The Simplest Explanation of Integrals in Calculus

Añadido: 2026-05-05

[00:00:00]But I want you to always have in the back of your head, I want you to have a really deep understanding for what an integral actually is. So that whenever you're calculating it, you kind of know, oh, what does that represent? What am I actually doing? You know? So we're going to take some time to make sure you have that sort of internalization of what an integral is right here.

[00:00:20]All right, so um let's take a pause here and just go back to something you probably learned in physics because what I'm going to do is tie something from physics into math and that's where we're going to learn integration. So from physics, you all have probably learned the definition of work. Work in physics, right? Remember that? Work is equal to force times distance, right? The basic definition of work. So let's go from that starting point and we're going to use that to end up with integration and I'm going to teach you how that works. So for work in the in the physics definition. And by the way, we're we're using physics because, you know, um math is useless unless it's uh it's my opinion. Math is useless if it's not used to do something, right? So we're using math to describe the world. Work is equal to force times distance is is an application of math, right? So that's that's what gets me excited about math is being able to use it for something.

[00:01:13]So the mathematical definition of work that you'll learn in a physical physics class is force times distance. Now let's let's draw a few things on the board to make sure you understand exactly what what I'm talking about here. Let's say you have a wooden block sitting on the ground. Here's the wooden block, okay?

[00:01:31]And it's sitting here and I'm actually going to push in this direction with a force F. And I'm measuring F in in Newtons, right? So it could be 10 Newtons or 5 Newtons or whatever.

[00:01:41]Right? Now sometime later, this box is going to move because I am pushing it after all. So the box is going to move over here.

[00:01:48]Right? Now even when the box gets here, I'm still applying a force F. I never stop pushing with a force F. I'm always pushing with a constant force and this box has moved a distance D.

[00:02:03]Notice I'm measuring from the front of the box to the front of the box, right?

[00:02:07]Notice I'm measuring from the front of the box to the front of the box. So this is the total distance the box has actually moved, right? So if I can measure the distance this box has moved in meters and if I know how hard I'm pushing with a constant force F, then I just take these two numbers and multiply them and I get another number that we call work and that's measured in joules.

[00:02:27]So that's a very useful definition in physics.

[00:02:29]Now for those of you who've taken physics recently, you might realize that this is a little bit simpler definition than we usually talk about in in a real physics course because you could have the situation where you have a box and you could have a situation where you're not actually pushing straight on.

[00:02:47]You might have a situation where you're pushing maybe at an angle.

[00:02:51]So the force, you know, you're pushing down at an angle, right? And that angle you could define here as angle theta or something like this.

[00:02:58]But no matter what, if you have any forward push in this box at all, it is still going to move forward a distance D.

[00:03:06]Right?

[00:03:08]So uh here we measure from here. This is distance D.

[00:03:13]Like this, right? And I'm still pushing with a constant force in this case.

[00:03:17]Nothing's changed, but I'm just angling this force down. So this is a force and it's exerted at an angle theta from the horizontal.

[00:03:26]All right, so when you have a force at an angle like this, it's basically the same thing, but instead of work being equal to force times distance, work is equal to force dot distance. The dot means uh you know, something you learn in physics. The dot basically takes care of this angle for you. The dot means F times D times cosine of the angle between the two. This is what you actually learn in physics one, right?

[00:03:51]And the reason that we have this dot here, this dot product here, this cosine angle, this this cosine of theta, it takes care of the fact that that that force is angled for you, right? So what it's doing is this F cosine theta, F times cosine theta, it's basically Let me rewrite it for you to make make sure you understand. So this can be rewritten as F times cosine of theta times D. F cosine theta times D.

[00:04:17]So what F cosine theta is is it's chopping this uh force and it's only looking at the force in the direction of motion. Cuz when you take the cosine of something times the magnitude, you basically chop it down and you only look at the X component, right? So all we're doing is we're looking at the X component of the force. That's what this is doing and that number is multiplied by the distance. So in general, what it means is when you push a block, it doesn't matter what angle you push it at. You always care about the direction of the force in the direction of motion.

[00:04:50]How many Newtons in the direction that I'm pushing that block, that's the force I care about. So if I'm pushing at an angle, I just chop it down to get the force in the direction of motion, then I take that force times distance, that's equal to work. In this case, I was already pushing in that direction, so it's just a straight multiplication. So in any case, force times distance, that's what work is equal to, right?

[00:05:13]So let's go ahead and plot this and see if we can learn something and and I promise you we're getting to the um we're getting to uh the definition of the integral uh with this line of thought. So let's just keep going here.

[00:05:28]So let's go ahead and plot this. What I mean by plot this is I want to do a a quick little graph here. This is X and this is the force in Newtons that I'm pushing and this is the distance that the block moves in meters. So here is zero. X is equal to zero. This is where the block starts and this is my force that I'm pushing on the block. So if I'm actually going to do this, let's say I'm pushing at 5 Newtons, right? If I have a block here and I start pushing it with 5 Newtons, then I push push push push push push the block moves moves moves. Let's let's say I go ahead and stop here at let's say 10 meters.

[00:06:05]This is what a force distance graph would look like. All it says is that I start with the block at zero and I push it with a constant force.

[00:06:15]And as I push it, of course, the block moves. So I'm moving this direction and I eventually I stop it at 10 meters. So if I take a block and I push it for 10 meters with a constant force of 5 Newtons, what would the work be? Well, we already said the work is force times distance. Force times distance, right?

[00:06:32]So the force is a constant 5 Newtons.

[00:06:34]The distance is 10 meters. Force times distance would be 50, right?

[00:06:39]50 50 joules.

[00:06:41]And then let me ask you this. What does this represent on the graph? So the work would be 5 times 10 is 50 joules. That's how you calculate it on paper, but what is it represented with in the graph?

[00:06:54]Well, 5 times 10, what this physically represents is the area underneath this rectangle to down to the X axis. That's what this means. So really, work is force times distance, but more generally, if you plot the force and the distance that the object moves, then the work graphically is represented by the area underneath this force curve down to the X axis between the points of motion, right? So it's physically represented by the shaded area here.

[00:07:25]All right? Cuz in general, an area is length times width, so we can easily calculate and see that the area under this curve is equal to the area that we always care about when we talk about >> [snorts] >> when we talk about um you know, for our work. We talk about calculating these things, right? So that is in general what uh work is equal to. It's equal to the area under this force curve. And that was really easy because the force we pushed was constant in this case.

[00:07:52]All right, so that's sort of the basic setup. Let's review everything you probably already known, but maybe haven't seen laid out quite so so linearly like this, right?

[00:08:02]Now let's see what happens when we change things a little bit. What if, you know, I still push a block, right? Just like I always do from A to B, okay? But let's say I'm not pushing with a constant 5 Newtons. Let's say I push with 5 Newtons in the beginning and then I start pushing harder, right? Which is going to accelerate the block. And then at the end I get tired, let's pretend, and I slow down and I push, you know, maybe with 2 Newtons or something with some low low amount of force. What would that kind of graph look like? Let's go ahead and see what that would look like.

[00:08:33]So if we had something like that, then we would have again a graph like this and this would be distance in meters and this would be force in Newtons.

[00:08:45]So just like we did before, let's say this is 5 Newtons and let's say this is 10 Newtons up here.

[00:08:53]All right? And let's say this is 10 meters. This is the same original setup we had in the original graph. Now let's pretend in this case, we're doing it a little bit, you know, differently.

[00:09:03]Let's say I start out pushing with 5 Newtons and then I run really fast and I push with 10 Newtons and then I get really tired and I finish up pushing less than 5. Maybe I don't know, this is 3 Newtons or something like this.

[00:09:16]Right? So if I draw a dotted line, this is where I end my journey. This is where I begin my journey. My question is, if I actually push with something like this, how do I calculate the work?

[00:09:28]How do I calculate the work on this? I mean, we would all agree that the block has moved from A to B. And we would all agree that I'm pushing in the direction that it's moving. So we would all agree that some work is done, but you can't do force times distance because force is not constant anymore, right? Force changes. Force is changing constantly during this entire journey. I get really high force and then I dip down and get lower force down here. So we all know that some positive work is done, but this definition of work, force times distance, it's a super simplified version when you have a constant forces that don't change. But in this case, I'm giving you an example when the force changes all the time. And not only does it change, it changes really smoothly and continuously, right? Well-behaved.

[00:10:12]There's no jumping around, right? It's just nice and smooth, but it's all it's always changing. So, how do we find um the work in this case, right?

[00:10:22]The answer is the work is going to be the area under the curve.

[00:10:31]And you have to take that a little bit with with some um you know, some uh some just some agreement that that this is sort of something you just have to accept. I'm not proving that to you. But what we have done over here is we've shown you that the work when the force is constant is the area under the curve. So, it shouldn't be too hard to swallow that when the work when the force is not constant, when it's changing all the time, that what you're looking for to try to find this answer is the area under this curve. This is what you're looking for, the surface area, literally the area from that curve down to the x-axis between the boundaries of when I start and when I end. If you could somehow calculate this area, then the answer that you get is going to be equal to the number of joules of work it took to actually do that motion. It comes from a direct extension of what you've already learned, work is equal to force times distance.

[00:11:23]All right. So, the question is how do you calculate this surface area, which is going to be the work done in this case?

[00:11:31]Um how do you calculate that? Well, you know, it's easy with rectangles. It's easy back here when we have a rectangles because we know, even if you didn't know force times distance, if I asked you, what's the area of the under this curve here, then you would be able to tell me because you know the distance and you know the width of the rectangle and you know the area of rectangle is uh you know, these two things, so you would be able to go ahead and calculate that. But in this case, it's not constant like this, so it's very difficult to just quickly bust out an answer and say, "Well, this is the uh area." Well, let me give you a hint and I think you probably see where I'm going with this.

[00:12:05]Um integrals in calculus, the concept of integration in calculus is used to calculate the area of funny-shaped curves like this, right?

[00:12:16]Let me say that one more time.

[00:12:17]Integration in calculus is used to calculate areas of things like this, irregularly-shaped areas. Now, if I just busted out and told you, "Hey, you know, calculus is used to calculate area, you know, uh under weird-shaped curves and integration can do that." You probably wouldn't be too impressed because who cares? I mean, how many times do you sit down at a desk and just want to calculate the area of some crazy-shaped curve? You probably don't care.

[00:12:42]But what I'm trying to show you by going through all of this talking is that nature, you know, I mean, things that you really do want to calculate like the work done on a box, basically boils down to finding the area of something, right? So, I'm trying to give you that motivation. If I just told you, "Hey, you can use integration to calculate area." You probably wouldn't get too excited. But if I'm showing you physical things in science and engineering that you'd want to calculate like work, it's a crazy important thing to know. And work is basically the area underneath a curve, then you would understand, okay, that is a really important application of integration that that I need to understand because that's how we describe the world, right?

[00:13:20]So, work boils down to being the area under the curve. Area under the curve basically boils down to calculating an interval. So, that's your motivation.

[00:13:27]So, let me take it one step further. How would you actually calculate this? How would you take a stab at it? Let me redraw this curve.

[00:13:34]So, what I'm going to do here is draw another graph directly underneath and we're going to label it exactly the way we did before and hopefully make it very clear. So, what we have here, this is 10 m out here and this is 5 N and this is 10 N right here. So, let me go ahead and draw the curve real quick. We started at 5 N, we got really strong, and then we finished really weak. So, we're trying to find the area of this this region here, okay?

[00:14:03]So, we don't know how to calculate exactly the area here, but we we can get an idea about how to pro- approximate this area.

[00:14:11]Let's just say you wanted to get a general answer, an approximation, not an exact answer. What you could do is you could create a little rectangle here right here in this guy, and you could find the area of this rectangle.

[00:14:27]Right? Now, we know how to find areas of rectangles because rectangles are easy.

[00:14:32]It's the width of the rectangle times the height of the rectangle. So, we know that we can actually calculate the area of this rectangle here because the rectangle goes up and it touches the curve, but notice how it's not going to be very exact because let me draw the next rectangle. So, we could draw another rectangle next door and make it like this.

[00:14:51]Right? And then we could draw another rectangle next door like this.

[00:14:58]And we could draw another rectangle uh actually, it's going to be more like this. We draw another rectangle here, let's say.

[00:15:07]Like this, and we could draw another rectangle down here.

[00:15:10]And these rectangles I mean, they're supposed to be equal widths.

[00:15:16]But for the sake of the drawing, it doesn't really much matter. So, we can do we can do another rectangle like this.

[00:15:24]And then the final rectangle will be like this.

[00:15:27]So, you see, by drawing lots of skyscrapers that just go up and touch the curve, we can get a pretty good approximation of it. But notice how it's never going to give you an exact answer because I have all of this dead space here that's not covered by any of these rectangles. I got a lot of dead space here that's not covered by a rectangle. A lot of space at the top of the curve that's not covered. I've got space everywhere. All the space that I'm coloring in black is not covered by any of these rectangles. So, I'm trying to show you graphically. If I wanted to get an approximation for the area of this graph, I would just have a rectangle that would go up and just touch the graph, but I'm going to have some dead space. An adjacent one that goes up and have have a little more dead space there and so on. So, I can get a pretty good approximation of what's going on.

[00:16:14]Um but unfortunately, it's not going to be exact. So, how do I improve on this estimation? How do I improve on it?

[00:16:21]Well, there's a couple things I can do.

[00:16:24]Let me go back here and draw this guy one more time.

[00:16:32]So, this is x, this is force, this is 5, this is 10, and this is 10 like this.

[00:16:42]All right. And again, I start at 5 N and I go up to 10, I get tired, and I get finished down here. So, this is what I'm looking for.

[00:16:50]All right. So, how can I improve on that estimate? Well, one way I can improve on it is I can make these rectangles much much nar- narrower. I can have rectangles I can have like three times as many rectangles.

[00:17:03]So, I could do something like this.

[00:17:08]And you can see how I'm climbing up the curve, right?

[00:17:15]All right. Now, I'm not going to fin- finish drawing all these rectangles in here, but you can kind of get the idea that if I make these rectangles really narrow, I'm going to have more rectangles, but is the narrower I make them, the more it's going to fit the curve and the less dead space I'm going to have up here. I don't have enough space to put black ink up there. There's little bitty little triangles up there that are all dead, but they're smaller than the ones from before.

[00:17:38]All right. So, basically, what's happening is if we make these rectangles super narrow and super small, then we can approach the true answer. We'll have more calculation to do, but we can approach the true answer of this guy.

[00:17:52]Right? So, just to kind of make sure you are on the same page as me with me. If we have one of these little rectangles here, right? Then if we say the width of this rectangle, let's just say the width of this rectangle is delta x. We use delta in calculus a lot with a triangle. We use delta a lot because delta x represents super tiny tiny distances.

[00:18:13]That's all it means. So, when you see delta x anywhere or delta anything, it just means a small little distance. And these rectangles are getting really small, so we're going to start calling it delta x. Now, let's just say that our curve, you know, touches this rectangle here and it kind of bends over. This is where we are, let's pretend.

[00:18:31]Then what would be the height of this rectangle? Well, these rectangles always go up and touch the um touch the curve, so the height of the rectangle is always going to be f of x wherever you're at, right?

[00:18:45]Because it's always touching the graph.

[00:18:47]The height of the rectangle always touches the graph. So, no matter where you are, the height of it is always f of x because f of x is your function. And the width of it is delta x, which just means small number. So, anytime you see this, just think small width, right?

[00:19:00]So, in order to find the area um the area of one rectangle, the answer to that is f of x times delta x. This is just the width of the rectangle times the height of the rectangle.

[00:19:19]All right. So, this is just for one little rectangle. All right. Now, let's go and extend this further. Now, to get the whole answer, we have many many many many rectangles all over the place. All right. So, if we wanted to get a pretty good approximation for the area, we have to add all of these rectangles up, all of the ones that appear everywhere, all under the curve. And the easiest way to do that is with a summation because that's the easiest way to add stuff up. So, I is equal to one up to the value of n >> [snorts] >> f of x sub i. I'll teach you what that means in a minute. Delta x sub i. So, here's where you start to get into math notation and people can kind of roll their eyes a little bit. All that's going on here is I have an index i and I'm summing up from i is equal to one to n. n is just however many rectangles I have under this curve. I'm trying to add them all up with a summation and each time I cycle through it, I'm looking at the ith rectangle.

[00:20:16]So, for the first rectangle, it's the width of the first rectangle times the height of the first one and then plus the width of the second one times the height of the second one plus the width of the third rectangle times the height of the third rectangle and so on and so on and so on all the way up until I get to however many rectangles I have, n rectangles under the curve. Right? So, that's what we have going on here.

[00:20:39]So, what what we're really building up to is >> [snorts] >> so far we've said that you can approximate the area under a curve by lots of rectangles.

[00:20:48]Here is sort of like a little formula that that teaches you how to add up the rectangles. Now, how do we get to the true answer? How do we get to the true answer? We said if we make those rectangles skinnier and skinnier and skinnier and skinnier, then we can approach the true answer. So, really we want to squeeze them all the way until they're so incredibly tiny that they're so thin you can't even really see them and in that case, we've basically arrived at the true answer for the area under that curve.

[00:21:15]>> [snorts] >> And the way you handle that, the way you do that in math is the following. So, let's rewrite what we have. Sum from i is equal to one up to n. I'm just rewriting everything that I had before.

[00:21:29]x sub i delta x sub i. So, what we want to do is we want to take the limit.

[00:21:37]Right? And we want n, the number of rectangles, to approach infinity.

[00:21:43]And we also want uh delta x, which is the width of these rectangles, to approach zero.

[00:21:51]So, these are both going to happen at the same time because if I put infinity number of of rectangles under the curve, then they, by definition, they have to get incredibly thin in order to fit them all under there. So, what's going on here? This is basically an approximation for the true answer.

[00:22:07]The true answer for the area under the curve.

[00:22:16]All right? So, this is it. You take the width of the rectangle times the height of the rectangle and you sum them all up and you take the limit as we say a number of rectangles goes to infinity, the width of these rectangles approach zero. That, my definition, that, my friends, [clears throat] is the definition of the integral. I just didn't tell you yet. That is the definition of the integral. So, what what happens is when we do all this limit business over a summation like this, then we define it as the integral. So, the work, which is what we've all been talking about in terms of, is defined as the integral from A up to B. That's the distance that the box moves from A to B of the function f of x over dx. I'll explain what dx means in a minute. Is equal to the limit as n goes to infinity and as delta x goes to zero of the summation i is equal to one up to n of f of x sub i times delta x sub i.

[00:23:22]That looks ugly, that looks intimidating, that looks scary, but I hope that I've broken it down in little tiny chunks so that you can really understand what this is saying. This is the definition of this thing you see in calculus books all the time. This squiggly line is called an integration symbol.

[00:23:39]Notice that if you look at it long enough, it kind of sort of looks like a summation. I mean, it kind of like it's kind of encompassing the whole thing here, you know, sort of anyway. This is an exaggerated version of this kind of.

[00:23:50]It's got a function under there just like I have here and instead of a delta x, it's got a dx. Basically, this is just notation. When you take the limit and you send the width of these rectangles to zero, then they're so incredibly tiny that you don't call them delta x anymore. You just call them dx.

[00:24:08]So, when you see a dx in calculus, it just means that it's an incredibly teeny tiny little bit we're adding over. So, what's happening is we're taking the height of the function times an incredibly teeny tiny little differential. That's what this is called, differential width, dx. This is getting the area of one of these incredibly tiny slivers of a rectangle.

[00:24:30]The integral is adding up uh from A to B all of these little teeny tiny slivers of rectangles and it's arriving at the total answer for the area under the curve. That's what that is. So, we've defined it all in terms of of uh work because we've derived the whole thing in terms of work.

[00:24:50]>> [snorts] >> But in general, the integral I'll just write sort of a conclusion here. In general, the integral isn't something that necessarily just applies to work, it's in general anything. A to B, you integrate a function f of x dx. That's That's what that is. And basically, what it is is the area under the f of x curve.

[00:25:20]The area under the f of x curve. Now, again, if I had just sort of started out this lesson and said, "Hey, integration is used to find areas," then you probably wouldn't be excited. But I'm showing you through a direct application that work is one of those cases when you're going to want to calculate the area of something and I'm trying to give you some motivation for why it's happening.

[00:25:40]So, let's recap really quickly and do a little bit more rapid pace and then we'll do a quick conclusion and sort of show you where we're at here. So, we started off by talking about work. We said work is force times distance.

[00:25:52]And we talked about plotting work or plotting the force distance here. If the box moves 10 m and I apply a constant 5 N of force, this is what the curve looks like. The work being force times distance is the area of that curve to down to the x-axis. So, it's a rectangle in this case. But then we said, "What if the force can change constantly as we move along this path that we're pushing?

[00:26:15]What if I push harder and then I back off and I push softer? I'm still moving forward, but I can change how hard I push." So, the graph would look like this. What would be the work done in that case? Well, it's hard to calculate with force times distance, but we say, by extension of before, that the area of this curve underneath that curve is going to be equal to the work.

[00:26:35]So, we say, "Great, we know we need to calculate the area, but it's a weird shape. How do we do it?" Then we say, "Well, let's take some approximations.

[00:26:42]Let's draw some fat skyscrapers in here, these rectangles up that touch the function, right? We can calculate the area of each one of these little rectangles easily. So, we could find an approximation for the area, but it wouldn't be right because we have all this dead space here. So, we would obviously have some errors." So, we say, "How do we improve that?" So, we say, "Okay, well, let's just make some skinnier rectangles." So, these are teeny tiny skinny rectangles. We can again find the area of any one of these rectangles because we know the width of it and we know the height of it because it always touches the top of the curve.

[00:27:15]So, again, we could find a better approximation, but it's still not going to be totally right because you have this dead space. But we're noticing that as you make the rectangle skinnier skinnier, the dead space at the top gets smaller and smaller and so we're approaching the true answer as we make these rectangles get skinnier.

[00:27:31]>> [snorts] >> So, we draw a blowup picture and we say, "The area of one of these rectangles is going to be the width that we're calling delta x times the height that we're calling f of x because that's the height of the function. So, the area of one of these rectangles is f of x times delta x. Delta x is just notation that means a really narrow rectangle." That's all that means.

[00:27:52]So, then we say, "How do we find the approximate area under the curve?" We take the width of each rectangle times the height of each rectangle and we sum them over all of the rectangles. And that's going to give us our nice approximate area. But it's not quite right because anytime you make these rectangles a finite width, there there's always going to be a little dead space at the top. So, then we take the limit of it the number of rectangles approaching infinity. So, we have infinite number of rectangles under the curve.

[00:28:21]And also at the same time, we shrink the width of these rectangles to zero. We're still summing over exactly the same thing. So, now we're basically summing an infinite number of rectangles because we're making the rectangles go to infinity and the width go to zero. So, we're summing an infinite number. If we do this calculation in a computer with 100 million rectangles, we're going to be pretty darn close to the right answer. If we uh if we take this limit like this, the answer you get is the total answer of the area under the curve.

[00:28:51]And then we say, "This thing that we just wrote down is the definition of what we call the integral in calculus."

[00:28:57]We're looking between A and B on the x-axis. We're integrating over dx, which means along the x-axis infinitely uh small uh distances in dx times the height of the function going to give you little tiny slivers of rectangles that are infinitely narrow and we're adding up. That's what this integral symbol means. We're adding up each and every one of those between A and B and that's how we're getting the area. So, this is the area under the curve, by definition, is the integral of f of x dx between A and B on the x-axis. That is an introduction motivation to what an integral is. And in this case, we've motivated it by talking about work.

[00:29:36]We've talked about work in physics and that it comes out to be an area under a curve and then it comes out to be, you know, important to be able to calculate areas. And that's basically what integration does. You're going to have lots of crazy functions. We'll talk about polynomial functions. We'll talk about signs and cosines. We'll talk about logarithmic functions and exponential functions and crazy, you know, arrangements of functions. But it's all going to boil down to one thing. How do you calculate this integral? How do we calculate the um the area under the curve? How do we actually do the math? I mean, this has been a motivation section to show you what an integral is, but we're going to teach you techniques of integration to get the actual numerical answers. Um and I want you to understand what that means. And this is exactly what it means. And finally, I'll close by saying that we've we've tied this all to work in physics, but integrals pop up you know, everywhere in science and engineering and math. I mean, they they really do. They pop up everywhere. Um and in general, it it pops up uh anytime you're trying to add up an infinite number of teeny tiny things. And that's what really an integral's basically doing. So, in this case, we've talked about work, but other examples would be if I have an electric field in this room, you know, electric field pointing in a certain direction. Um and maybe I want to go along a path and I want to add up that electric field. And maybe my path is not straight. Then I'm going to be adding up an a little tiny contributions of the electric field along my path to add up the total amount, right? So, that's adding up an infinitely a number of little tiny objects. That's what integrals are good for. Right? Adding up an infinite number of tiny little objects like these rectangles here. So, I can use it to calculate and sum up an electric field or, you know, through space, let's say.

[00:31:21]Something a little more practical. When you have a pipe, right? Just a metal pipe or a plastic pipe with with some kind of liquid flowing through it. You know, typically when we put water through a pipe, we we envision the water, you know, all flowing at the same speed along the cross-section of that pipe. Right? And that's a pretty good approximation. But if I put like syrup in a pipe, some really thick liquid or oil or something like that through a pipe, then if you if you could look at the cross-section of the pipe, the um the the the oil flow or the syrup flow in the center of the pipe is not going to be the same as what it's going to look like around the edge of the pipe.

[00:31:57]In other words, if I have a a cross-section of a pipe here and I'm trying to put maple syrup and flow maple syrup at 100 miles an hour through this pipe, then probably what's going to happen is the maple syrup is not going to flow too well near the walls because it's kind of sticking to the walls, but in the center, it's going to go a little bit faster. So, if you could look at a cross-section of maple syrup flowing through a pipe or oil or something practical more practical when you're pumping oil out of the ground, then because of the viscosity, the flow rate in the center of the pipe is different than the flow rate near the edge of the pipe. So, maybe you want to calculate how much oil you're pumping through your system, you know, knowing what this cross-section, you know, uh behavior looks like. And you can use integrals to figure that out because basically, the flow rate changes as you go across the surface of that pipe. It's not constant like it is in your kitchen with water.

[00:32:47]It's totally different as you go across the surface of uh the the cross-section of that pipe there. So, if you have a pipe 3 ft wide and you're sending oil down it, you'd really like to calculate how much oil you're getting. And so, you really would want to look at what the flow rate is in the middle, what the flow rate is at the edge, and you can use integration to add it all up, add up an infinite number of teeny tiny little objects across In this case, it would be across a surface area.

[00:33:12]So, I'm getting a little bit ahead in the applications, mostly to motivate you. Integration in general is to add up an infinite number of teeny tiny little objects.

[00:33:21]In calculus one, we're going to be concerned with integration only one direction along X uh of a function between A and B. So, make sure you understand this section. Make sure you understand the motivation for what we're doing. And then following on to the following sections, we'll start to really roll up our sleeves and teach you how to calculate integrals and how to get numerical answers and kind of give you more of the terminology so that you understand what's going on. We'll give you a lot of practice along the way.

[00:33:46]Make sure you can do each and every one of these problems that we solve by yourself. So, I recommend that you watch me work it and then you work the problem yourself on a separate sheet of paper to make sure that you're building your own skills. And if you do that, I promise you that by the end of this class, you'll be an expert at taking integrals in calculus.

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