01 - What Is an Integral in Calculus? Learn Calculus Integration and how to Solve Integrals.

hello and welcome to calculus one extra practice with integration in this set of lessons what we're going to do is really roll up our sleeves and get some extra practice with learning what integrals are in calculus 1 and also how to actually get a lot of practice in calculating integrals so as you know calculus is broadly basically broken up into two main topics that you learn in the first half of calculus 1 you'll learn about derivatives and that's all about how things change right how your rates of change of functions right and we've done calculus 1 extra practice with derivatives so we've covered derivatives in excruciating detail if you haven't done that please go back to get that material and really make sure you understand derivatives because what we're going to do now is what we usually cover in the second half of a calculus 1 course and that's called integration so really the big picture I'll kind of tell you right now integration which is what we're doing now is really the opposite of derivatives which is what we've done before so the two go hand-in-hand it's not that one's better than the other or anything like that it's just that they're opposites of one another so you really cannot understand integration very well if you have no idea what derivatives are and if you're not comfortable with that so if you don't understand that or if you haven't done that material go back and get my calculus 1 tutor which is the granddaddy course for calculus 1 and also go get my calculus 1 extra practice with derivatives set so that you can really really get good skills at derivatives so what we're going to do in this section is take a few minutes of probably a longer than typical section in this set of lessons to explain what an integral is there's not going to be any math in this section we're not going to actually start calculating integrals in this section but what I want to do is sort of take an approach that I really love and explain what an integral is because what we're gonna do is after this section we're going to spend section after section after section calculating these integrals over and over and over again and you're going to get pretty good with learning the methods to do it 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 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 all right so 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 why 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 and in the physics and definition and by the way we're going to we're using physics because you know math is useless unless it's well 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 an application of math right so that's that's what gets me excited about math is being able to use it for something so the mathematical definition of work that you 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 wood block okay and it's sitting here and I'm actually going to push in this direction with a force F and I'm measuring Evans in Newtons right so it could be ten Newtons or five Newton's or whatever right now sometime later this box is going to move because I am pushing it after all so the box is gonna move over here 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 notice I'm measuring from the front of the box to the front of the box right notice I'm measuring from the front of the box to the front of the box so this is the total distance the box is actually moved right so if I can measure the distance this box is moved in meters and if I know how hard I'm pushing with a constant 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 so that's very useful definition in physics 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 you might have a situation where you're pushing maybe at an angle 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 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 right so here we're measure from here this is distance D like this right and I'm still pushing with a constant force in this case nothing's changed to it I'm just angling this force down so this is a force and it's exerted at an angle theta from the horizontal alright 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 you know something you learned 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 and the reason that we have this dot here there's a dot product here this cosine angle this this cosine of theta it takes care of the fact that that net forces 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 so what F cosine theta is is it's chopping this force and it's only looking at the force in the direction of motion is when you take the cosine of something times the magnitude you basically chop it down you only look at the X right so all we're doing is we're looking 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 how many Newtons in the direction that I'm pushing that blog 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 so let's go ahead and plot this and see if we can learn something and I promise you we're getting to the we're getting to the definition of the integral with this line of thought so let's just keep going here so let's go ahead and plot this what I mean by plot this is I want to do a quick little graph here this is X and this is the force in Newton's 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 five Newtons right if I have a block here and I start pushing it with five Newtons then I push push push push push push the block moves moves moves let's let's say I go ahead and stop it here at let's say ten meters this is what a force distance graph would look like all it says is that I start with a block at zero and I push it with a constant force and as I push it of course the block moves so I'm moving this direction and I'm eventually I stop it at ten meters so if I take a block and I push it for ten meters with a constant force of five Newtons what would the work be well we already said the work is Force Times distance Force Times distance right so the force is a constant five Newtons the distance is 10 meters force times distance would be 15 right 50 me 50 joules now 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 within the graph well 5 times 10 what this physically represents is the area underneath this rectangle - 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 all right because 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 when we talk about you know for work we talk about calculating these things right so that is in general what work is equal to it's equal to the area under this force curve and it was really easy because the force we push was constant in this case all right so that's sort of the basic set up let's review everything you've probably already known but maybe haven't seen laid out quite so so linearly like this right 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 gonna accelerate the block and then at the end I get tired let's pretend and I slowed down and I push you know maybe with 2 Newton's or something there's 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 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 Noons so just like we did before let's say this is 5 Newtons and let's say this is 10 Newtons up here all right and let's say this is 10 meter this is the same original setup we had in the original rap now let's pretend in this case we're doing it a little bit you know differently let's say I start out pushing with five Newtons and then I run really fast and I push with ten Newton's and then I get really tired and I finish up pushing less than five maybe I don't know this is three Newton's or something like this right so if I draw down a 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 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 forces changing constantly during this entire journey I would 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 behave 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 the work in this case right the answer is the work is going to be the area under the curve and you have to take that a little bit with with some you know some some just some agreement 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 hand 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 all right so the question is how do you calculate this surface area which is gonna be the work done in this case 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 you know the area of a rectangle is 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 area well let me give you a hint and I think you probably see where I'm going with this integrals in calculus the concept of integration and calculus is used to calculate the area of funny-shaped curves like this right let me say that one more time integration and calculus is used to calculate areas of things like this irregular shaped areas now if I just busted out and told you hey you know calculus is used to calculate area you know under weird shaped curves an 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 shape curves you probably don't care 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 don't want to calculate like work it's a crazy important thing you know and work is basically the area underneath the 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 so work boils down to being the the curve area under the curve basically boils down to calculating an integral so that's your motivation 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 so what I'm going to do here is draw another graph directly underneath and we're gonna label it exactly the way we did before and hopefully make it very clear so what we have here this is 10 meters out here and this is 5 Newtons and this is 10 Newtons right here so I'm gonna go and draw the curve real quick we started out at 5 Newtons we got really strong and then we finished really weak so we're trying to find the area of this this region here okay so we don't know how to calculate exactly the area here but we we can get an idea about how to proximate this area 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 right now we know how to find areas of rectangles because rectangles are easy 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 can draw another rectangle next door and make it like this right and then we could draw another rectangle next door like this and we could draw another rectangle actually it's gonna be more like this you draw another rectangle here let's say like this you could draw another rectangle down here and these rectangles I mean they're supposed to be equal widths but for the sake of the drawing it doesn't really much matter so we could do we can do another rectangle like this and then the final rectangle will be like this 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 gret tangle that would go up and just touch the graph but I'm gonna have some dead space and adjacent one that goes up and have a little more dead space there and so on so I can get a pretty good approximation of what's going on but unfortunately it's not going to be exact so how do i improve on this estimation how do i improve on it well there's a couple things I can do let me go back here and draw this guy one more time so this is X this is force this is 5 this is 10 and this is 10 like this all right and again I start at 5 Newtons and I go up to 10 I get tired and I get finished down here so this is what I'm looking for 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 narrower I can have rectangles I can have like three times as many rectangles so I could do something like this and you can see how I'm climbing up the curve right all right now I'm not gonna 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 gonna have more rectangles but is the nail where I make them the more it's gonna fit the curve and the less dead space I'm gonna 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 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 will have more cow relation to do but we can approach the true answer of this guy right so just to kind of make sure you are on the same page as meet 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 rift of this rectangle is Delta X we use Delta and calculus a lot with the triangle we use Delta a lot because Delta X represents super tiny tiny distances 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 if we're gonna 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 then what would be the height of this rectangle well these rectangles always go up and touch the touch the curve so the height of the rectangle is always going to be f of X wherever you're at right because it's always touching to graph the height of the rectangle always touches the graph so no matter where you are in 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 so in order to find the area 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 all right so this is just from 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 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 1 up to the value of n f of X sub I'll teach you what that means in a minute Delta X so 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 here is I have an index I and I'm summing up from I is equal to 1 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 I threats angle 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 in rectangles under the curve right so that's what we have going on here so well what we're really building up to is so far we've said that you can approximate the area under a curve by lots of rectangles here is sort of like a little formula 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 basically arrived at the true answer for the area under that curve and the way you handle that is the way you do that in math is the following so let's rewrite what we have sum from I is equal to 1 up to n I'm just rewriting everything that I had before X so I Delta X sub I so what we want to do is we want to take the limit right and we want end a number of rectangles to approach infinity and we also want Delta X which is the width of these rectangles to approach 0 so these are both going to happen at the same time because if I put infinity number 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 the true answer for the area under the curve alright so this is it you take the width of the rectangle time the height of the rectangle you sum them all up and you take the limit as we say number of rectangles goes to infinity the width of these rectangles approach zero that my definition that my friends 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 0 of the summation I is equal to 1 up to n of f of X sub I times Delta X sub I 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 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 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 you don't call them Delta X anymore you just call them DX 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 a 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 the integral is adding up Oh 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 work because we've derived the whole thing in terms of work 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 D X that's what that is and basically what it is is the area under the f of X color 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 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 and we talked about plotting work or plotting the force distance here if the box moves 10 meters and I apply a constant 5 Newton's of force this is what the curve looks like the work being forced 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 what if I push harder and then I backed 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 so we said 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 let's draw some phat 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 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 so we draw a blow-up picture and we say 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 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 gonna be a little dead space at the top so then we take the limit of the number of rectangles approaching infinity so we have an infinite number of rectangles under the curve 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 a hundred million rectangles we're going to be pretty darn close to the right answer if we if we take this limit like this the answer you give is the total answer of the area curve and then we say this thing that we just wrote down is the definition of what we call the integral in calculus where we're looking between a and B on the x axis we're integrating over DX which means along the x axis infinitely small distances in DX times the height of the function is 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 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 gonna have lots of crazy functions we'll talk about polynomial functions we'll talk about sines and cosines we'll talk about logarithmic functions and exponential functions and crazy you know arrangements of functions but it's all gonna boil down to one thing how do you calculate this integral how do we calculate the 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 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 and physics but integrals pop up you know everywhere in science and engineering and math I mean they really do they pop up everywhere and in general it pops up anytime you're trying to add up an infinite number of teeny tiny things and that's what really an integral is 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 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 in a little tiny contributions of the electric field my path to add up the total amount right so if that's adding up an infinitely a number of little tiny objects that's what integrals are good for all 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 something a little more practical when you have a pipe right just a metal pipe or a plastic pipe 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 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 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 in other words if I have 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 gonna flow too well no it near the walls because it's kind of sticking to the walls but in the center it's gonna 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 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 it's totally different as you go across the surface of the the cross section of that pipe there so if you have a pipe three feet wide and they're sending oil down it you'd really like to calculate how much oil you're giving 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 and up an infinite number of teeny tiny little objects across in this case it would be across a surface area 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 in calculus one we're going to be concerned with integrational in one direction along X 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 follow me onto 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 will give you a lot to practice along the way 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 will be an expert at taking integrals in calculus

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Keywords: calculus, calculus 1, integral, what is an integral, integration, techniques, techniques of integration, what is integral, lesson, integral calculus, math, what is a integral, what is integral calculus, what is integrals, what is calculus, maths, mathematics, derivative, tutorials, antiderivative, indefinite integral, substitution, examples, integrate, trigonometric, antiderivative calculus, antiderivatives and indefinite integration, antiderivatives substitution

Id: BHRWArTFgTs

Length: 36min 12sec (2172 seconds)

Published: Wed Feb 03 2016