Okay, so I take my Z value here, my z apostrophe. I square it and multiply it by pie and I get the area of this disk here and in general, I can say that the area as a function off Z or excuse me, my area as a function of why would be pie times Z, which is itself is a function off. Why and all that would be squared. So pi times z squared. But Z itself is a function of why. Okay, In other words, my a as a function of why equals by times and my radius would be squared. But instead of radius, I have ze All right have c But instead of Z, I can write square root off. Why? So I can say I could say that it's like that. Okay? Or in other words, you can already see that I can simplify it. Square roots off. Why squared would be pie? Why? No. In order to calculate the volume off this entire soul. It we said that we have to sum up all these discs and remember, we're now in the differential world, right? So we would assume that this rectangle here that we have it's a super thin rectangle so that this d y so it's any infinite decimal e small interval, so it approaches zero. So that means that the amount off discs here brush infinity. So the amount of disks here approaches infinity. Because, my doctor, why approaches zero and my daughter, Why becomes d y becomes a differential? So since the amount of disks approaches infinity, the way I'm going to sum them up is again using intervals. So I can say that the volume off the solid equals the integral off the area off the disks that I have, which is a function of why times, d y. Okay, so you can think of it like that. So I have an infinite amount of disks here, and each disc has a different area, and then this integral operator is going to sum up all these discs together. And then I get my volume. So how did we right our A as a function of why b y So our volume would be no. P is constant, so we can put before the integral and then ah are in to grow Went from, like, zero up until I was four. And then I had Why d why? Okay, That means that my volume is p over to why squared from 0 to 4. So be or two. And now I would have four squared minus zero squared. So I would have this will be zero, and this will be 16. So would have 16 pi over two or eight pi meter cubed. Okay, so that's the logic of it. Remember all what we did. We took this graph here. This is equal squared off. Why? And then we went from Weyco zero open to Wyck was four. And then what we did, we just took each Z that we had for each why? And we rotated on both y axis. And so we got the area off the disk and we represented it. Uh, right here Pi Times Z is a function of why and all that squared. And then I got the area of disks and then my integral here summed up all the areas of the discs as I moved from. Why zero up until Wyche was four. But then my area I could write it down like this five times. Why? And then I took them to grow. And now I can say that the volume of my soul it is a pie meter. Cute. And one more thing that I want to note. He said, Be very careful. Don't think that you can just calculate the area of dysfunction, right? Let's say you know the area of dysfunction. I don't know it right now, but let's say you know the area of this function Z equals squared off. Why, right, this is why And this is he and then an intuitive Let's say sometimes people make a mistake and think that Okay, I know this area. So I'm just gonna say that I'm gonna multiply this area by two pi, right? And then I think that okay, you just rotate this area about why access? But that would give you a wrong answer so mathematically you wouldn't get the right answer is just like with a circle, right? In fact, with a circle we know now that our area off the circle was pi r squared. However, sometimes people think that it's intuitive and it kind of is intuitive if you think about it. Is that okay? The same conference off a circle is 25 are right. So if I know my circumference, then why can't I just, uh, multiply my circumference by our and then I should get the area of the circle right? But then the answer would be wrong because what you would get you will get two pi r squared . But we have already seen from integral, which is the right answer. And the right answer is that the area of the circus pi r squared. So you see, if you take the soup conference off a circle and you multiplied by radius, then you don't get the area off your circle. So mathematically, it would be wrong. So be very careful with that. So if you want to calculate the area of the circle, you have to think in terms off summing up The conference is like that, like we need in the previous year. And here you have to think in terms off going a little bit, taking the high, taking the easy and rotating it about the Y axis, calculating the area of that specific disk. The moving again, taking another Z, rotating it, getting another disc, calculating the area of the other disk and then using the integral to sum up all the areas of the discs. And that will give you the volume. All right. Thank you very much for thinning this lecture and see you next time. 17. integration by parts: Welcome back in this video, I will show you how you can compute more complicated into girls using a clever trick called integration by parts. So let's get started. So sometimes you're in a situation where you need to take the integral or for function that consists off two different functions so you can have an integral and then you have one function here as a function of X, and then you have another function here as a function of X, and then you have DX, and an example for that could be like this. So I have X times call sign X. Okay, so this extra would be like this function here and this closely next would be, like this function here, and this is a little bit more complicated, integral and and then you can calculated just by using this institution method that we covered earlier. But we still want a general formula that solves this kind of thing to grow. And let's see if we can do that. So let's imagine that we're gonna take a random function, and we're gonna forget now about this function here for a second. So we're just going to say that we have a function that consists of two functions. Okay, I have FX. That's GXE. And now it's not X times cause the next just for now, forget about this part. Just imagine that, Okay, we have FX times GX. Okay? No, If I know, apply this operator to this function. All right, then I'm gonna have to use a product rule, right, Because our first will have to differentiate this FX, and then I leave GX alone and then I leave my FX alone and then I'm gonna have to differentiate my g x. Okay, So this is our product rule. Now, what we can do next is we can take this DX, all right? And we can put it here so I can also write it down like this d f as a function of X times G is a function of X equals. And then I have d d X and then I'm gonna have f is a function of X g is a function of X, and I'm gonna have d x here. And then I have plus FX, the D X. She is a function effects, and I'm gonna have DX here and this because I took my DX and I could it here and then I distributed among those two terms. Now I could cancel out. My DX is here, but I'm not gonna do that. I'm just gonna keep my DX is there? Okay, so now instead, what I'm gonna do now, I'm gonna apply the integral. Okay? I'm gonna put an integral here, and I'm gonna put and into girl here, and I'm gonna put any two girl here. So if I do that, then this part here will be again F x times gx. All right, And now here. I'm just gonna leave it like that. Let's did like this, you know, to save time, I'm gonna write F as a function effects as F and G is a function effects Justin G. Then it's easier. So you're gonna leave it like this de de X if times g d. X. So I applied the integral on this side and on this side. If I apply them to grow on this entire site, Aiken distributed among two terms, so I'm also gonna have an integral here, and then I'm gonna f times d dx. And then it was applied to Gina and then I have DX here. Okay? And now what? I can do what I can do. I can simply take this term here, and then I'm gonna move it unto the other side. So why would I do that? Well, because now this term here will equal the following f times de d. X applied to G DX here. That would equal f times G. This is this minus now, since I took this term and put it on the other side, then became minus minus de d x applied to f times G, which is not touched by the differential operator times DX. So look, I have some kind of formula here now. Okay, So can this formula help me cackled into girl off this function? Well, let's see. So what I can do now is I can take this function here, right? The integral off x times courtside X and I can say that OK, let this x be this f and let this cool sine x be this so not g anymore. It's not g anymore. No, it's the derivative of G with respect to X. So my coastline x is the change of my G with respect to X, Okay? And I can do that in mathematics that can define things How I want. So I say that X equals F and coastline. Next is this term. So if I look at my formula, then what do I need? And what do I have? I have f which is my ex, and I have the derivative of G with respect to X, which is my call sign X. So I have my f But now what I need I need my G and what I also need I need my derivative off f with respect to X. Okay, So once I have g here, I can also have it here. So let's let's start with this part. So if I know that my f is X All right, then, to get this to get this term here, it's very easy. So d f with respect to X equals one all right, because it would be DX over DX, so that would give me one. So there you go. I have this term here, and this term is one. But now I also need my G. That means that I need to take my call sign X And since it's the derivative of G and that means that I have to take the integral off it to get my G extension, What happened was that I had my G and then I apply this differentiation operator and then I got my g g d x. Okay, and now we found Want to go back the g d X. If I want to go back to G well, I'm gonna have to acclimate. Integral. Right. So what's the integral off course Sign? X dx. Well, I know this If my sign if I have my sign and then I take the d X from sign then I will get call sign. That means that if I have my call sign then in order to get my sign and since integral is also anti derivative then I simply have to take my integral. So if I take the integral of course, and I will get my side So I will get here my sign, X And here I don't put my constant I don't put my constant because you only put your constant in the end. In the end of your final answer, your objective is to find your integral off dysfunction X Times Co sign X And once you have your entire answer for this interval then you would put your close An Indian here. You wouldn't do that. But now look, you have your sine x so you have your G right? So now you can just go along this farm line, you can just strictly follow this formula. So ex school sign X, the X And by the way, you should have DX here as well. That would equal f times G. What was my f my f was X right? If was X and my g waas sine x right, I got my g c. It was sine X minus. And now I have this integral here. Now, what was my DF dx No, my d if d X was one. So I have one here and what was my G by G was sine x dx and look, I have simplified my life now because now it's much easier to take the integral off this part than this part, because here one is a constant and you can put it in front of the integral sign. So let's continue. That means that I have X times sign X minus one into girl sign X dx But what would be integral off sine X? Well, if I have my call sign and I take the derivative of it the D X Then I won't get minus sign . So if last my minus sign and I take the integral of it, I will get my co sign, which means that you five my sign without the minus sign. If I just have my sign and I take my integral of it, it has to be minus call sign. Okay, so that means that it would be x signed X minus. And now I would have miners call sign X minus course Sign X like this. And now on Lee. Now I put my constant in the end, so my final answer is x times Sign X Now miners and minds will give you plus right plus call sign X plus c So that would be your final answer off this interval. Okay, off this into So you found your answer for this integral by integrating it by parts. You used this formula and here I showed you where this formula came from. That your first apply the product rule, and then you rearrange some terms. You get this formula, and then that's how you apply this formula for this kind of problem. And then you get your final answer. Okay. Thank you very much for attending this lecture and see you in the next one. 18. arcLength 1: Welcome back in this video, I would like to go back to our hangar. Example. If you remember, we once had an example about a hanger. I'm kind off building, and we were concerned with calculating the area off the side of the hangar. Okay, but now in this video, I would like to show you how you can calculate the length off this edge. Okay, so in meters, how many meters is this edge here? Okay, this edge here. So we modeled our hanger, remember? It was a parabola, and we're gonna do it a little bit differently now. We're not going to shift it. We're gonna have a probable like function, but we're gonna have it like this. So this is my why? And then this will be my ex. And now we're just gonna go from minus four up into four. And here this was 16. The height here was 16. So we had this Prabal here, right? And that was decide of the hangar. But now what we want, we want to calculate the length off this entire arc and again we can use Integral is to do that. And I'm gonna show you how to do it. So thinking went up droid a little bit of wider, so I'm just gonna say it's okay. Why? And then I'm just gonna have this is X. So this is for minus four. I think I like big drawings. Then it's easier to show things. So I have 16 here and let's chop it into well minus three minus two minus 10123 and four. Okay, so, look, I'm going to show you something now. If you think about it, then if you chop it into eight pieces, right? So you, right now you have eight pieces here. So if you say that, OK, this is my one point. This is another point. And then I have points here, and then I'm going to say that my minus four I'm gonna make it equal to X. I okay, So excited, cause mind for that's this point, OK? And here I would be one. And here this would be my X I plus one. This would be X I plus two. So it would be x two x three and now what I can do, I can say that. Okay, this is my this entire interval is my Delta X okay? And it's equal for all intervals for all intervals. So this is also my belt, I x here. And this could be my delta. Why I OK, so Delta wise would be different, though. So what I'm trying to achieve here, I'm trying to connect these dots. Okay, so I could connect these dots with straight lines. Okay. And if I do that, if I connect these dots with straight lines, I will have an approximate length off my arc. So let's say that this is my length one and these are straight lines here. This is my l two l three l four, l five, l six, l seven and l eight. So if I know all these lengths and I some normal all up, then I will approximately get the length off my ark. It's not gonna be the exact answer, but I will approximate it. Of course, there will be some kind of error. So here you can see that I have some kind of errors here, right? This little white space, some kind of errors. But that's okay. So let's take this first case and let's assume it. Let's zoom it and let's see what we will get. So if I have Delta X here. All right, So this is my Delta X. And then since we took this case here, I have my delta. Why I So this is my delta y I hear. Okay. And of course, that the why would be different for each case. So for this case, Del, thanks would be the same right there. Thanks would be the same. But then here built a Why would be different here it would be. Ah, dealt why I plus one. I guess so. But this would be my delta. Why I and then here I would have my straight line. So this is a straight line. And of course, this is not the most precise answer. Because in reality, I have an arc here, right? I have a narc, but then I don't care about it. I just take a straight line. So my straight line, I'm gonna call it l I So this is some kind of mark, right? This is the real thing, but I approximated with a straight line l I So how can I calculate? Ally? I can cook with leg dish. We look to Cingular and serum Delta X squared plus Delta Why I squared. Okay, so that's how I congratulate my ally. But can I rewrite this equation in a different way? How could I rewrite my delta? Why, for example? Well, what if I know the slope off my ally and remember the slope Waas Delta? Why I over Delta X. So this was my slope. Right? So what will happen if I take my slope? And this is the slope off Ally? Okay, so this is the slope the slope off l I I slope of allies Delta y I over Delta X. So what's gonna happen if I take this slope and I multiplied by Delta X? Well, then these two terms canceled and I will end up with Delta. Why I write, which means that I can rewrite Delta. Why? I like this slope times Delta X. In other words, I conveyed it down like this Delta X squared Plus And now I have don't know why I over Delta X, which is my slope and then multiplied by Delta X And remember all that has to be squared. Okay, all that has to be squared. That means that I can. Well, first of all, that means that here I would have this kind of expression Delta X squared plus. And then I would have until the why I that the X squared in Delta X squared. Now I can factor out my delta excess. So what I will have I will have Delta X Squared one plus don't know why I delta X squared. Okay, now, if I take my Delta X squared out of the square root, then it will become there. Thanks. So I can write this entire thing like this one plus built up Why I built X squared. And now this del tacks that went out of the square root. I can just write it down like this, though. Thanks. So I can write it down like this said, this is my ally. Okay. And well, I did it for for l one, but the same formula would apply for l two l three, etcetera. And now, in order to get the approximate length of the ark, I would have to sum up along these eight cases, right? So I would have to sum up l one plus l two plus etcetera plus l eight. And of course, I can just use this summation sign from I equals one up until eight. And then I have Ally here, which means that I can also write it down like this equals the summation sign from I equals one up until eight. And then I have this expression here. So I have one plus don't know why I over Delta X squared and then I have dealt X here, right? And now, as I go from Michael's one up until eight, the only thing that changes is this they don't know why. So in the first case, I would have dealt a white one. Then I would have felt the white do doctor. Why three, etcetera? So I would have eight terms here, and then I would sum up all these lengths and I will get the approximate length off the ark . Okay, but this is still the approximate length what we want. We want the exact length of the Ark. And to get that to get the exact length of the ark, where they know what we have to do, we have to chop my ex into smaller pieces. In other words, my del tax has to go smaller. In fact, my del takes has to approach zero. If it does, then the amount of lengths right will approach infinity. So in other words, the amount off. So when I say the amount off, then I used this sign here. The amount off there lies will approach infinity. So the more intervals I have here, the more seemly I chop my X domain. The more lengths I have And then if I sum them up, I will get closer and closer to the real length of the Ark. And in order to get the precise length of the Ark, I need to let Del Tax go toward zero approach zero. Then the amount off lengths will approach Infinity and I will get the total length of the Ark. And mathematically, I can write it down like this, that again, I'm gonna use limits for that. So well, first of all, if I say that this my summation sign and this my ally here and I start from Michael's one. But then I don't go up until eight. But I'm gonna go up until end, Whatever the number is all right, and N is the amount off lengths that I will have. So if any 16 then I will have 16 links, okay? And he's 100. I will have 100 links. Then I would have like a one plus l two plus up until l 100. But now, in order to kid the exact length of the ark, I have to use the limit. I think the limit and I say that as n approaches infinity then, uh, what will happen here? I will get the exact length of the Ark. So when n approaches infinity, then this summation sign. All right, well, I can write it down like this. That's clear limit as n approaches infinity off this thing I from one up until end one plus Delta Why I built the X. This is squared Delta X. Now, when n approaches infinity, then Delta X will approach zero. In fact, then del tax becomes DX and in fact, this part here Delta Why Delta X will become a differential as well. D y d x and the route. If so, when n approaches infinity, then I will have this think this summation signed, which is the discrete submission sign will become a continues summation sign. So I will have an integral And in this case, I would have then from a to B and well, in this case, you will be from minus four up until four. And then I would have one plus, do you? Why d X squared and I would have dx here. And there you go. If you applying this formula, then you will get the exact art Blank's. And in the next video, we're gonna apply this formula to this specific case seeing the next video. 19. arcLength 2: welcome back. So in this video, we will apply this formula that we direct in the last video to calculate the length off this edge here, the length off this hangers edge. And I'm gonna do it here on this small a piece of paper just so that we would see the formula here. And very simply, we would just take the function that we had. So our length equals We would take the integral. We would go from A to B. In our case, it waas from minus four up into four. And now, first of all, what was our why? As a function of X, right? What was it? So it waas? Why equals 16 minus x square. So that was our function, right? So this was our function. That was the function of this parable here. Now, in this formula, what we need, we need d y d x. So we have to take the derivative off dysfunction with respect to X. And we can do that Well, very simply, de y the X would be well, 16 would be zero, but then I would have minus two x. Okay, so that would be my derivative off this function. So this function here represents the parabola. And if I take, they're derivative of this Preval with respect tax don't get minus two X And then I put this minus two X here, so I'm gonna have square root one plus minus two x squared DX or since its squared this miners becomes plus So I will have minus 44 one Plus, in fact, I can say that this is four x squared DX. And if I calculate this integral, then I will get the exact arc length off this case off this hangar. Oh, this hangar here, However, this is a rather complicated integral and, uh, you would have to go to integral tables to see how this kind off integral can be solved. So if you go to integral tables in your textbooks, for example, then for sure you will see something like this that okay, when you have a squared plus X squared the X, then that would mean that the analytics solution for this would be X over to a squared plus X squared, plus a squared over to and land ex bless a squared plus x squared. Bless, see, So that would be the analytic solution for this carve. Integral. Now in in our case, in order to use this formula when we have to do is first of all, we have to go into this four. Which means that we have to get rid of this four here. Which means that we can factor this four out. And we could have something like this minus 44 and then square root. And then if I factor this four, then I will have 1/4 here, plus X squared DX. Now, this four, If I now go out of it, then I will have to hear okay. And now I have this. And by the way, this too can also go in front of the integral sign. So now I have this form, right. My a squared would be 1/4 and my X squared will would be X squared. So, in other words, I can now continue with my calculations like this. So I'm gonna go from here until here. Okay. So this was just my January formula here, but I'm gonna continue my calculations here on this line so my two could go in front of my integral then I would go from minus 4 to 4, and then I would have 14 plus x squared DX. I remember a squared here would be 1/4. Okay, this a scored here, that would be 1/4. Which means that my a would be 1/2. Okay, so I'm just gonna follow this formula now. So this integral equals no to I'm gonna leave it here. And now I'm just gonna have, uh, this case x over two. Well, in fact, this entire case, right? I'm just gonna say that this entire case will go here, right Will go in between these brackets. So this entire thing without mine ish constant okay, I don't include the constant, because now I'm calculating the definite integral, not indefinitely to grow. So I omit the constant, and I only take this part here, this entire thing. And then I just put it here inside the brackets here. And then I would go from minus 4 to 4. And that means that I will have something like this. So, first of all, I would have for over two so one off four plus for squared. So I'm just following this formula plus one four times to now. Four times two is because, remember, a squared is a squared is 1/4 and a squared over two is 1/4. And then my two would be here. Okay, And then Ellen and then four plus 1/4 plus four squared. And then So all that right minus. And then I have the same thing here. But I have minus for two. And then I have 1/4 plus minus four squared plus 1/8 which is this one over. Four times to 1/8 times Ln and then I have minus four plus. And then I have 1/4 plus minus four squared. Okay, so that will give me approximately continue here. That will give me approximately eight point 32 which is this part, If I haven't made any mistake. 8.32 which is this minus minus 8.496 Mine is 8.496 is what this it gives me right. This entire thing that gives me mine is 8.496 miners and miners will give me plus. So in total, I will have 16.816 meters But remember, we also had to hear. So after multiplied by two so two times. 16 point, let's say eight. That should give me 33 points. Six meters. So you're arc length is 33.6 meters. All right, so that would be our answer. And the next video will be the last video in which we're gonna treat one variable, integral. And after the next video, we're gonna going to multi variable intervals or multi dimensional into girls. But there is one more thing that I want to show you using one dimensionally Google's so see in the next year. 20. arcLength in space parametricEquations: Welcome back. The final thing that I want to show you before we going to multi dimensional intervals is how to get the ark lengthy in space. So if you remember, when we learned about Parametric equations, we had a spiral and the spiral we could represent like this. So this was our vector are as a function of t as a function of time. And let's say we have a spiral here. Call sign T in the union victories, I plus Sign t The unit vector was J and plus t. Okay, so whether it would give us it would give us a spiral. So if I have my Z axis here and then I have my y axis here and I have my exact is here now this formula, what it means is that well, first of all, let's assume that our time goes from zero seconds up until two pi seconds, so we don't consider more. We really consider from zero seconds up until two pi seconds. It means that this formula which gives us the spiral, will start from here, right Then after 2.8 seconds, the Z axis will be two pi meters. All right, All these are meters here meters, meters and meters. However, this is the expert and this is the white part. So since we start from here's when time equals zero, then this term becomes one. Okay, and then later on, why becomes one? And so as we start moving, we we start moving in a spiral, right? And we end up here. So this is no when you're Z is two pi So you see, it's a spiral. So we start moving in a circle. If we consider the X y plane, then if you look at cost 90 and sigh Inti than on the X Y plane we move like like along a circle. However, we also move up along the Z axis. So in the end, we have spiral And now we want to know the length off this path and the way we do that is similar to what we saw in the previous video. With the arc length, I would chop this path into small pieces into lengths and we can say that Okay, our ally equals the following Delta X as a function of time, right? And this would be now they'll Thanks. I plus Delta. Why, as a function of time plus Delta Z as a function of time, All right. And I have to specify that. Okay, this is Delta X. I Delta want I and Delta Z I because they are gonna be different for each length. So I take some kind off random l I hear right. And for each l i that I take on this path, you will have a different Delta X different Delta y and different Delta Z. The only thing that will be the same will be dealt a t, which will have some kind of constant interval. But now how can we rewrite all these three terms? We can rewrite them like this. We can rewrite them like this Delta X over Delta t Okay. And then I will have ah d I hear so at a certain time. So this is my slope, or this is how my exchanges with respect to time. And this is then ah, at what time it happens. So as the time goes from 0 to 2 pi, then I need to be able to specify at which time it happens. So based on that, I will know which length I'm dealing with. And based on that, I will know which Delta, X, Delta Y and the Z I'm dealing with. But okay, I can also Oh, I also need to multiply this because this is my slope. Okay, this is my slope. This is how my exchange with respect time. Now I need to multiply by Delta t. And if I do that, what will I get? I will get Mindel tax because thes to cancel out so I can write this built x i by simply taking the slope off. Still the x i with respect to delta t that time I And then I multiplied by delta t and the same thing here built a why I at at a certain time And then this is my slope and I multiply Bite off the tee plus Delta Z By that time I that the tea, remember, this is my slope How my Z changed with respected time. And then I multiplied by Delta T and of course, Aled. That has to be squared. Forget that. All that has to be squared because it's like a fag Aerin theory. Okay, so this is very important. All that has to be squared now, just like in the previous video, I could simply take those Delta tease out. Okay? I can factor them out and get them out of the square root. So when I will get then I will get something like this. I'm just going to write it down Safe, time built X. There's a T squared plus Delta. Why Delta t squared Bless. Dealt that Z Delta t squared. And then I'm gonna have built that t here. So what I did, I factored them out. And then I had built a t squared factor out. But then when I took it out of the square root sign than my that that the squared became Delta T. And now, in order to calculate the entire length off this spiral, I need to take the discreet summation from I equals one up until end. L I or the sum from I equals one up until an and this entire thing. Okay, this entire thing. And now if I use cankles again and I say that limit as n approaches infinity, then what will happen then? These become differentials, right? So if I take my discreet summation Eichel's one and l i Then I will have an integral from A to B. And then I will have a square root here where I would have dx DT squared Bless. Do you? Why the t squared plus DZ DT squared DT. And if I applied this formula now for this specific problem where my time goes from 0 to 2 pi, I will get the length off this park so I can do it very quickly because actually, it's very easy. So if I take the derivative off course sign with the respected time, then well, I'm just gonna use the upper apostrophe. If I take the call sign with respect to time, then I will get the minus sign t right. And if I take de sign the T, I'll get a call sign t and this is the same just too different notations I d over DT or or opera passer feet and then d t with respect to DT would be one. So what you will get you will get the integral from zero to two pi and then here you would have minus sign t squared, plus call sign D squared plus one squared now minor scientist squared. It will give you a positive number because minus something squared will give you positive. But this is a true econometric identity and it will give you one. So and remember So you also had DT here. So this regatta magic identity will give you one. So you will end up with this. You will have an integral off from 0 to 2 pi and then you would have squared off to here. DT So you will have squared off to time from 0 to 2 pi. So in the end, we just got put it here. In the end, you will have squared off two times two pi meters said that would be your length. Your arc length here. That would be your final answer. Okay, so by now you have already hopefully being convinced how powerful the concept of integral Czar. You can calculate many useful things with it. And starting from the next video, we're gonna look into double triple and even quadruple integral. I'm going to give you an intuition for that. So see you in the next video 21. multidimensional Integrals intuition 1: Welcome back in this video. I want to give you a brief intuition on multi dimensional Integral. You will see that the general logic of air is the same. When you compare it to one dimensional Integral, you simply need to generalize. You're thinking about it. So let's get started. Let's first talk a bit about visualization. Let's say if we want to visualize a mathematical relationship in other words, a function if we want to visualize it geometrically, we only have three dimensions at our disposal, right? Only three dimensions. So we have. We have height and we have with and we have lengths. All right, we have only three dimensions at our disposal. So we can only maximum visualize functions that I have two inputs and one output. After that, it is important to rather focus on the meaning off what you're multi dimensional function represents rather than trying to visualize it. So, for example, if I have a one the function where I have one input all right, and then I have one output so extreme input. Why is my output then? I have some kind of function here. Then I can use one dimension for my input and then another. Did I mention for my output? And then if I have this kind of function two inputs the function So this X y x y here and then my output easy then I can use one dimension for one of my inputs, another dimension for another input. And then I can use my third dimension to represent my output. But then, if I have a function where I have three inputs, let's say x, why and Z then my function here has X. Why and Z then my output is let's say Kay, then no longer I can completely visual s my entire function. So I can only visualize my inputs now because all three dimensions we'll go to represent my my domain. So here my domain is a line here. My domain is a plane, right? So X y said this is my domain. So it would be a plane and here, in fact, it would be a cube. All right, but here I cannot visualize my out anymore because I'm out of dimensions. And of course, in this case, when I have four inputs and then I have my fifth output, that here I can't even visualize my domain anymore. So starting from here, up until here, this function and this function, we can visualize it. And here we visualized it with with some kind of line, and here we typically visualized it with some kind off surface. All right, But here, starting from here from this line, we cannot visualize are functions geometrically anymore. And so we have to focus on the meaning off our functions. So let's not visualize our functions. But let's just think about the context of what my function represents. What is my output? What are my inputs? And then just think in terms off. Okay, My output depends on my three imports or on my four inputs. If you remember, in the past, we had a small a simple example on mood that your mood can depend on many things like weather outside or whether you have had coffee or etcetera, etcetera. Okay. And now, if you remember from the previous videos, then, uh, when we take this case and this first case, then if this is my function here and I have some kind of line, then we represent it are integral is like this, calculating into gross in this case meant calculating the area under the curve. But what did it exactly mean? So I want to explain this thing now, using another type off point of view. And that other type of point of view is also suitable for this case and this case in this case and for a while, dimensions. So let me give you another way how you can look at it. So I'm gonna draw a graph here, a function in my output, his velocity or, as you already know, velocities the same, like the change of displacement with respect to time. Right. And then on this axis here, we're gonna have time. And on the other axis, we're gonna have our actual displacement. And here we're gonna have time. And let's say that the your graph here is like this. Okay? And here this point here is T star. And here this point is T star, divided by two. Okay. And this is our velocity graph. And you know that that the velocity is the change of displacement with respect to time. All right, so I have some kind of graph here, and if I take the derivative of it or this operator D d t. Then I get this graph. No, I want you to think about the word change, right? Because this graph here, it represents the change of displacement with respect to time. So what does it mean? Well, let's see, here in this part, you can see that the changes positive, and here it's negative. Right. So we're gonna assume that a time equals zero. Displacement is zero. Okay, The initial condition is that when time equals zero, the displacement is zero. And now, as we start going, remember the domains here, the same as we start going from time equals zero towards Time Star. If we know, Go and look at this value here. Right. Well, first of all, we look at this value at Tom equals zero. Then we look at this value than this value. But what are these values? Well, there changes right there. Changes of displacement with respect to time. And now, if we take the integral of dysfunction, then essentially what we do we some all those changes up, right? We sum them up, we some them up on top of each other. So we take this change, then we summit with this change. Then we summit with this change this change, etcetera, etcetera. So as we go along this timeline, right, we see how our velocity evolves as we go along this timeline, right? So as the time goes, we see how velocity involves. And we want to know how displacement evolves as a function of time. In order to do that, if we know that our starting point is displacement equals zero. So in order to see how our displacement evolves over time, what we need to do, we need to take the changes. And we need to some those changes up one by one. And then we can reconstruct our graph for displacement. So we see that at the beginning, the changes positive, right? Which means that my displacement has to go up because the change is positive. Therefore, my displacement has to go up to the positive region. Right? But now here, for example, here the displacement is too positive. But it's less okay here. It's less. Therefore I still go up, but my slope goes down right now. If I'm here is still positive. But then the value here smaller, my changes smaller. Therefore, my slope is smaller, it's still positive, but my changes smaller. So in this point here, there is no change. Therefore, here the slope is horizontal. Here I am time start over too. And now here I have a negative slope, but it's small. So my slope is small and here again my change is negative but becomes more negative. So my slow becomes sharper and etcetera and etcetera and its center, and you can see what we're doing here. We're taking thes changes off displacement with the respect of time, and we are adding them up on top of each other. And if we do that, starting from this point, then we can reconstruct our displacement graph. And then here we're back that Time Star. And therefore, since this area is equal to this negative area, then we're gonna end up a displacement equals zero again. And so this addition off changes on top of each other is the same thing, like calculating areas under the line in this part and above the line in this part. So why have I brought up Aled these different explanations, like area and then the summation off changes. That's because calculating the area it's only valued for this case. When we go and start working with this case, then there is no area anymore here. It's like calculating volume under the surface and in this case there is no area either. And there's no volume either. Here you can visualize it. And in this case neither, however, this kind of thinking where you add up changes and then reconstruct your regional graph. This kind of thinking can also be applied here and in this case And in this case, the only difference is that in this case, what you have, you have Ah, one dimensional output and a two dimensional input. So x here and why Here you have Z here. So now this is your domain, right? Your domain no longer is a line, but your domain is a plane. It's an X Y plane. And here what? You do well here? If you remember, we divided our input into small intervals, built a T and then as it approached zero, then it became DT. Well, here our interval is not a lying anymore. Here. It's an area. It's a small differential area. So we take this area here. Delta A Okay, and that consists off built X and delta y So Delta a equals delta x times delta y And as they both approaches zero your delta a approaches zero and then becomes d A equals d x times d y all right. So you can think of them as infinite decimal e small squares. And then, of course, you have some kind off some kind of surface here, no longer a line but a surface. And then here, as you went along, your input and you send up changes. And then you got your reconstructed original craft than here. You some those changes up by going first of all along the squares, then on these squares. And then, well, let me do it with red along the squares like this. So you do the same thing like here something those changes up like you go through t like this and you also go through t like this. But then, since you have two dimensions now you have to do this operation. But many times. So let's see if instead off x I have t here right then Then it would be like this. But then I have to go through my time many times because each time I scan through my time, I keep my why Constant. And then, as I have scanned through my time, I increase my why and then I scan through my entire domain like this. And as I scan through my entire domain like this in an organized manner, I always sum up all these changes. And then I have some kind of other output here. Well, it's not easy anymore. Let's say it's B K and my domain is still the same. Why an X? But then I will have some kind of other surface here that is a result off summation off these changes. 22. multidimensional Integrals intuition 2: So when you have this kind of situation where you have two inputs ex and why and then you have a function f x and why and then you have Z as your output and then this is your output Z why and X again, this is our domain and then we have some kind off surface. So this is our surface here. So the way we calculate, let's say our double integral are like this. We take our domain, all right? And then our differential is D a. Right. Let's say right now we have small square here says Delta X and Delta y, and that gives us Delta A. Now, if we take this Ah, small square right and we multiply it by the height that reaches up until the surface. So this is our height here. So what? Well, what will we get? We get a cue, Boyd, we get a cue. Boy, that looks like this. So we get something like this. So what is a que boy two but is like cube. But all the sides are not equal. So let's say this is my cube. But then this could be my cue. Boyd and This is what I will get over here and then as my delta A approaches zero then, Mike, you Boyd becomes thinner and thinner. Okay, You see, it becomes thinner and thinner. On the other hand, the amount of these Q boys will increase, right? In other words, I could also think of it as my So if if I take my delta and I multiplied by age or by my height, I will get my delta volume. And as my daughter a approaches zero, my Delta William approaches zero, and it becomes my differential value. And this differential volume is an infinite decimal. Small volume, it becomes very thin, and I will have a lot of cute boys here. I will have a Cuba here, here, here, everywhere that will cover the entire domain and the entire volume under the surface. So the Q boys become very thin, but the amount of them will approach infinity. So the amount off que Boyd's amount of Q boys approaches infinity. But the volume off the Q. Boyd's approaches zero, and then the multiplication of them or the internal will ultimately give us the exact volume under the surface. All right, so mathematically it would look something like this. So I have my double integral. And then I have my function here, which is Z as a function of X and y And then I have my DX and d y. And then let's say that my ex here is to and my why here is also to right in this case, this integral this inner integral would be four DX, so it would go from 0 to 2. And let's say that our why is three just to make it clear, are y street are excess to and why is three So this part here is for DX, So this integral is responsible for scanning from zero to X right? Keeping why constant and then the altar integral is for D y. So it goes from 0 to 3 c. Why is three who will go from 0 to 3 and therefore this differential is for this integral. So you can see that as I scan through my X this inner integral scans through X I keep my wife constant, but then I move along. Why a little bit and I have since I have moved along, Why a little bit Then again, I scan through my ex entirely. Then again and move along. Why a little bit? And then I scan through my ex entirely. And if I do that, then I will get the volume under the surface because my differential. So this my differential is infinite decimal or so my delta area. So this is my the area, right? And that's because my delta area approached zero. So let's make a simple example, just ill make sure that we understand it. So I'm gonna create a small function where the Output Z then my input, his ah is an X y plane. And this is my domain. And I'm going to say that my ex his to and the Koreans is 20 So, zero why? This one is for why in my why is three therefore, my coordinate is zero and three. So my wise three and my x zero. Okay, so this is my domain. And then I'm gonna assume that I simply have a plane and this plane has a value of two. So to so therefore, I can do it like this. And there you go. I have a plane here and now I want the volume under this plane. Okay, so what is my function here? My Z as a function of X and y equals two, right? Because it's a plane, It's constant plane. So I'm gonna take my double into grow, and my outer integral is reserved for Why? So my outer integral goes from 0 to 3 from zero till three. So this is from my why, and it goes here. Do you? Why? And now inside I will have my Xing to girl. So d x, That goes from 0 to 2 zero and two and then the function itself is this which is to so I can put two here. Okay, so now we're gonna take this first part, and then So you take this, you can put two in front of the integral sign you have DX. And were they know that that would give me two times two equals four. Okay. And now this four goes back here. So what happened here? I scanned through my ex right from here. It'll to buy while keeping my Why, Constant. Okay, so I have done that. I have calculated the area off this plane. Oh, this small plane here off this plane on the xz plane while keeping y constant right. So right now my wife is zero. So have calculated this area. And now this outer Integral will take this plane here and it will scan ID through the entire why region. And then I will get the volume under the surface. So let's do that. So I'm gonna take this result. So this is for right. This entire area is for this red area and then I'm gonna have zero and three four and d y, which is four. Why? From 0 to 3 and it will give you four times three equals 12 cube meters. And there you go. The voice you underneath the surface is 12 cubic meters. Of course, if you're function is more complicated, let's say if my z x why is something like two X plus three? Why, right that you would take your inner integral and what you would do you would go from 0 to 2 and then two X plus three. Why? And then you would put d x here and then In this case, you first of all, do like this zero to two x d x plus zero to three. Why? And the X Now here we can keep our Why Constant. Just like three seconds. Go out of the integral sign. Right? So you can write it down like this. You will have to over to x squared from 0 to 2 plus. And now, in this case, you would have three. Why? And then you would have X and again you would go from zero to to and then you would have. In this case, Well, this would be one here. So you will have, uh Well, this will give you 42 squared minus zero squared will be four plus three. Why? And then you will have times two. So you your function will be four plus six. Why? And notice now this inner integral will give you not a number but a function of why only So you have gotten rid of your ex, right? You've gotten rid of your ex, but now you only have. Why here, which is good because now this red area here it's not rectangle, but it's some kind off area as a function of why. And now you can just scan this expression through your y axis. So I'm gonna take this function, and I'm gonna put here and I'm gonna write it down here. So my outer function now is from 0 to 3. And I was four plus six. Why de y So I can write it down like this? Four 03 d Y plus six. Why? My wine needs to stay inside the integral, do you? Why? From the little three. And you really know how to calculate that. It will be four times three here, plus six over to. And then this becomes Why squared So it will be three squared. It will be nine here. Okay? And so your final answer will be 12 plus three times nine, which is 27 and 12 plus 27 is 39 cube meters. So that's how you do it. But again, it just meant that. Okay, let's assume that this function is this, okay? It gives you this kind of shape. In fact, it it doesn't, but let's just assume that it does. So when you first calculated this inter function and you went through D X, you kept your why constant. And then you had some kind off area on this plane. Right? But then you have a wire variable here. And this wild variable means that as you go through your why, then this area will change. So in other words, when you are now here when you're wise here, then you have some kind off intermediate area, right? Let me just enjoy it again. When you have your output here. You know why in your ex and, uh, let's say you have some kind of surface. And now when you're why is somewhere here, then you see now this intermediate plane that you have, right, it's different than here. So you see now, because you have this why, in this expression as you go along, why than this area changes? Because the surface is not just a plain, it's changing surface and therefore this area as you move along, why this area changes. Okay, so here you would have some kind of other area. So as you go through why your area changes and that's why you have this. Why here and wants to use your outer integral with the differential the white. Then you sum up aled these different areas and you get your volume under the surface. So that's the logic of it. And now in the next video, I'm gonna talk about triple and quadruple into gross. 23. multidimensional Integrals intuition 3: welcome back. Before we go into higher dimensions, I very quickly want to go through one more example with double integral. You see, sometimes you can have a situation where this is your output and then these air your inputs . Why an X And then you want to calculate the integral over the domain. But then sometimes you don't have rectangle boundaries. Before we had boundary like this. Right? So that was our boundary. But let's say that now I won my boundary to be like this. That my ex goes from zero till two. But then my why is a function of X Okay, so my y equals X. If that's the case, then I will have some kind of line here. So X as my ex goes in this direction then my why for for my why I would have some kind of line here. And if I do that as I go from X equals zero up until X equals two and then my wide depends on X, then that means that I will have a line here and my boundary now is this triangle OK, so here I will still have some kind of surface of course, I will have still have some kind of surface. On the other hand, however, when I calculate my integral I don't go through this entire rectangular domain. I only consider this domain here. Okay? And that means that when I calculate the volume under the surface, I only consider this part. So my surfaces here. But then, because of this domain, let's say if I project this entire triangle up, then it will be reflected here on the surface. Right? So I will want to calculate the volume below the surface. But Onley in this area where I have this triangle OK, so I will omit everything outside of this triangle. So I will not calculate the volume here in this part I will only calculate the volume as long as I'm in my triangle. So how would I do that? Well, we're gonna do it like this. So first of all, what is our function? Our function in this example is e equals X plus why someone off do some kind of easy function. And of course, this surface does not represent this function. But that's OK, in fact, now I will change the order now My interview to girl will be reserved for why, And that's because now my wine depends and X So let me show it to you. So my why equals zero here in my y equals X here. Okay, so I will have my function X plus Why? And then I'm gonna put de y here. So now this is my inner integral. Okay. And now this entire thing I can put in the parenthesis and then I can have my X equals zero and X equals to here. And then I will have DX here. So what will happen now is that my inner integral will make my wife into X. And then I can use my alter integral to calculate the volume underneath the surface. But on Lee, within this domain. Okay, so again, I project this triangle up. And then on the surface, I only considered this part of the surface that lies directly above my triangle. I guess this is the best way to formulate that from this surface that I have. I'm gonna choose the surface that lies directly above this dry angle. Okay, so let's calculate my entering to girl now, so it will be zero and why and then I have X. Do you? Why plus integral zero? Why? Or my mistake? Y equals X in here as well. Y equals X. Why and d y so in this case since I have X here. But I have the why here than X is like a constant. So it goes out. So I have X zero now. Why? Cause X I can just put X here and then I have d Y plus zero x. Why do you Why? Okay, so I don't have to put Wyche was X here everywhere I can just put x and X because I know that it's y equals X. And since I have d y here, I could put my ex out of it. So what will I get? Well, here, I'll get X Times X because, uh, well, if you think about it, then ah, when I have Ah, do you? Why? From zero to X Then I will have Why, in the parenthesis from zero to X so it will be X minus zero equals X and I already have one x here. Next times X. So it's it will be X squared. Bless. Now here I will have 1/2. Why squared in the parenthesis from zero till X So I will have an expression like this. X squared plus 1/2 x squared Minor. Zero squared. Right. So I'm gonna continue here. OK, so this one goes here X squared, plus 1/2 x squared. Excuse me. Three half x squared. Okay. And there you go. I have gotten rid off my why? And now I only have a function, but thanks. And now I'm gonna take my outer integral, which is this one. Now my ex goes from 0 to 2 and then I have 3/2 x squared d x. So it's 1.5 and then ah, zero to x squared. The X equals actually is very for a living in this form, because I have three over two. And now since I have to here I will have 13 over here, and I will have excused from zero till two. Now the three skansen out and, uh, I will be left with 1/2 and then too cute, minus zero cubed and look. I will have 8/2 equals four four cube meters. And that's the volume under the surface that is directly above this rectangle. Okay, so the surface here, it doesn't have to be a plane, and it's not a plane. It's some kind of curvy surface. But this part of the surface lies directly above the triangle. And then I simply calculate the volume, using the double integral underneath this surface. Okay, but now let's talk about tripling to girls. 24. tripple integrals intuition: my tripling to girls would have a situation like this where I have my X, y and Z as my inputs. I have some kind of function that has three variables and I will get my output. Let's take a okay. Now you can't any more visualize this kind off function or if you have triple integral now tripling to girls are now for this case. Okay? And you can't visualize that anymore like you could with doubling to girls. There are no areas there, no volumes. So now you just need to think in terms off. Okay, If I had my double integral then as I scan through my ex while keeping my white constant Then I summed up the changes along X while keeping my wife constant. Then I changed my y a little bit. And then again I went through X while keeping y constant and I summed up old changes. And like that, I could reconstruct my original function. And here I think the best way to show it is using a table and using at discrete example. So let's say that with my ex can have two valleys, Onley one and two Onley two values my Why can also have only two values one and two and my Z can also have only two valid one and two. Okay. And now I'm gonna create a table. I'm gonna create a table where I have X and Y and Z. Okay, this is my X and Y and Z, and now here. But we hear the differential is d X de y and easy. So here the differential is not an area anymore. Like it was here. Here, the differential is an infinitesimally small cube. Okay? And that will give you de voyeur. That's your differential here. And so what you do, you have a lot of infant decimal small cubes and then your triple integral mathematically is formulated like this. So okay, as a function of X y and Z times dx do you? Why and d z So that's your mathematical formulation. So your first integral would be reserved for your DX, your second integral for D y and your third the most outer integral for DZ. Okay, but the logic is still the same. You scan through X while keeping y and z constant. Then your change of why and you scan through X while keeping you a y and z constant. And then you change. Why again while keeping your z constant. And then you scan through X again and then you change your C. And I think the best way to show it to is using the stable. Where you have here you have K, X, Y and Z, right? And then your accent, Why? And the all of them had only two values, right? One and two. And this is a disc readings apple. Remember that? So what I'm gonna do, I'm gonna do like this. So let's say that I have this kind off situation. My ex can be one and to and here is well, one and two one and two and one and two and ah, in this case, my wise are one. And in this case, there to in this case, there one. And in this case, they're both too. And in this case, Z is one. And here is also 111 And here's these are 222 and two. Now, if you look at it, you have eight combinations here consisting of three variables and all these a combinations are different, right? here You have 111 Here you have 211 1212 to 1 112 212122222 So what is me? It means that you see, I keep my y and Z Constant and then I scan through my X one and two. And now I change my why and I have to. But my disease still one, right. So I keep my y's and Z's constant. But now my wise to my Z still want. And then I scan through my ex again one and two And then again keep my wise constant But now I'm gonna go back to Why cause one? But now I'm gonna change my Z Now my Z is to And now, as I keep my Z constant but at two. And my wife Constant at one. I scan through Xa get one and two. And now I changed again my Y 22 But now my z's also to and I keep both of them constant and I scan through my ex one and two. And for each combination I'm gonna write my answer here. So my function I didn't write down. Yeah, my function is K X, Y and Z equals X plus. Why? Plus z. So this is my function, citizen. My output. These are my three inputs. Okay. And now for each combination, I'm gonna some them up. So what will be my function here? One plus one plus one is three. Here. It would be four, four and five, four and five, five and six. Okay. You see, for each possible combination off X and Y and Z, I'm gonna have some kind off answer for my out, and then what I'm gonna do, I'm gonna some of them all up, and that will give me 36. 25. quadruple integrals intuition: with quadruple integral. You have a situation like this where you have four inputs x, y, Z and K and then you have some kind of function and then you have f is a function of X. Why Z and K and then your output is That's called H No mathematically, it would look something like this. So you have four integral right? And then you have your h as a function of X. Why Z in K and then your differential here would be the x de y DZ and D K. And of course, you're entering tickle would be for DX, then you're second integral would be for D y. Your third would be for DZ, and your fourth would be for DK. So first you calculate your inner into girls, then your second outer be 1/3 outer and your fourth alter. And even if you have five internals or six into girls, the logic is still the same. You can expand it the way you want, depending on how many inputs you have in your function. But if we use the table again, well, then the table will be very similar, right? You have, let's say, you have X here and you have. Why here And then you have Z here and then you have k here. And then you have ah h as a function of X, y Z and K. Okay, And now again, since your ex is your inner to grow, it varies the most and your outer integral is K and therefore it's most stable. It varies the least, but we can go through it very quickly. So now the table would be two times bigger than the previous table and then five and 67 and eight. Okay, Again, since being entering to goes DX, it will vary the most. 1212 12 12 So here kay as well has only two values one and two remembers. It's a discreet example. So one and 21 and two, one and two and one and two. So here wise are one and 12 and two one and 12 and two 11 to to 11 to to. Here's these are one and one and one and one, 222211 11 to to two and two in case our most stable one and one and one and one and one and one and one in one, 2222 to to and to and to you See, now we have a lthough the combinations here, gay off X, Y Z and K All right now, one thing that I want to specify is that well, first of all, in this example, we don't have a differential. It's discreet example here was simply have dealt out X times Delta. Why times, Delta Z times, Delta K. And since each of them is one, you will have one times one times one times one all that equals one. And now here you would have four and five, five and six. And by the way, that the function here would be H equals X plus Why, plus Z plus k. All right, this is where the numbers come from. Then again, five and six, six and seven. I just put these inputs into this function and I get my age. So I will have five and six six and seven, six and seven, and then seven and eight. And now, if I some them all up, then I will get 96 off whatever units I have. Okay. All right. And now I want to give you one example about how you can calculate the mass off a cube were the density varies. So we're not gonna going to math here, But I'm gonna show you conceptually what you need to do to do that and how you can apply triple into gross for that. 26. tripple integrals application: welcome back. So in this video, I want to show you how you can use triple Integral to calculate the mass off of soul it in which the density off the soul it is not constant, but in fact, it changes with respect to position. So what do I mean by that? So our soul ID is a cube. All right, So let's say this is my Z, and this is my why. And this is my ex. Okay? And now I have a cube here. So my ex, he's one my wise one as well. And my Z is one as well. So you find Drama Cube here. Then I will have something like this. So this is my cube, okay? And this is my soul. It so I have material inside. There is no air inside. There is material inside. Okay, that's any steel or Iran or whatever. And then I want to calculate the mass off this cube. Now, we already know how to calculate mass if you know the density. The general formula for mass is the density of the solid times. Devore youm off the cube. Right. So you have to volume of the cube and if you know the density, then you can calculate the mass and you know the units. Like for the density, the units would be kilogram per meter cubed, and the unit of the volume is a meter cube. So it's logical that if you multiply density by volume, then in terms of units you will have kilogram over meter cubed times, meter cubed. Then you cancel out meat cute and you will be left with kilogram and kilogram is the unit for mass. Right? So mass is measured in kill graphs. Okay, but now it's easy to do. If you're densities constant. Let's say if you're dense, d equals five kilograms. Uh, meter Cute, right? It's easy to do that. However, it's a lot harder to do it if you're density is not constant. And in this example, we're going to say that our density equals five Bless X plus why and plus Z now what does it mean? It means that when x and y and Z when they're old zero When I'm here, my density is five kilogram over meter cubed. Okay, But then, as I go in whatever other direction, either along X axis or along the Y axis or along the Z axis or a combination of them. My density always changes. For example, if I'm here in this point, then my X equals one Y equals one and Z equals one. So my danced at this point, right? At this point, my density would be five plus one. Bless one bless. One would equal eight kilograms per cubic meter. And if I'm somewhat here, then well, my Why is one in my Z? It's one but my ex zero. So here I would have something like five plus zero plus one plus one equals seven kilogram or meter cubed Okay, seven kilograms per meter cubed in this point. And you can explain it like this that maybe you have some kind of soul it and then some kind off material scientists They, uh, put different kinds of other Let's say materials inside this soul, it maybe they put more thence material pieces here. Let's see, Maybe they If this is food just for the sake of example, this is would then maybe as we go towards this point, they put more and more nails inside my wood. So yeah, maybe this is a good example, I have would. And then I have no nails here. But then, as I moved towards this tip, then I put more and more nails inside, so my average density increases. Okay, but if this is my function, this is my density function. Then I still need to calculate the total mass off my cube. Right. And we can do that using triple integral. So the way we would formulate that would be like this. That Okay, if my mass equals road times for you, then I can write it down. In terms of Integral is like this, I have my three into girls. And then I have my density as a function of X. Why? And z right? And then I have my differential, which is dx de y and easy. Okay. And then this is my entering to girl. And this is my seat, my first altering to grow. And this is my second outer interval. So essentially dx de y and easy what it means it gives me a differential warrior. Right? So that means that I take a small, tiny volume here. All right, so this is my small, tiny devalue. I infinite decimal e small devalue and what I do, I take this volume in a multiply it by this density, and then I get a small mass here. So if I take my d volume and I multiplied by my density, then I get my D mass. So Omar Infante decimal small mass at this point. And then at some point, I'm gonna have another cube, and I'm gonna do the same thing. I'm gonna take my for you. My differential volume multiplied by the density at this specific point, and I will get my mass at this specific point. And then if the volume off my little cube approaches zero, then the amount of my cubes, the amount off my de volumes, right, approach infinity. All right, so this is the amount of my differential volumes approaches infinity. And so as I go through the entire soul, it I will perform this operation using the integral us some old these answers up and then get my toll mass. So essentially my integral will sum up all my differential masses as I go through the entire solid scan through all the differentials. And then I will have my mass off my soul. It So let's try to do it mathematically as well. Let's see what we will get. So we always go from 0 to 1 in all dimensions. Okay, so let's take our inner integral. First of all, our inner integral would be this. So I have my ex that goes from 0 to 1. So this is X okay? And then I have my five plus x plus. Why? And plus see and d x. So what does it give me? Well, it will give me this five times one Okay, Because my DX goes here and I take the integral of it. Plus now I have my integral off x dx. Let me said, I have 1/2 one squared because, well, I can do it here into grow x dx. It would be 1/2 x squared, right? And then here I go from 0 to 1, which will give me a one square plus why? And plus Z. Okay, So Okay, maybe it's good if I go through it step by step so that everybody would understand it. Create some space for myself. So what I did here I had five so integral from 0 to 14 DX bless into no 0 to 1 x DX plus integral 0 to 1. Why DX plus integral 0 to 1 and then z d x. Okay, so now I had five. This gives me five. This gives me 1/2 times one squared, but it will be one so so 1/2 times, one will be 1/2. So here my wife's constant and I have DX here. My white can go out of the integral. So I will have plus why Times X from 0 to 1, which will give me one again. And the same thing here is Z equals X from 0 to 1. Okay, So in fact, what I will get, I will get 5.5. So this is this Plus Why? Because this thing will give me one game one minus zero. Here's what One minor zero. So plus Z. So look, I have gotten rid of my ex, OK, Minor Integral Got rid of my ex and now I'm gonna take my second outer Integral, which deals with why so again, from 0 to 1. 5.5 plus why plus z de y. Okay, so that will give me five point five. Okay, I'm gonna do it, just like I did hear. OK, it's the same technique. Plus, And this could be also your exercise so you can try it yourself and see if you get the same answer. Like I have gotten it 1/2. Okay? Plus. And then I will have the sea and therefore I will have six plus z. So my first ultra interval got rid of my d My second outrunning to grow got rid of my wife . And now I'm gonna take my final integral, which is the most outer integral. And I'm gonna have from 0 to 1 six plus z and it will be DZ and I have six plus 1/2 and then Z from 0 to 1. This will be one. So I will have six plus 1/2 so 6.5. And since it's a mass because I multiply density with my warrior, I will have 6.5 kilograms. And there you go my total mass for my cubic solid. It's 6.5 kilograms. Okay, so this is it for the integral part. And now in the next video, I will give you a brief introduction and intuitive introduction for differential equations because in the future you will probably have a course on differential equations. And then from my experience, when I had it, I learned how to so differential equations. But I didn't have. I feel for it. I didn't really understand what differential equations represented or why I needed them and what it exactly meant when they said that okay, solved the differential equation and didn't really understand what it really meant. I just knew how to automatically calculate it. Now I do understand it, and I want to share it with you. So let's, um, meet in the next video and I will show you the meaning Off differential equations, what they represent and, uh, also what it means when someone tells you so. A differential equation to see in the next video. 27. differential equations intuition 1: welcome back. So in this video, I'm going to give you an introduction to differential equations so that when you start taking your differential equations course, or maybe you have already taken it so that it would make more sense why you do things which do there. So, like I said in the end of the previous video, I'm not gonna go into deep math here, but I want to give you an understanding off differential equations What they're used for, What it means when someone tells you that soft a differential equation so that indeed when you have to work with them, you would I understand why you're doing the things that you have to do in order to solve them. So I'm gonna use a typical example where you have a system. Okay, I'm gonna show you a system. So So this is a spring, okay? And this is called a dapper. So this is a damper here. And this the sprink Okay. And now I have some kind of mass here. I have some kind of block. Okay, Just some kind of block. And this block has some kind of mess. And then let's say that the it's some kind of solid road. Okay? It's a solid road, so I can pull on the mass and I can push the mass. Gagan pulled the mass and I can push the mass. So there is some kind of force right? Applied on this block. And for now, I'm just going to say that this forces called Force applied and this force applied is a function of time and in fact, this force Later on, I will show you that this force is a sensible function. So that means that as my time goes, then sometimes I pull on the mass, I pull in the book and then I push it and then again I pull it and I push it. And then I have my spring here and I have my damper here. So this spring it also gives you some kind of force, and we're gonna call it F s. So this is a spring force. And for physics, we know that the Spring Forces minus Carrie X and then X is a function of tea as well. In fact, we can establish that this is our ex and it's positive. So here, if we go in this direction. Our excess positive. Okay. And now this K is a constant that characterizes our spring. And why do we have this minor sign here? Well, this minus sign is here is because if our X goes in the positive direction, then what happens when you pull on the string? Then if we go in this direction, then we pull on the spring. And what happens when you pull on the spring? The spring wants toe. Go back. Right. So you pull in the spring. But the spring wants to go back. So there is a resistive force. So if my ex goes in this direction in other words, if my block goes in this direction, then my strength force will pull the block in the other direction. Okay, so my spring force will pull the block in this direction. And if my ex goes in this direction, then I apply my force on the spring. But now is the compressive force. So if I tried to compress a spring, then what happens? The spring will try to push you back, right? I compress the spring, but spring pushes me back. Okay, so therefore, if my ex goes in this direction. My spring force will go in this direction and therefore you need the minus sign here because the minus sign says that if my excess posit if But if I multiply my positive X by negative then my spring force will be negative. And if my ex is negative, right then I will have negative X. But the negative times negative will be positive. So my spring force will be positive. Okay, so that's why you need the minus sign here and the same thing with the damping force. You also have a damping force here. The damping force is like this D x DT or, in other words, the velocity off the block and time. See and see the damping constant cape also a characteristic off a damper. And here we also have to put minus okay. We also have to put minus so again f d equals minus dx DT times. See, it means that the dam pick force depends on the velocity and on this constant. And since it's minus, if the block moves in this direction than my damping force will resisted and it will make the motion in this direction slower. So I have to have some kind of force in this direction. And if my ex goes in this direction, then my damping force will go in the opposite direction. Okay, so this is my system here and now I'm gonna use the Newton's second law. What is the Newton's second law? We covered it also in the earlier lectures. But in general, it's when you take away the forces that are applied on your system from Ike was one up until end and you sum up all the forces, right? Then what you will get You will get mass times acceleration. But what is acceleration? It's the double derivative off X with the respected time. So you can awesome write it down like this D squared X DT squared. Okay, so this is your second mutants law. And now what we will do next, we will just follow this formula. We're gonna sum up all the forces that we have on our system. But first of all, let's do it like this that we're gonna agree on a convention that when I write a face function of time, I'm just going to write it down like this. X is a function of time. We're gonna write it down like this. And now dx d t as a function of time where he's gonna call it eggs dot and same thing thes squared x DT squared as a function of time which is gonna call it X double dot Okay, so we're gonna take our force applied. Plus our four spring plus our damping force and all that will equal mass times d squared X DT squared. Okay, so this is our second Newton's law. But now let's rewrite it like this. F A plus, what was our spring force? It was minus. Okay, X plus, what was our damping? Force minus. And then we had ex dot and times. See? Okay. Equals I m well, and I'm just gonna write it down like this double dot okay? So we can just simply write it down like this f a minus k x minus C x dot equals I m x double dot And now what we can do we can put all the ex is an ex dots on one side and f a on the other side. So we're gonna do it like this. We're gonna have m X double dot And now, Plus, we have C x dot and plus que x equals our force applied. And, uh, this is our differential equation. Okay? Differential equation. And what is a differential equation? Well, differential equation is just a mathematical formulation that describes our system. Okay, so it's just an equation that describes our system. But what makes it different from all other equations that describe systems because we have covered other functions as well that describe systems. Why is this equation any different? Well, for that, I'm gonna explain to you like this. So first of all, what is our I'll put here? What's our I'll put here? Is X okay? And what are inputs? Well, one of our inputs is apply force. But we also have two other inputs. We have our X double dot or acceleration, or ex dot, which is our velocity. So you can think of them as inputs as well. And you can see it if you rewrite this equation like this. So if you just right down que x equals f a And now I'm gonna have minus c x dot and minus m x double dot And now if I just divide everything by K and I just remove K from here. You can see now that this is your output X and these are your inputs. But no what makes different from other types of equations? Well, because in other types of equations, your inputs were all independent, right? So you had some kind off rent a function where you had three inputs X, why and Z Then you had some kind of function off X, Y and Z, and then you had some kind of output. Let's take a But here the situation is a little bit different here, not only Bazaar independent because x dot in x double that they depend on the output they depend on X. In fact, these two inputs are the changes off X with respect to time and exit double. That is the double change of X with respect to time. So let me use a blunt dagger to show it to you. So let's say that you have one input, which is f a right and then you have another input, which is eggs not and then you have another input, which is X double dot and then you have this function that you had here f a divided by K minus C over k ex dot minus m over k X double dot And then you got your output, which is X. Okay. And now what happens is that your ex Goche here and a differential operator is applied to it. The over DT and what you get. You get your ex dot right? And then your ex dot goes here. But then this eggs dot Another differential operator is applied to it. And now you get your ex double dot and then your ex double that goes here and here. So you see, now you have one independent input, but you have to inputs that are not independent, but they depend on the output. So you take your output and you differentiated with respect to time. And then you use that differentiated output as an input that goes back into your function. And then that differentiated input you're different. You again and again. Use it as an input in your function. And in our case, we differentiated our output with the respect of time. Maybe in another example, you differentiate Europa with another variable. He doesn't have to be time all the time. But in our example, it's time, and in many cases, in real world examples, it's time. So that's the main difference between normal equations that represent systems and differential equations. And so in the next video, I will show you what it means when someone tells you that you have to solve a differential equation to see in the next video. 28. differential equations intuition 2: welcome back. So in this video, we will conclude with the differential equations. And in this video, I will show you what it means when someone tells you sold the differential equations. So what does me? So in the last video, I showed you the difference between differential equations and normal equations where you had all inputs that were independent but differential equations. You can see that some of the inputs are not independent. They are dependent on the output. So you had your X, and then you differentiated them. You got your ex dot eggs doubled out, and both of them went into the function as inputs again. Now, remember one thing, Uh, our variables both outputs and inputs, all of them were a function of time. Right? So you're applying force was a function of time. Your ex dot was a function of time. Your ex double dot was a function of time. And your output X was also as a function of time. So all my variables are a function of time. Now, when someone tells you that solve differential equation, it means that you take this system and we want to represent it in this way. my input this time I have some kind of function as a function of time and my output is X. Okay. In other words, I have my output. And then I have my input, which is now independent, and I want to find an expression that is able to give me my output is a function of time. So X is a function of time. And then I want some kind of expression, so that if I put my time excited, I will get my ex. So I put my time inside my function and I get my ex. So solving a differential equation means that you go from this representation off your system to this representation of your system. Okay, So let me just read with red that I take this representation of my system. And then I want to get rid off all these dependant inputs and even this one independent input. I know that all of them depend on time. So why not find an expression that just takes the primary independent variable, which is time? And then I put it inside this expression and I get my output. And then I don't have to worry about my eggs, dots and x double dots. So this is what it means and solving a differential equation, because this is a blood attacking way to represent the system. But Soling differential equation means that I find a mathematical representation to represent this instead of representing my system like this. So, in other words, I want to replace this mathematical representation, right This equation, I want to replace it with some kind of equation where I have my tea in my function and I get my ex. And now, in our specific example, if we do that, if we solve our differential equation and get this form, then okay, I'm not gonna go into specific math here, but I do know the final answer so I can give you the final answer, and then you will see and our X really east it position of this block, right? Because I have my spring here. I have my damper here, and someone is applying some kind of force to this book. And if you put all these forces combined, you want to know how your block moves as a function of time. Okay, You want to know your X as a function of time. So the answer for this differential equation if we solve this differential equation, the answer is another number. And remember, when you solve a differential equation, you don't get a number. When you slow for differential equation, you get another function, and that other function relates time with the output directly, and it cuts the middleman. Alright, It cuts those three middle men here. FAA X stopped in x double dot So the final answer would be Well, first of all, you have to know that are applied force as a function of time. It's, um, some kind of constant force f zero times call sign two pi F zero, which is the frequency and time. In other words, this is the amplitude here, and this force is also a Tory. So sometimes it points in this direction. Sometimes it points in this direction. So it's a coastline function. Okay, so you have some kind of coastline function here. And so if you replace this FAA with this expression and sold the differential equation, your final answer will be X as a function of time equals some kind of constant that we're gonna call, see one times go sign two pi and now you have frequency. But it's different kind of frequency. It's the block frequency. Okay, It's It's not the force frequency, but it's the blocks oscillation, frequency times, time plus si two, which is another constant times. Sign two pi again. The block frequency times Time plus and I'm gonna continued here. Plus if zero which is this divided by mass off the block. And then you have to pie. Then here you have the block frequency You where this entire thing. All right. And then you have to pie. And now you have the force Frequency will also be squared. And then all that in the brackets times mass and then the entire expression I multiplied by call sign two pi f zero, which is the force frequency and time. And there you go. If you look at this equation, then you can see that you have only one input, which is time. Then you have your function, which is the entire function here. Okay, is a function of time and you get your ex right? So you have solved your differential equation. And remember, you have to frequencies here. One is how this block Ossa lates. If you forget about this force and you say that this force doesn't exist, right? If you just don't apply any force and you let this look toe Ossa late, then you have this frequency. It can also ate at this frequency F B. But then you can apply some kind of force to this block at the completely different frequency. And it would be at F zero. And there you go. This is your expression. And now one final thing before we finish with this video is that sometimes if your FB, which is your block frequency if that truly equals the force frequency. In other words, if you apply, you're forced to this book at exactly the same frequency like this bloke Ossa Leight's. Then what happens is that the this denominator here, right, it would be zero. So, in other words, maybe it's better to write down like this. If f block approaches F zero, then this term approaches zero, right? And if this term, if the denominator approaches zero, if this term approaches zero, then this entire term will approach infinity. In other words, this term becomes very, very big. In other words, this entire oscillation off this block will become very big and it's very dangerous. And it's called resonance. It's very important for for structural engineers resonance and it can lead to structural failure. So many, many things, including bridges, have failed and collapsed in the United States. There was a bridge many decades ago because of the wind. The frequency at which the wind was hitting the bridge was the same, like the frequency off how this rich was oscillating. So the bridge started oscillating more and more and more until collapse. So this is something you wanna avoid. Okay, Well, this is it with the differential equations. And, uh, in the next video, I will cover one more thing. In this course, I want to include one bonus lecture about transformations. There is one special transformation that I want to cover. We have really covered transformations in the past when we went from petition to pull accordance. But there is one specific transformation that I want to show you, which is when you go from time domain to frequency domain, which is used in the engineering quite a lot. So see you in the next video 29. 156 time frequency domain: Welcome back. Well, we have reached the final part of our course. And this is a bonus lecture. This is not part of calculus, but I still want to include it because I have seen it a lot in engineering and it seems to be very important. So I want to give you a heads up and I want to give you a small intuition about this concept. And this concept is about going from time domain to frequency domain. And at the beginning when I started with my engineering journey, when I first heard about it, I was confused. But now I am not. So I want to share it with you. It's a very short lecture. I just want to show you one thing. I want to show you two different ways. You can represent a wave. So I'm going to have a wave here, okay? I'm going to have a wave or a sinusoidal function. So, and let's say that I have my time here now. Okay? This is my time axis, not Sita, not angle, but time. And I have some kind of sinusoidal function. Then I will say that this is one and this amplitude here is minus1. And I'm just gonna call it w, which is wave amplitude. And I know that the frequency of this wave is one hertz. Is there another way how I could represent this wave? Well, there is another way. You see, I can represent it like this. I can change my domain and instead of time, I can have frequency. So if I have my frequency, then get time is measured in seconds, but frequency is measured in hertz. And so if I have my amplitude here, my wave amplitude, and now it's going to be magnitude. So it means that minus1 here will still be one here. But now, if I say that, okay, this is my frequency line, then I have hurts here, right? I have one hertz, I have two hertz, three hertz. So another way to represent this wave is like this. This is my amplitude one. And since the frequency of my wave is one hertz, then I can represented like this C. So that's how I can represent it. And now if I have another wave here, which has doubled the frequency. So like this, you see now my frequency two has two hertz. And so on this graph, when I replace my time with frequency, I have this kind of graph. So you see there is a difference. You represent the same thing but in a different way. But this way might be easier, right? So here you have to read an entire wave. And here you just say that aha, I have my frequency and this is my amplitude. And then you have an idea about the wave. And the same thing here you have something that has double the frequency. And already from this graph, you see that, okay, this wave has doubled the frequency like this way. So you can imagine in your head that this wave will oscillate twice as fast. The red one will oscillate twice as fast compared to the black one. And this kind of change from time to frequency, it's a very common thing in engineering. Alright, we have reached the end of this course. I must say that I really enjoyed working on this. My objective for this course was first to give you a truly intuitive feeling for all these hard to understand concepts in calculus. And secondly, I wanted to train your skill in applying calculus in other courses or at work on real-life projects. In other words, I want you to have a feel for calculus. Thank you for following my course. And I wish you best of luck in life. Thank you.