Transcript for:
Overview of Calculus Chapter 4 Concepts

all right we start chapter four chapter four is like I said a switch from chapters three and everything before where we used derivatives before we're going to be using something different chapters four and five which is going to end our course in chapter four we discussed the second of the two main problems in calculus the first main problem in calculus where we spent the majority of our class I mean more than half of it was can you find the slope of a curve at a point what was that dtive the second question in calculus wasn't about the slope of the curve at a point it was can you find the area under a curve do you remember that question I introduced it I think like the first day of real calculus I said can we do this given some sort of continuous function from A to B can you find out what the area is between that curve and the x-axis basically the area under the curve from here to here and the answer is yeah yeah and we actually have two ways to do it I'm going to introduce to you what's called the rectangular method today uh we will deal much more with it in section 4.4 4.3 4.4 uh they're kind of tied together on how to actually do this with rectangular method but I'm going to introduce it to you today and then we'll talk about what's called the anti-derivative method uh which is a more concise way to do it are you ready for it so the this I'm going to kind of go in a weird Loop I'm going to introduce this with the rectangular method we'll talk about anti-derivatives anti-derivatives then I'll use rectangular method to kind of prove what we just learned about anti-derivatives you follow me so we have to wait till 4.3 to prove some of the stuff I'm talking about here today I'm going to give you kind of an intuitive approach on it though hopefully it makes sense for you so first method let's talk about the rectangular method and at least the idea behind it so rectangle method rectangle method says well why don't you divide your interval into n some number and equal subsections equal as far as their width goes subsection sub intervals whatever you want to call it and create a rectangle in each sub interval in each subsection hence where we get the the term rectangle method or rectangular method so here's basically the idea let me redraw the picture for you let's say that I make this a little bit bigger too here's my graph here's my interval the idea behind the rectangle method is can you find the area under this curve the way it is no not really because we don't have a formula for this and even if we did it doesn't have straight sides and only way we know how to take an area up till right now in Geometry is if a figure has straight sides right whether we can break it up into rectangles or triangles because we know how to take the areas of those figures or perfect circles you know how to take the areas of those figures this this is not a regular shape that we know of it's just some random function in fact that even been specific it could be any function but the idea is let's go ahead let's break this up we'll say okay we're going to split this up into however many parts we choose let's just assume those are equal widths that's about right it says if we have some sections here what we're going to do in each section since they have equal widths so what I'll I'll tell you why we need equal widths uh later on in section 4.3 4.4 but right now let's just assume we have some equal widths that's so our bases are all the same our rectangles equal bases means they're easy to add together that's basically it so for each each interval sub interval let's make a rectangle now there are several ways to do this I'll show you one way the problem is now you got to determine how you're going to have the height of our rectangle so one way to do this is to use what's called Left End points another way is to use right end points and a third way is to use midpoints and here's what that means you go to the right end point of your interval here's my interval I go to the right end point I'd say I'm going to go up to where that touches my figure does that make sense to you a left end point would say I'm going to go up to where my left Point touches my rectangle a midpoint would say go to the middle see where that touches the figure and then make your rectangle that would determine your height so we have left end points where that touches right end points where that touches or midpoints let me show you the midpoints and here's how you do it okay from my midpoint i' go up here where it touches that F ofx whever my function is that's where I cut it off and that makes my rectangle does that make sense to you why do that well it's actually not going to matter when we when when we finish 4.4 we'll talk about what's called the definite integral it doesn't matter whether you use left right or Center uh but for us if we are always using one end point sometimes you can systematically over underestimate the area and that would be a problem for us then this one would go okay let's go up where that touches I'll draw the top of my rectangle finish it off and you would do this for every interval that you have how do people feel okay about the idea here the idea of drawing rectangles there do you see where the midpoints coming into play here midpoint gives us the height of our rectangle that's going to be useful when we get down to 4.3 and 4.4 very useful now how would I find the area of this figure the approximation of this area what would I do with this rectang together if we know the area of a rectangle which we do right a rectangle is just a geometrical figure if I can find this area and this area and this area and this if I can find all the areas of all my rectangles all I have to do is add them together and I'm going to get at least an approximation of the area of my Figure You Follow Me Now is it going to be exact well you can tell it's not I mean sometimes I've overestimated sometimes I've left out some area sometimes it's about right but sometimes and I don't know whether I'm I'm perfect or not so I'm not going to be exactly right it it'd be a miracle if it was exactly right it's not going to be a miracle this isn't Harry Potter mathematics so magic doesn't happen uh in math sometimes but anyway that's another topic I suppose uh it's not going to be exact it's going be approximation now the question is how can I make this better Tak number so would you agree that the more rectangles I have the better approximation this will be so I mean here I had 1 2 3 4 5 I had six rectangles if I had 100 rectangles for the same width 100 rectangles mashed up in there I'm going to over underestimate by a less margin lesser of a margin right so the more rectangles I have the better approximation becomes so as the number of rectangles n goes to infinity or as the space between the rectangles goes to zero right does that make sense as space as this goes really really close to zero that's the idea of a limit as that goes to zero then I have a very very good approximation effect so good that it's exact so the idea here is cut your region into rectangles every rectangle you're going to take the area of that add all the rectangles together and you're get a good approximation of your figure how do I make it better you let the number of rectangles go to Infinity the way you do that is let the space between them go to zero for a better approximation make the space between each interval notice that in this context the space between each interval that's the width of a rectangle approach zero and again the idea is well think about it if you have a finite distance A to B and you let the distance of each sub interval go to zero how many how many rectangles you going to have if you have a finite space and you let the distance between here and here go to zero you can pack in an infinite number of rectangles right if you pack in an infinite number of rectangles you will have no over underestimate it'll basically be like finding the area under a distinct point which is impossible you can't do that but if you say really really really small and you can't put even put a paper between it then it'll be exact and that's the idea that's the limiting factor same kind of basic concept of a derivative right we took the distance and made it zero and we're taking the distance and making it zero very similar idea only now we're talking about an area instead of a slope how people understood the idea of this good that's fantastic uh more on this in section 4.3 and 4.4 this is all you get for right now just a little taste this is appetizer are you satisfied no of course you're not that's why you need more of a meal this is the meal okay different method anti-derivative method that no it's up th head pretty good H I suppose if that's the appetizer this is like the salad and the main course and we tied it all together the dessert and then the chapter five is icing on the cake baby so how to use it if you like cake answer your favorite is right here you don't put icing on a crem bluet burn it burn you do you do Burn It Anyway anti-d driven method this I'm going to go a little bit past where the book takes you the book has you do just what anti uh how to do anti-derivatives what I'm going to do is tell you what an anti-derivative stands for and then how to do them you're going to find out now I'm going to prove this to you after we do 4.3 and 4.4 so I'm going a little bit beyond that if AMX here's kind of the special deal this is what you find out the end game of uh section 4.4 is this basically if a ofx stands for the area under a curve for some function if a ofx is the area function for some f ofx some curve like this like this for f ofx on some interval I've got to give you an interval otherwise you can't find an area does that make sense you have to have an interval areas are a definite thing uh we'll talk about definite integrals later on but areas are definite you have to have points have numbers then here's what I'm going to tell you this kind of a very cool thing uh this is where we're going to end up again I I said five times now this is where we're going to end up after section 4.3 and 4.4 you get it we're going to end up with this statement which is a fantastically marvelous statement it's magical right now because you don't know the concept behind it but it's magical what it says is um this thing right here is an are this thing is an area right it says if you take this is kind of crazy to think about but if you take the derivative that's sense for derivative right you take the derivative of your area function it will give you your original function it will give you the line the curve so basically it says this if your area function represents the area take a derivative and you get the function back does that make sense to you it shouldn't really nothing all that much I'm going to prove this to you later I'm not proving it yet what you need to know right now is that if you are given an area function take a derivative of that thing and it will give you your function back well now that's that's cool but we're never going to start with this we're always going to start with our line right that's what we want to get to so well this implies this thing this says if we want to find the area under a certain function we basically need to be able to undo this we need to undo a derivative for some interval let me go through this one time for you so you kind of get the the interplay here here's what it says it says the area of a curve is a ofx what I know for sure is if I take the first derivative of the area that's equivalent to my curve itself so if I want to find the area under my curve this statement is true right the first derivative of the area is the function so if I want to find the area of the curve under the curve what I'm going to do is undo the derivative that gives me the area so basically here's the idea you pretend you pretend that your function is a derivative if you undo that thing and treat like a derivative you're going to find What's called the anti-derivative that is the area under the curve that's the idea of the anti-derivative method we'll get more into it in a second I'll prove it to you uh well I'll prove it with an example I have to prove it to you with 4.3 and 4.4 later on or prove it a couple examples so if we're going to find the area under the curve under F ofx we must undo a prime of x to find a ofx the area would you like uh an example that you can actually kind of see I was hoping you'd say that have I lost you yet no halfway you get this right yes oh good good now this is not what we're talking about today so it's good for 4.3 do you get this no you don't yet trust me you don't you don't you won't get it fully until 4.3 4.4 uh the idea is can you find anti-derivative I'm going to teach you that in a second but I want you to understand the concept behind it as well the concept is we're trying to find the area under a curve that's why I gave you this this introduction that's the idea so I'm trying to get that in your head remember how I always kept saying what's a derivative what's a derivative you're like oh my gosh SL c a point you know that right good because I pounded it in your head and that's when to do the same thing here what we're doing is we're finding the area under a curve that's what an anti-derivative is what I've told you and I've done it without proof I'm not going to prove it till later I've told you this I said the first derivative of your area is your function so if I can undo my function by treating it as a derivative I will find the area function so basically we need to go backwards from a derivative did you get that if the are if the uh if the function is the first derivative of the area that's what that says the function is the first of the area if I can undo the derivative I get my area that's theidea that's the anti-derivative method in a nutshell very quick statement now let me do some of some of this with geometry um to show you that it does in fact work then we'll get on into some anti-derivatives and how to do those things so example if we have a function f ofx = X+ one I give you something easy to graph because I'm going to show you the graph in a second I want us to find the formula for the area on 1 tox now this is would kind of be stupid if I give you just 1 to three because we could just use the triangle and find the area but what I want to find is 1 to X and I I'll show you why this Works according to that first off can you graph this jez I hope so what's your Y intercept what your slope would you agree that that is your function we're trying to find the area of that between what I say negative 1 oh you know what I want negative one let's make that negative much much more interesting negative 1 to X now I don't know where X is I made it positive but it doesn't have to be positive what we're talking about wherever this x is is the area of that shaded region now keep in mind that this function is x + 1 what shape is that can you find the area of a triangle yes of course you can yes let's find the area of a triangle the area in terms of X notice I'm using the same terminology I used over here right that's an area function the area in terms of X wherever X happens to be is how do you find the area of a triangle again very good2 BAS side so I'm going to write 12 base height so that would be oh my gosh what's the base oh let's think about that what's the base is the base x xus one oh close how do you find the distance between two points between two numbers subract you subtract them you subtract the numbers right you subtract then so this would be the distance between here and here is xus not one not one this would be the distance between here and here would be x - one right there going to be a difference here so it' be x -1 yeah that's x + one do you see where we're getting x + one well I mean you can see it you can just see it on the graph really just intuitively this distance is X right yes that's one more unit that's x + 1 not x - one that would be over here so we have x + one SO2 okay that's x + 1 okay uh oh my gosh what's the height what's the height F okay f of x h no H's you don't want any hes you're thinking I like the thoughts I good thinking you have good thoughts none I'm happen to be right right now but that's okay all right uh tell me something if this was a three if that was a three what would the height be very good if this was a seven what would the height be eight eight you all see SE the the eight if this is an X what is the height x + one you're taking the number you're adding one to it that's what the height is so this height whatever this x is right doesn't even matter if it's here if it's a one then it's 1+ 1 it's two so this height X+ one you okay on the height of X+ one the function says the height right that's what the function means so take the value the value is X plug it into the function by plug in x i I just get X x + 1 that's what we do we don't plug it in to get X squ that's not what we do we're not squaring the number and then adding one we're just plugging the number in I said if that's a three you get 3 + 1 if that was a seven you get 7 + 1 if that's an X you get x + 1 whatever it is plus one how many people understand orig feel okay with that good then our height is also x + one you okay with this so far I'm just kind of proving this for you with an example it's not a genuine proof but I'm I'm showing you that it does work if I distribute this I'm going to get let's see 12 that's x^2 + 2x + 1 do you believe me you should cuz that's right come on now if I distribute the 1/2 even further I get x^2 / 2 + x + 12 you follow me with that so far okay now here's a cool deal would you agree that whatever I tell you for X you could find the area of that triangle if I said X is n could you find the area of a triangle yes you could you just plug in nine this is in fact our area function because we made it up geometrically it's just the base times the height divid two that's all we've done so far so good now what I want you to do is do some magic this is the area function I want you to take a first derivative of that would you do that for me I hope you got magic hope so is not that very magic I told you it was going to happen but what's the first derivative of a and that we we denote that a prime right that's the first R of a what's this X very good because you bring the two down the twos simplify you get X what's this what's this Z aha does that look familiar yeah in fact this is your function that's true every time right the derivative of your area is going to give you your function every time that's going to happen that's just basically one example showing that this is is true for this example now it doesn't prove it at length I will do that later uh but it shows you that you can do it shall we do one more would you like to see one more okay I'm going to kind of start simplifying this a little bit okay you know X squ looks like right okay now this is only between zero and one so I'm not going to draw the left side of this I'm just going to draw that do you all see that that's the area we're talking about right there yeah no is that a triangle is it an exact triangle no no okay so we can't do the same thing but we're going to use the idea the idea is here's the idea if F ofx = X2 and FX also equals the first derivative of our area function you believe that right right mhm then the first derivative of our area function equals X2 that's a basic substitution it says if our function is X2 and we know that a function is equal to the first derivative of its area function you follow me on that then we say all right well if FX is X2 then X2 equals that that's that's our jump there can you make the jump okay so basically we're asking this question can you find a function whose derivative is x s that's the question you ask can you find a function and you know what we're going to call it a of X CU that's going to be our area can you find a function a ofx whose derivative is X2 let's try a couple let's just let's just have some fun and try a couple right we know how to take derivatives pretty easily let's try one let's try uh well if we use some common sense or some critical thinking I should say would you agree that you need a power bigger than two probably because when you take a derivative that power decreases by one right so let's try X CU does it work no because this is going to give you a prime of X = 3x^2 right that's not going to work okay so we need to get rid of the three somehow so if I can somehow undo the UND divide by three then maybe oh let's try that let's try X Cub over 3 would that work let's try if I take the derivative of that thing what I would do is I I'd bring down the three right the three and the over three are going to be gone I'm going to get X squar okay does that work yeah this one not so much no this one yes worked how many people feel okay with that so far so you're undoing the you're going backwards so is it related to derivative absolutely but you're going the other way and we had something something unique right we had to have this over three somehow now let me ask you another question it's going to blow your mind up like a mind grenade what's the derivative of that what about the there's another one works this very cool actually what about that any Conant any constant any con no matter what it is any constant is going to go away when you take a DP right so what you actually get here is a family of Curves Each of which would work as an area function so here's what we've done we've started with this function and we've said okay I know the function is actually a first derivative of an area function whatever that happens to be so I'm going to set my first derivative of the area function equal to it I'm going to undo that treatent like a derivative and undoing a derivative that's why we call it the hey anti-derivative method okay you're undoing anti it's opposite of derivative means derivative so undoing opposite of you're undoing a derivative that's what we're doing here now unfortunately for us this is a this is situation right we have one two and we have an infinite number of possibilities for the area function that it could actually should be and in general that's what you have to put you have to put that some your zone out you have to put that plus C the plus C is important it represents a family of Curves Each of which could be the area function now I've also said some specific things about this I said between 0o and one which is going to help us out a little bit but if I didn't say that this would be the general area functions of X2 of X that's that's what it would be you'd have to have more spe specific information in order to give me an actual verified area do you follow me on that you have to have that so do you mind if I go over here do you have any questions about this one are you with me so far that there's more than just one of them right now was the first for you you actually asked before say what you actually asked before erasing I know right I was polite today oops so a ofx = 1/3 or X Cub 3 + C C is a constant C is also the Y intercept of a ofx if something's the Y intercept of a function you should be able to find it by plugging in zero true so what this means is that a of 0 if I plug in xal 0 can you verify that that's going to give me just the C 0 Cub over 3 is0 plus C so it's going to give me c now I don't want to blow your mind up too much but for this for this context do you understand that this this area curve is from 0 to X wherever X is in this case that's what that stands for so this interval that interval is 0 to X this one might throw you for Loop but I need you to stick with me okay if this is the area if this is the area from 0 to x from 0 to X do you follow me on that it's from 0 to X and we are finding the area from zero to what do we plug in for x one no we didn't plug in one to find the to find this what did you plug in you plug in zero when you plugged in zero What's the area from0 to zero What's the are that's basically saying what's the area of a single point can you find the area of a single point point a point is that which has no breadth so that means you have no width there so you can't find an area 0 times whatever the height is but you still get zero so I know this is this is kind of awkward to think about the the area of a single point but that's what you're doing here you're saying all right I'm going from0 to I plugged in x equals 0 that's how I got this that's zero so what this says is this is basically the area of a single point when you set x = 0 your interval is from 0 to Z that's the area under a single point the area under a single point is always zero if you took any statistics at all you know this uh with continuous distribu uh if you ask what's the probability of an exact number it's always zero you can't get it because there's infinite number of possibilities that's the same same idea the area is zero so when you said xal 0 the interval is 0 0 that is the area under a single point and that's zero so what this says is from here here's what this says it's two parts I I sneeze oh it didn't come out oh don't you hate that were you all ready for it and then nothing oh I hate that then your nose tickles like yeah I hate it nose tickles oh bummer that actually look like you sneeze for a second I tried I kind of pretended just so it look good on camera value effort thanks so here's what this this part says this says if you're if you're going to plug in zero your area goes from Z to zero that's area under single point that is zero this part says well I know when I plug in zero to my function I get C therefore if a of 0 equals 0 and a of 0 equals c how much does c equal Z yeah right what that says is though even though we had the C we had to take care of the C the C doesn't always have to be zero it can it can be numbers too we're talk about um initial value problems later on like later on this section you might have some C's that are numbers this one you get a zero so this this is what you'd be able to do from this part after you find your C you go okay then the area of that curve is X Cub over 3 now you can start plugging numbers in now that you dealt with a C on the interval of 0 to 1 the area is well let's fill that in our area went from 0 to X can you tell me what is the x value of our interval there where we're stopping you can now plug that in now that you have that c you know the C is zero right you you be able to plug in the x of one you go okay well then the area that stops at one is 1 Cub over 3 or 1/3 square units now that's our idea did you follow the idea I've given it to you in terms of geometry right now I've showed you some interesting ideas with area to kind of Flesh this thing out a little bit right now we're going to take a jump away from the actual area I just wanted to introduce it to you so it was a very long introduction do you understand the introduction how many people do you feel okay with it so rectangular method is you cut it into rectangles we'll talk about that later anti-derivative method is you go okay the first root of the area is my function let's undo a derivative that gives us an anti-derivative we somehow have to have to mess around with the c we have to figure out what C's are I'll show you better ways to do this as we go on by the way better way yeah don't I need you to understand the idea you need to get the idea I'll show you some short cuts and and ways to work around that you you follow but you have to understand the idea now we're going to practice most of the rest of our time in this section on how to actually do the anti-derivatives this was fun but I haven't showed you any techniques on how to do that yet I've said basically you need to be able to undo it and we kind of guessed and checked over here would you like some better ways to do this yes absolutely so we're going to start talking about that this is called the indefinite integral indefinite means that I'm not going to give you boundaries yet I'm not going to give you zero to one yet that be would be a definite integral that would be the area between the points that's what we were able to find right an actual number what I'm going to lead you with is how to get down to here that part this would be indefinite it say you have a plus c you don't know where you're stopping you don't know where you're starting it's just an indefinite integral because I haven't given you a range of numbers therefore you can't find the definite area we're just having the function of the area do you see the difference there if you didn't catch what I said rewind that and repeat that some of you guys kind of like weird looks you you don't quite get that I'm I'm going to leave it here I'm not going to give you numbers and start and stop at I'm going to leave you here show you how to get this and later on we'll talk about how to do the whole thing together so remember this is an introduction so indefinite integral means that you're not going to be able to find the area definitely that's why we call it indefinite oh yeah I guess I should call it this would be like a subsection so there's a new word in the board integral I'll bring it I'll bring down to what an integral actually is is it's not a hard concept um given a function let's call it f on some interval it's call it I I know this doesn't start out so fun right now okay it sounds sounds boring it's it's not that bad stup given a function f on some interval I capital F is called the anti-derivative please make a note here this is also the area function okay in general we call the anti-derivative just the capital letter of whatever the function is so if we have a function F the anti-derivative is capital F that's all we're saying here it will stand ultimately for the area function of your curve did you follow me on that that's what we'll stand for f is called an anti-derivative if this is true if you can take the first derivative of your capital F your anti-derivative and it equals your FX you'll notice I used the word an instead of the word thee think back to like 10 seconds ago maybe like 30 seconds ago when I said you had a plus c you can actually almost see it in the purple writing that doesn't quite erase that well remember how we had several actually infinite number of functions that you could take a derivative of and it will give you your original function back again that's why I say and that plus C gives you a whole bunch of them a whole lot of them and so what we're doing here is we're saying f f should be some function that when I take a derivative of it it gives me my function I'm looking at that's what happens that's what an anti-derivative is known as now we just talked about it like this right same stuff a was the anti-derivative f is the anti-derivative same stuff okay okay let me give you the example that we just talked about so capital F is the same say a of so same same as area okay you remember this example right in fact it's uh it was on the board it's on the board right there here's what we would say for this we would say that F ofx has an anti-derivative of capital F ofx and capital F ofx should be 1/3 or X Cub over 3 you okay with this so far is this the anti-derivative or an anti-derivative one it's one of many it's one of many this is an anti-derivative this is one anti der I'm going say anti-d why is it anti-derivative why well look it if you take a derivative of this does it not equal this why because the derivative with respect to X of this thing is this thing which is FX that's why it works this is all I'm saying here folks an anti-derivative is this property says I take a derivative of it and it gives me a function that's what it is this is one of those are there more there are many more there are infinitely many more I gave you I give you another one um another F ofx would be X Cub over 3 - 4 is that another one that's another that's another one of those anti-derivatives that's why in general all of them will have this form that's why we have the plus C to say okay you know what I'm kind of tired of writing every constant over here I don't want to write minus 4 I don't want plus three I don't want to write plus whatever I don't want 0 plus pi minus one2 that's all garbage I don't want to write that I want to represent this this represents all of the possible anti-derivatives do you follow me on that I can't have any other X's because that would alter the derivative but constants don't because when you take a derivative of a constant it gives you zero Z so this right here is all anti-derivatives the family of all anti-derivatives and that is also the interr I'll talk about that next time this represents the family of all possible anti-derivatives for f ofx = x^2 all possible anti-derivatives will other functions have different anti-derivatives of course they work of course they will other functions have different derivatives so they will have other anti-der is uh what I will do next time is I will talk to you about what an integral is what uh integration is it's basically you're find out the same thing as doing anti-derivative so I'll show you some notation that will be where we continue our study of calculus has brought us to this point where we're not talking about derivatives So Much Anymore in fact we're basically doing the opposite of a derivative from last time we learned that if a function has an area under its curve that's the area function or in other words that's the anti-derivative so basically it's left to us to undo a function by pretending it's a derivative and saying how can I reverse the process of taking a derivative that process is called anti-differentiation or finding the anti-derivative we we left with this example the anti-derivative of X2 is 1/3 x Cub + C why the plus C good yeah the point of anti-derivative was if I take a derivative of it in other words the derivative of3 XB + C it will give me back my original function that's the idea of an anti-derivative I take a derivative of it it gives me my original function back does that make sense to you the plus C is there because well any constant is going to go away when I take a derivative so that is the basically a family of Curves a family of area curves for which when I take a derivative of it it gives me my my original function back did you understand that from last time that's basically a last time concept now the process of finding anti-derivative is also known as integration so when you hear the word integration or find the integral it means find the anti-derivative does that make sense to you so anti-derivative and integral or anti- differentiation and integration synonymous they mean the same thing this process and I mean the process of finding the anti-derivative is also called integration integration let me show you something here here's what basically our definition in a nutshell this this is really it our definition of an anti-derivative in a nutshell says if I take the derivative I even have it on the board already if I take the derivative of my anti-derivative it gives me back my original function do you follow that if I take the derivative of this thing it must give me this thing do you follow okay there's another way that we're going to write this you see this is kind of confusing because this doesn't let us go from here to here right right this basically just says if I do this thing it will give me this thing you follow so we're going to write this a little bit differently we're going to introduce a symbol called the integral remember an integral means find the anti-derivative or perform integration which is to find the anti- derivative so this and what I'm about to write mean exactly the same thing this says if I take a derivative of the anti-derivative it gives my function this says if I find the integral of my function it will give me my anti-derivative plus C because we don't know the constant right we talked about that already it's a family of Curves by the way is that c important that constant yeah that's two points on a test or every time so you have to have the C that represents I don't know the exact integral as far as the constant and it represents a family of Curves for which I take a derivative and I get back my original so this this constant must be there and this right here is called your indefinite integral I think I referenced that before I said it's indefinite an indefinite integral because you don't have any boundaries you don't know where you're going basically right now all we're doing is finding the area function it's very similar to this you remember when I had had to find derivatives you ended with things in terms of X right so while that represented a slope it wasn't an actual slope until you plug a number in do you follow me this is the same thing this is going to represent an area but it's not an actual area until you plug something into it and that's why it's called indefinite and not definite definite would be I'm able to plug numbers in and I get areas out this is called the indefinite interval the indefinite integral and these things say exactly the same thing interchangeable this one says take a derivative of your anti-derivative you get the function this says find the integral of your function what this says is this it says find the anti-derivative of this there it is that that's the whole idea do you understand the notation we will do why is the DNX there that's a good question I'll give you that in just a second um that DX tells you what you're taking the integral with respect to so what your variable has to be so this DX right here says I'm taking this with respect to X so that variable has to be X if it's not we can't do it does that make sense it's got to be there and we we'll talk about this a little bit more as we go on and on uh basically I want to get down the the names the notation what's that called that's integral symbol or integral that's right what this means is find the anti-derivative or undo this as if it were a derivative of something so give me some function that I can take a derivative of and it will give me back what you start with that's the whole idea right here you okay with this so far one note this is also true hope this doesn't blow your mind but I'm I'm basically restating things a few different ways so you see them a couple ways let's see if you can manage to do this this is kind of critical thinking on a a lower level what's that stand for integral or what else was it called anti that's an anti-derivative right what you suppose would happen if I took a derivative of the anti- derivative what do you think it's going to give me basic function the F ofx this is true that's true statements well that DX I'll make the note right here that DX tells you what variable we are integrating with respect to so this right here says which variable to integrate for instance as why when we did this um now can actually have symbols for it let's go ahead and try to do an example that we had like probably five times but just so you get the symbols down we've actually done this problem I think we've done it I've had on the board several times we've actually done it twice the integral of x^2 DX says I want you to find the anti-derivative of this function what's the function with respect to the variable X that that should match up that has to match up for you to do it okay can't do it it doesn't at least not with our our single variable calculus that we're doing so if we're doing the integral of x^2 DX we're going to find the anti-derivative of X2 what's the anti-derivative of X2 actually already have that yeah that's the 1/3 or X Cub over 3 is 5 that doesn't quite do it yet we also have to have it's crucial we got to have that plus c ohus c right canus C well if you had minus C you could write that as plus negative C right so you're always going to write plus C so C is just some arbitrary constant that can stand for anything that's how you do most of these problems right now you find it says anti-derivative of this function with respect to that variable and we know how to do at least one of them right but we don't know how to do any other one so right now what I'm going to do I'm going to give you the basic integration table listen if it doesn't fit in the Bas basic integration table you cannot do it right now okay you can't create your own calculus here you if you do you're better than Isaac Newton or godfre lies who invented the stuff way back in the day that day was 1600s so uh probably not going to happen so don't invent this just follow the table you do have to memorize it I'm not going to put this on the board but trust me you will the only tough ones are really the uh some of the trig ones that I'm going to give you right now some of the ing integrals are are hard to memorize but think of it this way you already know the derivatives right we're just undoing them so that's where all this stuff is going to come from in fact the table I'm about to give you is going to have the derivative and then the integral that's associated with it you ready for it okay so this is called integration table Yeah basic integration table to see table would look like you know there's a lot of them there's a lot of integration tables yeah most of the back of your book has just all those weird pages right the look what are all those the curvy things someone had fun just scribbling no no no those are integrals and they mean something right they all have different formulas uh by the way this this long what is that thing integral it's integral but what does it look like what letter it's an S and what we're going to come across in section 4.3 is that that's actually a sum it's going to blow your mind but remember how I told you that you can add all the rectangles together that's what you're doing it's a sum it's a sum sum with a limit basically that's where the DX comes from I'll show you all this I'll make it very clear to you as we get there but this is what we're doing this is a way that you're adding all the infinite rectangles together and getting an area you get it kind of cool kind of cool it all makes sense in a little while so basic integration table like I said I'm going to start you off with the derivatives and we're going to come up with the integrals that that are corers of them because we know an integral undo a derivative right so if we know the derivative we should be able to somehow manage to get an integral out of that so let me ask you this what's the derivative we'll start very simply what's the derivative of X with respect to X very good one dtive of X is simply one derivative of 2x would be 2 4X would be four so the Der of X is is one what that means is this hopefully you find out what I'm going to do here what's the integral of one what do you think if the derivative of x is one the integral has to undo a derivative that's what this says here a derivative undes an integral an integral undes a derivative so they have to go in reverse there are like inverse operations like division multiplication addition subtraction square and square root same idea so if the derivative of x is one the integral of one is X sure why think about it you're undoing this like it's a derivative right that means if you take a derivative of this thing it should give you back that does it now we're missing something here plus C you okay with this so far okay you know let's do this in general what was a derivative of x to the r power what do you do with that what do you do okay we add one to the I'm I'm sorry I'm talking about derivatives here derivatives how do you take a derivative of that- Rip r okay multiply by R first and then subtract one from the r yes so like X Cub you'd have 3x^2 so this would be r x to the r minus one you follow me on that okay that means that an integral should go in completely the reverse order so where we took a derivative we multiplied and then we subtracted right the reverse order of that is ADD and then divide by that new number does that make sense to you so here we we multiplied first and then we subtracted for an integral that goes reverse that means you're going to add and then divide add and then divide so our integral says I'm going to take oops not our X I'm going to take my X I'm going to add one going to add one to that exponent and what do you think am I going to divide by r r minus one or r + one R look what you do here look we actually did this right how to you get from here to here you add one but you don't divide by two you divide by three why because when you take a derivative of it the three has to match up with that go away right so you're you're matching up the new exponent with that denominator do you follow that so this isn't R it's not Rus one it's r + one C basically it says this in English you add one to the exponent and divide by the new exponent that's what that says in English you add one to the to the exponent and divide by the new expon shouldn't that be RX R yeah this one yeah nope no I'm I'm kind of changing here I don't want to have this over here I don't want to do the same thing I'm just kind of using this as a a template to what you're you're doing you're doing this in Reverse right so this says you're multiplying and subtracting this says you're adding and dividing by the new exponent I don't want to make it look exactly the same in this case and I know you can but it's going to give you back x r what I want to give you is things so you can do this actually F cuz not you're integrals aren't always going to look like this this would be way too easy they're not always going to like that okay not not going to do that yeah know this sure that would be X to 5th too easy but what happens when you get something like this well then you have to know I'm going to add one to that I'm going to divide by the new X I'm going to put a c and that's the correct thing so when I take a derivative of this it will give me back that does that make sense to you that's the way this has to work so yeah I know that these don't match up exactly I know that also one more thing I do have to mention before we go any further notice that sometimes you're going to have this as I even have a one because really we don't even write ones most of the time in mathematics so if ever you see the integral DX you go what's in there it's a one it's a one or integral of 2 DX well that that would be 2x but this this that stands for the one between the uh between the function want integrated and the DX do you need parentheses there you don't not unless you have um a few terms have a polinomial in there or something you may or you may not um often times people don't put parentheses in there because it's implied that you're going from the integral to where it says DX okay which by the way you must have the DX if you don't have that DX I mean that says you're not even integrating anymore you don't you don't have respective which variable you have to have that it also closes off your integration does that make sense to you it's kind of like the limit remember on the limits how you had to keep writing limit over and over and over and over again if you didn't I went ha I laughed at you and I crossed out your papers and I burn them well not really but you had to have it it it's a must because that's the same idea are you okay with these so far some of you still don't like this I know get used to it if you don't like this you're going to hate me on the next view not this one uh can you tell me what's the derivative of sign yes very good with that in mind what what's the integral of cosine sin plus C sin plus C good yeah very good do you feel okay with the the integral of cosine it's it's nice to think of these as I'm looking for the reverse of a derivative I'm looking for something that when I take a derivative of it it gives me back that thing do you follow that's what you're trying to do the reason why we have it this way is because if you take a derivative of this thing you bring down r + one those cross out you subtract one from the r + one and you get R that's why we have't written that way how about some more got about five more to do function please don't freak out with this but I'm going to do this a little differently I told her not to freak out she uh the derivative of cosine what's the derivative of cosine now stop stop I'm going to change this because you know what we don't want to do we don't want to have derivatives sorry integrals of weird things I don't want to know what the integral of negative sinx is cuz are you going to see negative sin x no probably not you're going to see see things like sinx do you get it so I'm trying to make this thing this thing into the thing I have over here The Only Exception is that one we don't write the minus one we have it in in terms of R so you see it more clearly but this if this gives you Nega if the the derivative of cosine gives you negative s what's the derivative of negative cosine give you s do you get me that's what I want here I don't want to know NE integral of negative sign I want the integral of sign that's why I changed that okay you agreed that the derivative of negative cosine was positive s you agreed with that so what's the integral of s cosine are these going to take some practice to memorize you bet you yeah you bet you you go in a reverse of derivatives the opposite thing backwards OPP no no you're you're lucky there's no such thing as that there's worse things and where do we get lucky at you don't learn them in this class that's next semester it is 4 okay derivative of tan X what's the derivative of tan x f s x okay this should be pretty pain I mean painless now this one's pretty easy that means what's the integral of^ 2 x 10 x I like the plus C I heard it plus C by the way uh look at the board here real quick is it plus C like this it's actually plus C like this okay whatever the angle is and then the plus C is the very very end okay three more well let's see if you remember this one um do you remember and I'm going to change this so don't get too excited y the derivative of coent yes cose if if tangent is secant cotangent is cose square with a negative with a negative in front of it do you know what I'm going to do I bet you can you can see it it's the same thing I'm going to do over here or same thing I did over here if I have coent the derivative of cotangent gives me negative cose what I want this to be is positive cose so look what I can do I can say okay if coent gives me negative cose negative Cent should give me positive cose s are you still okay with that one so I'm just moving negatives around basically and the reason is because I want to have my cosecant here to be positive I don't want to have the negative cosecant in there so what's the integral of cose s x that's that the integral has to undo that don't forget to Plus two more SE see do you know see someone on the right side do you know SE right side right side right side middle yes that's some work to do side by the way don't forget that X it's not see tangent X it's see X tangent X see X time tangent X so this is like this well then what's the integral of see X tangent X yeah plus C if you know this one you can pretty much figure this one out has a negative but you can pretty much figure this one out yeah make flash cards up make flash cards up or something you need these you ever done flash cards before there's these like cards that you flash not like blink or Flash gra the back side and go integral integral of s tangent second squ oh yeah no but make sure you have them you also probably know what I'm going to do with this one I don't want that negative there I want the negative here so if Co the derivative of cose gives you negative cose Cent then the derivative of negative cosecant gives you positive cosecant Cent so I'm going to change that around that way I have the integral of cose xent x DX don't forget the DX is cose X plus C a't it fun isn't it cool that's not too bad here's the here's the bright side the bad side is this stuff sucks right I mean this hard to memorize especially these four these ones not so bad this is easy this is easy this is pretty easy because you deal with s and cosine more often this is okay you know what you know where people mess up I guarantee you you know where people mess up where you're going to mess up if you make a mistake where you going to mess up signs the signs the sign positive negative all the time because you're used to doing this just think about it think about it for a second if the derivative of s is cosine the integral of s can't you can't have the same thing both ways right the derivative of sign can't be cosine and the integral of sign be cosine something has to change there so here the integral of sign yeah it's negative cosine the reason is the derivative of cosine would give you negative sign so the derivative of negative cosine would give you positive sign you can think through it but it's harder people make most of the mistakes with signs almost all the time that's where where all of it happens so do that stuff the bright side is if it does not fit in this table you can't do it right now I'm going to show you one exception to that one just one uh it's a it's called integration by substitution it's very easy but again what you get must fit in this table if it doesn't fit you can't do it that's a good thing and a bad thing good thing because if it's out of our realm you won't even come up with it so you know you're going have an answer it's a bad thing because sometimes it's hard to make things fit and so I'm going to show you some examples on this uh but memorize these tables I'm going to need to erase them are there any questions before I begin did you see how I got all of them basically just undoing derivatives is the whole idea of integrals it's why it's called an anti-derivative all the T you right down the board uh no nice try good TR okay I'm I'm not getting that one anymore we we dealt with that one four times how about this one the integral of x to 4th DX does it fits your table yes that's something to a power that fits your table this is why I gave it to you this way what are you going to do with this subtract or add multiply or divide you're basically then doing a derivative so instead of multiplying and then subtracting you are adding and then dividing so on our answer here you're going to have x to the what five yes that's right you add one you and then divide by what is it write as X sure yeah in fact I think your book often writes it as 1/5 X you know it's the same thing though right x 5/ 5 or 1/ X and then don't you forget that plus C oh let's try this one notice how I have to always write my DX right if you don't have that it's not a good way to write an integral you're missing something does that fit your table yeah not quite here's what you can't what you can't do you can't do this you can't go oh not so much oh yeah now you're right no problem now yeah could you break the quotient rule by just doing the D to the top D to the top D of the bottom you can't do that you can't do you had a quotient rule for that right well now we don't have things like the quotient rule but note this is what I said to you earlier it must fit your table precisely I don't have anything in my table one nothing there's nothing one over that is there make it a make it X to the3 that would fit my table this new no oh dear this is true yes make it fit your table that's what you got to do so X to the3 hey now we have X some power now we can do it just be careful when I say add I do mean add I don't mean get X to the4 because that would be subtracting right if we're adding to that you're going to x to the -2 very good over what over2 that's what you're getting over2 yes of course because you take it over your new exponent -2 + C you might make this a little bit prettier you can do2 you could do uh 1 over 2x^2 plus C which is probably the most appropriate way to write that do you see that this goes down that stays there negative goes up you okay with this so far any one of those would be acceptable this is most acceptable yeah now by the way could you check your work hey what checks your work duh I mean yeah you should never get the wrong answer you can always check your work unless you do it really hard over the rules but most of the time you can check your work right all the time you can just gets more difficult but here with these easy ones man you shouldn't ever get one wrong shouldn't ever get one wrong somehow some of you Willow we will find a way okay is that in your table make it fit your table what do you do so we got it x to the 1/2 what do you do with the exponent you add or subtract with integrals you add so we're going to get what's 1 +2 is it one it's three so x to the 3 over this is why I do the over so you can really see this over 3 right over 3 you got me that's x 3 1 / 3 if you I'm I'm never do this again but you can see it this way x^ 3 over 1 over 3es reciprocate and multiply what you end up getting is 2 X 3 over 3 I ran it a room sorry plus C you notice a couple things about this firstly did I put the plus C right when I did the problem not really we only need the very end just the very end you can ignore it as long as you don't forget about it all together put the very end also could you check this with the derivative let's see derivative says you multiply by that hey there goes the 2/3 subtract one from it there here's the 1/2 the C's gone because C is a constant right derivative of a constant is zero that's the correct answer that's what we need to be getting ready for one more I'm not supposed to do this one I'm good mhm what do you do first yes now play along when you do this you get I'm going to do every step for you you get x to the -1 + 1 over1 + 1 right you get x to the 0 over 0 you get 1 oh crap you get oh crap why why why crap you can't so when you see this can you actually do it this right here is something out of the out of the realm of this class we we do late transcendentals we do uh functions like this later in 4B which you're you're going to get this I'll give you a little hint U what this gives you is actually the integral of this is actually lnx that's the integral of 1 /x so when you see this when you see 1/x you can't do it in this class right you're not going to be able to do it here I just want you to see it that yeah that power rule only works if your exponent is not ne1 if it is negative 1 you divide by a new exponent that is zero and you can't do that so this one's off the table for us we can't do that one this one if you really want to know this is can't do that you're familiar with Ln right natural laog l x let's cover a couple properties before we go any further and then we'll start just practicing some of these things getting more and more used to taking an integral finding those anti-derivatives uh the properties you're going to find out are very very similar to the properties of derivatives I'm just going to State them for you you'll see them uh because they're exactly the same as derivatives oh yeah that makes sense first one is if you take so properties if you take and have an integral of a constant time some function that has an X in it what do you think you can do with that constant we can and typically you see this a lot more with integrals with derivatives we didn't really do that remember we said we could but we ended up just multiplying with with integrals you really do you do it a lot you can take constant outside of your integral just like you could with a derivative possible yes I'm going to put two and three together here's two and three two and three say let's suppose that you had two functions being added together within your integral like that with derivatives we could separate those couldn't we we take the derivative of the first one plus the the derivative of the second that means here that works as well you get the integral of the first one plus the integral of the second what we couldn't split up was that right we still can't we still can't do that so you can't split up two integrals like that either it's very much like der because you're going backwards from a derivative so you can't do that either but this you can and this you this is two and three this says you can take the integral of f ofx DX plus with plus minus for minus G says I can take the integral piece by piece basically that's what that says this one is not true not still okay let me caution you about the C's you know we have to have plus C's right let me caution you on on how to do the C's I I think I mentioned this just like 20 seconds ago but I want to make sure you see an example let me give you a real quick example on this um if you had the integral of let's just do 2x no that two is a constant two I should be able to pull out in front of my integral and I can so this would be 2 * the integral of x remember how I said your integral has to fit your integration table this does what's the integral of x oh dear are we doing derivatives think think think x two is correct yes oh my gosh I eras the integration table you all forget IM start bra I did two two 2 2 x^2 / 2 + C does that make sense to you you have the two you have the x^2 over 2 plus C now for those of you who have written it like this for those people who think oh I need the plus C immediately well that gets confusing because what are you supposed to do do this if you do that what do you get what do you really get you get x2 right you get x s plus plus 2 C however what is plus 2 C two two two times a what two * which gives you another Conant so does it matter there 2 C no no just C that's why I said there there's you have options you can put the plus C immediately and be careful on it or really just ignore it and put at the very end of your problem can you do that for me here's another example about why this is kind of useful I'll give you one more integral of 1 + x DX can I split split that up this no that's the integral of 1 DX plus the integral of x DX here's what I mean about the C's just watch just watch okay watch for a second you write down later the integral of 1 DX is X true if you said oh x + C you have to call it C1 cuz it's going to be another constant plus this is x^2 over 2 yes we just did that plus C2 but then when you add them together you get x + C1 + x^2 / 2 + C2 and you get x + x^2 / 2 + C1 + C2 by re recom meting those things follow that how much is C1 plus C2 it's constant plus constant another constant who cares what it is it's going to be a constant does that make sense so here ignore the C1 and the C2 that's too hard to deal with so basically you'd have x + x^2 / 2 + just C it's just going to be a constant there you're not adding together different cing go whole crazy it stands for any constant so any constant could be C1 plus C2 but who cares integrals you have to line up you just always end it with plus C you going to have one plus C at the very end of your problem that's for indefinite integrals no matter what now one last thing you remember how it derivatives when I said I want you to take the derivative of two 2x 2 don't write this down just watch - 4x + 3 you didn't split that up did you no you did 4x - 4 and then zeros for derivatives you can do the same thing with integrals so here could you split it up yes you can but you could you get directly from here to here sure you can what's the integral of one what's the integral of x x 2 where's the C coming from you need a plus c that's the same thing I'm just saying that this is possible just like deres this is possible therefore you can go directly from here to here you with me on that what you can't do is this again you can't break the multiplication thing you can't do integral of this times integral that that doesn't work so when you have functions of X multiply together like this one is there a special for that no yeah kind of you can go direct to there there but something like this what you could not do you can't take the integral of this thing times the integral of that thing that doesn't work doesn't work at all what could you do distri dist you could distribute that would save all your problems right now you couldn't do that with that all right you can't do that that wouldn't be possible but I'm going to show you ways that on certain certain cases like this one I will show you how to do that particular example all right some integrals are going to be off limits for us in this class you get that 4B uh but a lot of them we're going be able to do but they must absolutely with all else with everything else failing fit your integration table they have to do that so whether you need to distribute uh whether you need to do something else with it make it figure table somehow that's the only way this stuff works still okay shall we try a couple more and then call it a day so that one we can't do yet integral of 4 cosine X DX let's see what do you do with the four constants we often do that we really do we say this is four * the integral of cosine X DX that makes it fit our table that's easier to recognize just don't forget about the four you got the four the four is going to be there being multiplied by whatever you get next what do we get next what's the integral of cosine think about before you answer what's the integral of cosine X DX negative or positive it should you should answer this way when you take a derivative of it does it give you positive cosine X because that's what you're looking for was it sinx or negative sinx positive sinx remember you can always check your work no sense in getting the wrong answer here the derivative of sinx is cosine positive cosine that works that's what you want to get out of that okay if you ever see parentheses around that it doesn't really mean anything special it just means you're taking the integral of the entire thing all right so this this basically means the same thing as what I've been writing as this those parentheses just say everything's included in that do you see the difference there there's really no it's just grouping it it's just saying this integral the entire thing it's not making anything special no special multiplication or whatnot can you take the integral of x do that now on your own take the integral of x write that out just do you don't have to say it just write it out integral of x is one true or false false what is it you can write 12 x^2 or you can do x^2 over two plus because we have a plus say we can separate them by addition that means we can just do them piece by piece if they're being added or subtracted not multiply divided but added subtracted integral of X2 what's that I'm going to assume your mumblings were X Cub over 3 I think some of you said that actually and then what no two C's just plus C that's that's all we want that's pretty much it how many people understand the basic idea of integrals we're in practice a whole lot more of this I'll show you things we can and can't do next time all right so we're still doing some integrals our anti- differentiation going backwards from a derivative backwards of what a derivative is so let's go ahead and talk about that problem that's what we're going to start off let should get our brains a little motivated when it says an integral that means find the anti-derivative of whatever function in terms of this variable that I give you so this says with respect to X so our function has to be in terms of X that's what we're integrating with respect to so when we have 2X 5 - 3x^2 + 4x - 8 do I have to necessarily split that up term by term or can I just find the integral of each little piece with addition subtraction each little piece is fine with multiplication division each little piece is not fine we knew that integrals just like derivatives were separable by addition and subtraction what that says is it allows you to do the integral of this piece minus the integral of this piece plus the integral of this piece minus integral of this piece and then jam it all back together you follow me on that let's go ahead let's take this integral what's the integral of 2x to the 5th can you tell me x 10 10 oh no not quite okay is the two still going to be there well you're going to reduce it out okay but is it going to be there when we start let's do that this little piece that two is a constant right when we take an integral just like a derivative the constant doesn't do much in fact if you were to break this up you'd write your constant in front of your integral now we're a little beyond that we just do the integral as we did derivatives we look at it we do it piece by piece so we say this is a two no problem the X to the 5th what's the X to the 5th become x to the 4th or x to the 6th x y six yeah over what yes the new exponent is what you divide by so we're going to get 2x 6 over 6 before we start simplifying are you foll me on this so far again you know what it's probably useful to check check this with a derivative at some point if you took a derivative here notice you're going to get 2x 5th that's nice you can check that make sure you have these right minus let's see the next piece how about 3x^2 what's the integral rule of 3x^2 okay I like the way you said 3x 3 over three because I'm going to simplify in a in a moment I going to write out the integrals just like they are and then I'll simplify just just a bit we okay so far plus plus what very good you sound like my little drones today 4x^2 2 that's good no emotion but at least you're right fantastic and then the minus 8 goes to zero right because all constants go to oh wait when do constants go to zero we're not doing derivatives we're going backwards of that so if a constant goes to zero in a derivative a constant goes with an X in an integral so this goes Min - 8 x and that's it right uhhuh over one I guess you could if you really wanted to we don't need to show the over one but we are missing something very crucial yeah plus C is important again what we're finding is a family of curves or I'm sorry a family of area curves basically from this this is a a family of all the curves such that when I take a derivative of it it'll give me that that's what we're talking about are you with me by the way if you take a derivative of this will it give you that certainly that's how we this in the first place that's why it's called the anti-derivative now we're going to simplify a little bit some of you already told me this is 1/3 x 6 or x 6 over 3 - x Cub + 2 x^ 2 - 8X plus C that's our that is our integral I show a hands tell me if we okay with this example kind of a basic example right not not too bad I am going to show you some ones that you you do have to do some work to this one was nice because everything fit in our integration table right away there are going to be some cases where that is not true so we'll take would you like to see a couple of those examples okay I think I gave you one last time that I'm not going to give you again but I want to talk about it like that we're not going to go all the way through this but I want you to to look at it I I gave you an example similar to that before can I just take an integral of this and integral of this and multiply it together does that work no no it only works Separation by addition subtraction so if I gave you this problem what's the only way you have to do it right now distri distrib distribute make it look like this and then you can do it does that make sense you get like X 4th minus uh 5x^2 + 6 at least I hope so I guess we can go all the way through it's not that hard but once you do you see the distribution there once you distribute then you can take integral it's it's fairly easy after that and that's what we're going to get you you okay on that one so even if it doesn't look exactly like this sometimes you can make it look like that by some distribution or some working with our fractions or manipulating some trick functions we'll see that on the next ones are there any questions on on this particular one you guys alive today out there yes okay you're freaking me out do you remember me telling you that with your integrals they absolutely must be exactly like on an integration table or otherwise you cannot do it that was not a lie okay that's a true statement so when I give you something like this is that right there on your integration table then you can't do it the way it is that means you have to manipulate it somehow so when we look at this what we're constantly trying to do is somehow figure way to make this fit something on an integration table that's the idea these fit because they're pols they're very easy to do with tri functions they got to match exactly well you can't even be missing a little piece we got to make them match up so when you're deal with things like this it's often nice to break them up and see if you can manipulate around those trigonometric identities that you know and love maybe not love but at least hopefully know so for instance let me let me give you a a method on how to do something some of these one method would be okay let's say I split this up as 1 / sinx time cosine X over sin x are we okay on that one do you see it's legal to do that's still cosine X over sin^2 X right now why am I'm doing this well firstly because I'm pretty good at it right I know what this is going to be but secondly because I'm trying to get some other identity because I know right now cosine X over sin^2 X that over you see the over that's nowhere to be found in your integration table is there there's no over there's no fraction on your integration table anywhere right but what there is is when you have something over sign and something over sign you can often times translate those into some other trig identities that might be on your integration table so if you see an over that's not a good thing for you right now okay if you see an over right now and later on I'll show you some other ways to to go about doing this method one is well manipulate into something you can work with if you have something over sign over sign translate that back into a different trig function and then see what you have what is one over sign that's cosecant times not tan I'm pretty sure I gave you a nice one if you haven't memorized that that uh integration table which I'm guessing you probably haven't in one day uh why don't you take out your notes and look at it find the integral of cosecant Cent cose xent X I want you to find it get it in at least read it on your own find it reading something like that reinforces the idea even more besides just hearing it from somebody so I want you to find it look at it and identify that they're right I wrote them on the board they're on the video and they're also in your book did you all find it okay so cose cang I give you a nice one because that's actually in there if it wasn't you wouldn't be able to do it it has to be there it has to so that means you're going to manipulate these around until you find something that's exactly in your table which I guess is kind of a nice thing isn't it if it's not there you can't do it sometimes making it get there is fairly hard though so this one you said negative cosecant X yes plus very good feel okay with that one let's take a look at one more that I want you to see now it looks kind of nasty but firstly before you go any further can you tell me something that is wrong with that integral right now respect to say that louder for me it's with respect to X but the variable T do you see the discrepancy if I'm asking you to take an integral with respect to T oh sorry with respect to X my my variables have to be X if my variables are T that needs to be a t if it's not then you can't do it the way this is all right you have to have that in the same variable so that's correct that's fine don't forget your DT that that's important now oh OMG is that in my integration table Heavens no no we don't have anything over anything in our integration table so this can't possibly be now one more question can I take the integral of this the integral of this and integral of that does that work no is there a quotient rule for integration no make it fit make it fit in your integration table so manipulate this somehow someone on the right hand side of the room give me an idea about how I can manipulate this to make it look maybe better say what okay I can't quite just move that up there whenever I want to because it's being subtracted but you're on the right track I like the thought can you factor stuff out Factor stuff what do you mean Factor stuff out will say that again break it apart when you when you deal with things sometimes it's often nice occasionally to be able to do this to it right which is a true statement notice that if I do that and this is a very big hint when whenever you have one term on the denominator that's possible Right two terms not possible but one term you can always do that do you see what I'm talking about if you break up each individual numerator term over the denominator Factor you get all these different fractions Each of which will be simplifi each of one so yeah absolutely correct let's make this I'll show you every step here let's make this t^2 over T 4th - 2 T 4 over T 4 and there's still a DT by the way don't o try not to overthink these things do the simplest thing that will work not the hardest thing do the simplest thing that will work that's the simplest thing that's going to work why is that the simplest thing that's going to work I don't know you tell me can you simplify each of those fractions absolutely and they're not going to have any denominators anymore so when I have t^2 over T 4 well I know that we subtract exponents when I divide like like bases so that's going to give you t^2 or t to the -2 minus how about the last one what was it oh that's great I love that does this look a whole lot better than that one yeah for sure because this fits our table and that's what we needed to do so let's do the integral of 2 sorry T2 minus 2 do it on your own see what you get out of that okay for I'm putting a dot dot dot because I'm going add more to that question after we solved the first part uh before we get there let's look at at this part uh we're going to do the integral of T2 - 2 well hopefully you got T the is it3 or netive 1 you're adding one to it -2 + 1 is 1 over1 did you make it that far minus what 2 2 T 2 T cool maybe plus C at the end now you can pull off the plus C until the very last step if you want to as long as you have the plus C we found out that plus C that's a constant doesn't matter where you have it as long as you have it at the end it's just a constant now I will make this a little prettier I'm going to move that exponent to the denominator that'll be -1 over t - 2T + C we have negative variables we should the end down negative xox yeah typically yes are there cases where you don't yeah there are but typically yeah would you if you got that one good for you fantastic integrals versus derivatives so far easier or harder harder harder hard you'll get used to it now let's look at the next question find the equation of a well you better used to it ha uh find the equation of a curve such that the slope of this point is x s let's think about what this means find the equation of a curve such that the slope at each point the slope at each point is x s what gives you the slope of an equation so basically it asks this give me an equation so that when I take a derivative of it the derivative is X2 do you got me that's basically asking for the anti-derivative that's what this question asks says give me an equation so that when I take a derivative of it when I get the slope at any point that's a derivative that's going to give me X2 so this question says okay here's what you know the slope at each point recognize that's a derivative right the derivative means a slope at any point that is dydx the slope equals X2 true the slope = x^2 how do you undo that well how do I find the equation that's going to give me that take an integral of X2 DX you're going to notice some weird things and this is this is kind of the way it works it's interesting I I can show it to you a little bit later where all these symbols come from but do you notice something this is weird but you know something about this if I treat this like a variable and I multiply I get d y = x^2 DX do you see where the DX is interesting huh if I want to undo a derivative an integral undoes that an integral undoes that on both sides integral of Dy is y aha look at that that's 1 Dy right with respect to Y 1 Dy with respect to Y will give you y does that make sense to you did I lose you yeah do you remember I lost you yeah what was this no no what was this clearly with respect to Y it is 1 equals the integral of x^2 DX which is exactly what we have right there that's very interesting it says undoing a derivative gives you an integral crazy crazy very cool anyway that was a side note Beyond a little bit of our the realm but it's interesting how that works right cuz some of you were asking where's the DX or maybe thought where's the DX from coming from why do we have that it's from doing that doing this idea undoing a derivative says well you work it this way integral of 1 Dy that's y take integral of both sides you get integral X2 DX oky dokie that works cool okay now back on to this did you need that did you want me write that back for you you have it right hey it's on video you can watch it again what is the integral of X2 DX so you saying that that's the equation right that's the what does that represent right there the family the family of curves or function the family of functions I guess function family so that when I I take a derivative of that it's going to give me X squ is that true for sure any constant now can I be specific can I give the equation in this terminology can I do that with this problem right now no not right now not right now because I haven't been specific on the C but if I change the problem just slightly and here's what I'm going to add on to it I'm going to say let's find the equation of the curve such that the slope at each point is x s and that's why I add the dot dot dot given that it passes through the point 21 given that it passes through 21 you know what it doesn't make it any harder it really doesn't what this says what this says right here is this family all the X Cub families you can think of right says any of them that are shifted up and down along the Y AIS do you notice that there's no other X's this is just X cubes shifted up and down the the y axis so where your your Y intercept is different what we're saying is now ah choose a specific one and now tell me what it is so the way to go about this is you say all right well I know for a fact what we just found out is that y = x Cub over 3 plus C not e c right that's what we have hey do you have an X do you have a y plug them in you're going to be able to find c that gives you a specific equation not a family but a specific equation so when I plug in my one for y my 2 for x let's just solve for C If I subtract 8/3 from both sides what do we get Negative 53 something like that yes no so when I do that I get C = -53 can you see that down there how do I finish out the problem well it it's not that hard it's you now know what c is right here we were stuck without a c so if I want to finish this off I go back up to here I say all right I know the equation so that when I take the derivative of it it gives me x^2 when that happens and it goes through this point it's y = XB over 3 and then what that's right let's verify that for a second is this going to go through the point 21 sure that's how we found C when I take a derivative of this does it give me X2 then this is the only equation so that when I find the slope at each point it's x s and goes through this that's that says it's not a family anymore we've narrowed it down to one of these graphs how many people feel okay with that one that's the idea this is very similar to what's called initial value problem that we're going to study right now uh we're going to talk about differential equations just an introductory not anything too drastic they have a whole class on differential equations right I'm giving you like you know when you get a new thing ice cream and you open the lid and you get the ice cream on the top you're like oh I better get that stuff cuz it melt real quick you know that the ice cream is differential equations I'm giving you the stuff off the top that's it that's the the skim tasting of ice cream has really sweet but uh you get more of the good stuff in the actual class okay turns bitter no it turn bitter it's fine what kind of ice cream you're talking about consider are fine you had garlic flavored ice cream it's actually pretty good it's not like that it's an acquired taste yeah so don't go home and say oh I learned all about differential equations no you didn't all right this this is just a little little teeny bit of it but the idea is is there uh and we're star with a basic idea let's pretend for a second and this this was basically it I'm just kind of going to illustrate a little bit more let's pretend that I have a derivative that I know like that the idea is this it's it's basically the same thing we were doing here it's can you find a function an anti-derivative so that the derivative gives you this it's the idea is um you know what may be right a different way I don't think I need that that'll make it easier for us the idea is basically the same thing if you know the derivative IS F ofx can you find an anti-derivative basically a function so that when I take the derivative of it it gives me our original function back again that's the idea it's essentially the same as this let me show you how this works with what's called an initial Val problem after we do one basic basic example let's say that Dy DX = x 4 can you tell me what y equals if the Dy DX = x 4 can you tell me what y equals that's what it would be it's the same it's the same question here do you see how it's the same question it's interesting it same thing I asked you here find me an equation so that the slope is this that's what it says find me an equation so that the derivative is this instead of slope I now have the word derivative but it's same idea because slope is a derivative so here it says well I know that Y is going to be the integral of x to 4th DX if you want that proof again here's the proof again well not even a proof but here's a manipulation of these variable Parts if I separate them I get Dy = x 4 DX yes to find y I have to undo a derivative on both sides undoing a derivative is called an integral the integral of 1 dy I can put a one anywhere that I want to the integral of 1 d y is y the integral of x to 4 DX is exactly what I have here and that's why we're we're coming up with this that's how you find anti-derivative that's that's that does that make sense to you can you follow that have i blown your mind like a mind grenade are you with it yeah all right and Joe is Right X 5th over 5 or 1 x 5th plus C over to plus C so what's a solution to a differential equation it's actually not one it's Solutions differential equations just like we have solutions for a family of Curves before I put this little piece into it the reason why I gave you this first is because this problem is this problem is a differenti of of equation it's as the same concept this right here is a differential equation that's a family of Curves you got it this right here would be an initial value condition and that's what I'm going to introduced to you over here in the terms of differential equations so both questions are the same give me the slope give me the derivative I'm sorry sorry give me the slope of uh if the slope is this find me a function so that the slope is what I've given you find me a function so that the derivative is what I've given you it's the same exact question then I can add pieces on to it here's a condition that lets you solve for C I'm going to give you an initial value problem that let you solve for C would you like to see one of those totally that was funny I'll leave that this is basically when you only need one solution based on some initial condition so this is the family of solutions right here we're going to say well we only need one of them we need something to fit a specific purpose and that's what we're going to do here so basically here's an example if you have dydx = 1 2x to the quantity Cub an initial condition looks like this it says why of some initial x value equals some initial y value that's what that says and here's what I'm going to give you this looks pretty complicated but here's what it says in in real life it'll give you like something y of one is zero that's what it would be can you tell me what is your x coordinate here what's your X the zero or the one the one the Y would be the zero this says when you plug in one you get out zero does that make sense to you that's what that says so can you first undo the derivative secondly can you use this to solve for C that's the two parts of our problem so firstly can you do that for me can you do that go for it if the derivative of y is 1 / 2x 3r we need to undo that derivative that's the integral that's the anti-derivative idea so what we want is to undo this thing Tell Me Maybe the first thing that you would do make itable make it fit the table to the top you could you'd have to be very careful with that I'll show you both ways here okay be very careful on how you handle this option one is bring the bottom to the top right now right right okay yes that's right if you do this you can't take the integral yet you cannot do it what you would have to do is split that off but you're going to get not 2 to the 3 23r over x to the I'm sorry times x3r DX do you see that how it doesn't go to eight that's actually 1/8 do you see what I'm talking about it's not8 it's 1/8 so be careful on that you get 18 x to the3 that you have to leave because that does fit your table what can you do with the 1/8 probably a good idea to do right we don't want to mess around with fractions too much because we create more fractions it's easier if I write this as 1/8 X3 DX what is the integral of x to the -3 DX can you tell me X good out of line okay well that is -1 over 16 x^2 do you see the1 over 16x2 out of that can you make it that far that's our our algebra coming back at us we're going to move our xon it down -2 * 8 is -6 but I'm going to put the negative on the top makeing that netive 1 and then and say it again very good you'll find maybe the better way than moving that up to the top maybe a little bit easier to see would you like to see that one different option here since you can't really do anything fancy yet you don't know how to deal with parentheses at all do you there's been no such thing as parenthesis maybe deal with that first and go oh well this is 1 over8 X cubed yes you follow then does the eight move to the top no it's just the X the 3 that's the integral of 1/8 * X3 and that gets you right to that step can you follow that one as well so either way you want to do that I really don't care but sometimes it's easier deal with those parentheses first off before you start moving all these pieces around because you have to end up moving it back down to the denominator anyway in either case though we're going to get -1 over 16x2 + C tell me what I found what is it it is an integral yes it is an family a family of what functions functions Works family functions that does what for us what does this family functions do solve your initial what do you mean solve for someone over here tell me what what H what does this family function said wrong Weir functions represent for us come on help me out we've we've had two ways to represent this way number one was what's a integral or an anti-derivative represent represents areas we could consider that as an area because that's an integral yes in this context this one this is a family of Curves such that when I take a derivative of it it gives me that exact thing do you follow that's what we're represent a family of Curves how we find one specific curve or one specific function is we use the initial value now this is basically the same idea as me giving you a point before only I'm not giving you a point explicitly I'm giving you a point right here like this I'm saying I'm not saying what x is but you should know what x is what is X here and Y is z it says y y of one is zero so what am I going to plug in well let's see it says the initial value for y the initial value for y is zero the initial value for x is one plus C that lets you solve for C in our case we'll have 0 = 1116 this is kind of nice and easy in this particular case you add 116 and you get 116th equal C substitute that back in and the exact equation that we're looking for to satisfy not only the differential equation but the initial value is this by show hand some we feel okay with that one so far let's try one more just to make sure you really grasp this I'll show you an example of uh how you can do it in actually real life uh how this stuff works for you and then we'll be done for the day so last one let's say Dy DX is cosine X initial value is y of 0 = 1 this one's going to be nice and quick for we know quick through it um if Dy DX equal cosin x y has to equal the integral of cine X DX we we know that we know that that is going to undo that derivative so hey can you tell me what's the integral of cosine think about it before you answer I don't want anybody to answer in the first two seconds of what you think uh think of the integral of cosine X what is it ah s or negative posi plus C the plus C is important right hey if you forget the plus C that initial value is meaningless isn't it you got to have that plus C to solve for something so you had to have that do you see why it's sign by the way take a derivative of sign you're going to get cosine now we plug some numbers in uh what do we plug in for the Y in this case that's my initial value is y z or is y1 which one y of Z gives you one the other zero that's your X this would be like y of X that's why why I had it right here y ofx equal y sub Z so our X is zero our Y is one how much is sign of zero oh boy some of you are killing me today how much is sign of zero it's definitely zero yeah 1 equal C no you you guys are right 1 equals c so our function that we were looking for is sin of x + 1 is it true that the derivative of this is going to still give you cosine X is it true that it goes with that initial value absolutely being the x is zero that means that y it would be the Y intercept so whenever we have the Y intercept would C always come out to be the same should it should yes yeah absolutely because you're plugging in zero for X it's got to be on that Y axis special case I suppose it's kind of nice when you think about it when you do an integral right you're always going to have something in terms of X or whatever your variable is plus some C which does not have your variable so that naturally if you're plugging in zeroes everything else go away except for that would you like to see a real life example are any of you in physics are any of you going to take physics oh good do you know that you can do catapults yes do a catapult example this for that Reas it's the main reason to take physics build the catap that's what I would take it I hate physics I mean I loved it I love my teacher was not very good not like the teachers we have here they're good here is that so catapult you know they're originally intended to shoot cats that's why it's called a catapult just kidding but that would be pretty fun I don't want cats especially when you don't like there's no cat I like now let's say you really suck at building a catapult and you shoot it accidentally straight up in the air right above your head is that a good thing no not with the Catapult possible it's possible seen it happen with a really badly built catapult straight up in there so let's say it accidentally shoots straight up in the air with an initial velocity of 128 ft per second I want to do a few things I want to find the position function of this for the projectile basically for the height I want to know what the maximum height is and I want to know when it's going to hit you in the head actually when it hits the ground so hit your foot when will it land on your foot when's it going to hit the go oh let's say one more thing about this uh let's say that it actually shoots straight up with an initial velocity of 128 fet per second and you're launching it off of a 16t platform so you're up 16 ft and then you launch it so maybe not straight up but just slightly over so it's going to land on the ground okay does that make sense to you or you move the platform real quick or maybe your catapult is 16 feet high so when it starts throwing it it's at 16 ft so maybe that the top of it launches it straight up something breaks I don't know you mess it up but it launches straight up from your huge catapult and it's going up there and you're like oh crap it's going to hit me and so you just watch it and you count let's see when it's going to hit you that's the the idea here cuz we would all do that naturally um let's talk about a few things that we should know firstly let's call our position function s that's typical for a position function so s of T according to time the initial height what is the height of this thing at tal 0 right when you launch so s of Z is 16 do you follow me on that 16t so right when it's launching is at 16 ft High CU that's when it's letting go what's the velocity uhuh yeah it's exactly right going positive second now here's a good question for you what is velocity in relation to a position function it's the what the what velocity is the first do you remember that's the first r so velocity is the first derivative of a position position function and it said initial velocity so the initial velocity would be this the first derivative at Time Zero because what's going to happen to your projectile is it going to stay at 128 ft per second forever no probably not cuz it's going out of space it's going to slow down immediately after you launch it out of a catapult right gravity is going to affect it we're not going to do wind resistance but gravity is going to affect it it's going to come up and then it's going to come down true so the initial velocity at Time Zero is 128 ft per second I don't know if I spelled that right probably SP it wrong not get it spell I think I have too many L's do I yeah acceleration there we go if you have acceleration like that uh what's acceleration I didn't say anything about acceleration but think outside the box here or actually think outside this globe accelerating at 9 now very good he did you hear what he said so it's going to be de accelerating decelerating it's still form of acceleration it's just in the reverse Direction so you're going to be accelerating negatively cuz he's pulling it towards the ground and he said 9.8 m/s squ which is true but I'm talking about feet oh so not 9.8 do you know 32 yes accelerate now oh wait another interesting question what's acceleration as it relates to a position curve do you remember all this rates of change position function was a position the first derivative gave you velocity the second derivative gave you acceleration so here's what we know we know the second derivative actually that's going to be a constant the second derivative is 32 ft/s squared Why is it a constant well gravity change for us so it's going to be pulling down a constant acceleration of -32 ft/s squ we will feel okay with our setup here I'm going to show you what to do with this next time we'll start here we're going to continue talking about our catapult problem and here's what we knew so far we're launching from a catapult and as the projectile leaves the arm reaches up and reaches a maximum height of 16 ft and then let's go and this thing shoots straight up in the air because you built it wrong it's going to have fine on the ground right in front of you got it it leaves the Catapult with an initial velocity of 128 ft per second now what what's going to happen as soon as it leaves the Catapult what happens to that projectile de acceleration it starts slowing down that's right the velocity starts slowing down because gravity is affecting it now what our goals here are to do is find the position function that's position according to Time find the maximum height we can do that we already know how to find maximum and minimum and we're going to find out when the projectile will hit the ground these two this is going to be based off our position function the maximum height we're probably going to have our first derivative for that you with me on this so what we what we know initially with our problem well we know that it's starting out at 16 ft so I said it starts from height of 16 ft that's because our catapult arm reaches a height of 16 ft and then let's go so the position at time equals zero isn't zero it's 16 it's like a it'd be like a pitcher throwing a ball if a pitcher throws a ball the ball doesn't start from zero right he's not going to go that'd be the worst pitcher ever he's going to throw it from up here right it starts from like 8 ft and then goes up it's not going to start from the ground same thing with our CLE so the position at 0 is 16 ft the first derivative gives us velocity so our initial velocity is 128 ft per second that means that our initial velocity at zero that that's where we start right here the velocity at zero has to be 128 ft per second right when it's leaving it's traveling at 128 ft per second straight up are you guys okay with that all right now acceleration how we came with acceleration isn't Magic it's just something to do with the Earth's gravity it says that acceleration - 32 ft/s squared means that gravity's pulling everything with that acceleration if you drop something from here it's it's going to momentarily be at a velocity of zero right right when I drop it and then gravity is going to take its effect and start moving and speed up and speed up and speed up and speed up until it hits the ground you follow me and that acceleration that's acting on is a constant -32 ft perss squared that means that the velocity changes at 32 ft per second every second so drops it's after 1 second will be at 30 per second then further and further and further and further and further so do you see where we're getting all these numbers from now the interesting thing is using just this information we're going to actually be able to work backwards and get all the stuff that we want to find so let's try to work on that first thing right now what we know is basically we only know the second derivative notice how this is S Prime of 0 and of Z we don't have a position function we just know something about the position at zero we don't have a velocity function we just know something about the velocity at zero at time equals zero are you with me on this what we do know is the acceleration function that's all we know as a matter of fact so if we know that we've got the second derivative the acceleration function in terms of t can you explain to me how I could get maybe somewhere else besides that not taking a derivative because that would give me a third derivative I don't want that but is there a way I can get from here to maybe one derivative higher if we integrate the second derivative it must give us the first derivative does that make sense well it's kind of interesting concept and that's going to work for us so here's what we know S Prime of T is going to be the integral of S Prime of t T the first derivative is the integral of the second derivative and of course you need a DT are you okay with that one all right now why are we doing this again well we don't have the Velocity function we just have the Velocity at zero so this is actually going to help us remember the initial value problems that we did earlier this is basically what that is so this says all right my first derivative is going to be the integral of whatever my second derivative was in this case it's just -32 well that's a pretty easy one what's the integral of -32 DT can you tell me that2 T why not X our it says T might as well use T 32t wait and and you said what plus is a c important yes okay you know a student asked you today why is the C why do you even need the C and the the answer is well if you take a derivative let's say this and this and this you don't have to write this down just kind of watch if you took a derivative of all these things what's the derivative here what's the derivative here what's the derivative here now if you take the integral of each one of the these are all the same right if you take the integral of all these things you're still you're going to get this right is it necessarily this one no I know that all three of these came from three different sources so I can't take the integral of this and get one thing back again it has to represent all of them how we represent all of them is there could be a constant over there that's where the plus C is coming from that's why we have that so anyway we certainly do need the plus C here so good catch on that now tell me something I can do someone on the right hand side of the room since we have S Prime of t = 32t plus C tell me what I can do now initial velocity okay so initial velocity hey I have that it says Sime of 0 equal 128 where does the zero go where's the zero go that that stands for T right so S Prime of t or Sime of 0 says I know the first derivative equal 128 when the t equal Z does that make sense to you are you sure S Prime 0 equal 128 right so Sime of 0 the 0 meant the T the 128 was the velocity at zero this means the velocity this means a velocity at zero that's 128 can you solve for C this one's really nice how much is C so with a little substitution we have the S Prime of T is -32 t + 128 appr feel okay with that so far we now have a velocity function says velocity at any time T now here's something interesting which question can we answer right now the first one the second one and the third one can we answer the position function yet we don't have that on the board can we answer the maximum height yet kind of almost almost we can get close to this one we'll talk about that in a minute but we're very close to this why where would a maximum occur when the slope is zero do we have a formula for the slope yes we do it's right there that's the slope right so when the think about that oh I broke my pen state no I just recap last time when the Catapult goes up and it stops that would be where the velocity would equal zero right then it's going start moving negatively so right where the velocity equals zero is where we should have maximum or in other words where the slope of the tangent line the is zero it's going to reach its peak and come back down so that's going to be what we're going to be using in just a bit now before we can actually find the maximum height though because we're going to have to plug that number in we need to figure out the position function right after that so let's go ahead and do that next part if we know S Prime of T is there a way you can figure out s of T our same our same basic idea s of T equals the integral S Prime of T and integral will undo that derivative so in order to go from our first derivative back to a position function let's just take an integral look so that's how we write it this would said take the integral of the first derivative of your position function this is the first derivative of your position function I substitute that back in I'm going to take an integral of this can y'all help me what's the integral of -32 T what would you do 12 okay I'm going to do this a little bit longer but yeah you guys are right I'm going to do -32 t^2 over 2 just so people can see where that's coming from I'm taking integral I'm adding one to the exponent dividing by the new exponent plus don't forget about the 128 that doesn't go to zero this is not a derivative it's integral so 128t plus C good plus C now of course you guys are right this is going to be -16 t^2 in fact if you've ever seen position functions for uh projectiles they always have that -16t ^2 in front of it why why gravity that's why it's not magic it's not magic someone actually did the work and said oh yeah yeah if we're accelerating downward 32 ft per second squared if I integrate that twice itive position function that's going to be in my problem okay now tell me people on the left hand side of the room what would be the next step that I would do here Sol for C good now back to your initial your initial function since we have that this is my position at zero my position at zero is 16 my position at zero says plug in zero to your T this is my position at zero and I knew that had to be equal to 16 it says if you plug in zero look at this what it says if you plug in zero it's got to be equal to 16 since I plugged in zero it must be equal to 16 you you okay with that how much just see clearly everything else goes zero so that means our position function is that should make sense says think about this diagnose this what's this from that part what's that from what is it from though where does it stem from well I know it's integral but is it from the height is it from the velocity or is it from the gravity oh dear goodness where did thetive 16 come from is from integrating this twice right I sure hope so this part is from Gravity we just talked about that come on people some of you are zoning out right now stop zoning out you with me that's some gravity you integrate -32 you get 32t integrated again you get -32 T ^2 over 2 that's -16 t s that's gravity what's this what's that come from ini that's your initial height in fact if you plug in time equals z you get 16 right so you start from 16 ft what's this from that has to do with your initial velocity when you start out what are you at so basically I'm shooting something off and then letting gravity take its effect on that that's what that says feel okay where that stuff is coming from at least now with any type of a projectile problem like this could you just dbe the answers your initial value is the same there's a definite pattern between the value your initial oh yeah absolutely all you need to know is your initial velocity uh gravity and your initial height and you can do basically the same thing with projecti functions now of course this doesn't take into account certain things this doesn't take into account wind resistance doesn't take into account like a propellant so you know how South Korea launched a rocket or North Korea sorry launched a Rocket yesterday and it went up and came down well we couldn't model that exactly right because of wind resistance because it probably didn't go straight up unless they're really just not very bright uh because from what I know you don't launch rockets straight up that would be stupid U but if they even if they did right we won't be able to model it great because if they have some sort of propellant that's launching it you have to take that into account how the propellant burning off things like of that nature this just basically says I'm launching letting go or maybe the rocket took off and then the moment it took off it just blasted real quick and but then it's not firing anymore that would be a different story this just has an initial velocity and we let gravity take its course as a difference there now we're almost done we have the first part we got this we know that this is equal to -16 T ^2 + 128t plus C in our case C was 16 okay part one done we can do part two and three at the same well not the same time but anytime that we want how would you do part two let's talk about that when would part two happen oh you set it to zero set what to zero not the position right no that would be a height the first derivative though gives us the velocity that says when it's zero it's going to be at its peak so if we set this equal to zero 32 T = 128 what's t equal if you divide 128 by 32 exactly four I love it four what seconds cuz T is a Time second so what what's your maximum height is it four no no that's 4 seconds how would you figure out your maximum height here's what you know your maximum height occurs after 4 seconds so after 4 seconds you're at the maximum Point uh of your height and then you start dropping how would you figure out the maximum height plug it in where your position the position function yes that's why we had to figure out the position function first so that we could actually find the maximum height we could have found out when it is the maximum height very easily uh as soon as we had that we could have done it but to figure out the actual maximum height you got to have something to plug it into so maximum height occurs at 4 seconds can someone please plug that in for me into my position function and find out exactly how high we get 784 784 ft someone else double check that for me 784 ft please it's probably going to be a whole number yeah 7 so what this says is that your catapult you built is going to launch something from a height of 16 ft it's going to go to 784 ft that's after 4 seconds and start dropping Like a Rock when is it going to hit the ground acceleration zero acceleration is never zero it's a constant hey let me ask you this question uh what is the height when you hit the ground you'll understand the height when you if you fell on your face your face would be at zero when you hit the ground right right now my face is at six six feet and then it's going to go and I'm going to get zero feet off the ground does that make sense now which one of these represents your height is it the position the first derivative or the second derivative position your position so what this says is when does this equal what number zero when does it equal zero this is actually a math SE an intermediate algebra question we just did this today I just taught this in another class it says take this set it equal to zero solve for T that's going to give you a time in fact it's a quadratic right it's going to give you two times do you see the two times I'm talking about one will be positive one will be negative you're going to take those two times which one is going to make sense for this problem the positive T or the negative T oh some of you guys are not with me today what makes sense if you're going to hit the ground after some amount of time a negative time can you launch a catapult and it hits backwards in time that'd be pretty sweet I'm sure North Korea would have loved that I launched it and it hit them in the in the past they have won the Korean War they would have won yeah so that doesn't work so what we're talking about is in your quadratic when you do quadratic formula or whatever you want out of that thing uh you're going to set this equal to zero that's not right 74 is not right it's too you both give me that then I forgot two people it's 27 you didn't do 16 oh 272 I can't it's been driving 2 how do you calculus people make simple mistakes like that I'll never get it I don't understand we have good intentions letter come on you know what you match up with good intentions and bad follow through a bunch of crap bunch of crap you have all the good intentions of the world I intended to drive well when I was drunk perfectly did I ran over 37 people but fully intended to do well yeah okay D can just suck anyway okay so make a note on your paper I don't do the math I trust you guys I should never trust you guys apparently 272 ft instead of 784 make sure you have a negative my goodness coration this is ridiculous um any I'm not going to do the rest of that you I'm going to trust you to make your own mistakes on that one uh use the quadratic formula figure out the time one will be positive one will be negative the positive time works the negative time says you launch a catapult and and and it lands in the past that doesn't make any sense at all your positive time would be when it hits the ground but you feel okay with our example we got to move on all right