now the next thing we're going to talk about it it's honestly going to take me about 2 minutes to really go over it uh then we'll start start practicing it later but what we're going to do is make the transition from here to integrals here's what you need to know it's what I've already told you but I'll reiterate it oops what we're going to talk about is the definite integral these these things before the integrals you were doing were called indefinite they had a plus c because you weren't quite sure about the area it was an area function but you didn't know what the area was now we're going to be able to find the area wait for it there we go here it is again last time I'll show it to you area right now you know area as this you know area is that right on some interval a to there's another interpretation for it it's very very similar to saying this remember how you calculated derivatives with limits and then we found a pattern for derivatives and you said to heck with these limit things this is easy I like this right derivatives are great because they don't deal with limits which is in fact a fallacy because derivatives are all about limits you just don't expressly calculate the limit you understand same thing with an integral an integral is all about limits just like this this is called a ran sum or Ryman sum however you pronounce his name incorrectly that's what most people say uh is the remon sum is taking all those rectangles and adding them up together that's why it's called a sum it doesn't really matter where that point is we typically use left right or midpoints but it doesn't matter it's an arbitrary Point what that says is that when I add up all these these rectangles I do basically the same thing going from a a limit using a derivative using limits to a derivative using the shortcuts that we basically do and that's what we can do here as well we say you know what the notation here really meshes up very nicely I've showed this to you one of the time but now I'll show you exactly what we're going to have on top of this when you take a a limit of a sum you basically add an infinite number of rectangles adding an infinite number of rectangles is denoted with this that's an integral adding an infinite number of rectangles based on some function in terms of X where does the C subk go we're not talking about a specific point we're talking about any arbitrary point that goes throughout the entire span of your your x's on an infinite number of of inter sub intervals and it's based on whatever variable you're talking about DX the width of each rectangle if the width of each each rectangle goes to zero because n goes to Infinity that becomes that idea of a limit limiting the x width that zero remember with with our derivative we took the x distance and we made it zero do you remember that and we taking the distance between two points and making it zero and that's what we're doing here the same EXA idea only now we're adding together a whole bunch of little bitty rectangles instead of finding uh slopes between two little tiny points you get it that's the idea now where's the A and the B come in here's where the A and the B come in on an interval A to B on interval A to B this says this basically add up all the rectangles with the height of f ofx that have a width of a very infinitely small value from your starting point a to your ending point of B can you already do integrals awesome I'm going to show you next probably next time in the time after that how to deal with this it's not hard I think you'll really enjoy it okay so we are still learning about something called the definite integral and our idea was that we can basically do well the idea was we had a summation right we took a limit of it that was the area we also made that into this integral and said well what the integral stands for is basically a limit of a sum it's adding up all the rectangles where the partitions are actually you know what for that to work I have to make the statement I got a qualified a little bit you know how we made equal partitions those don't have to be equal you know how we made left end points and right end points and midp points they didn't have to be there they could be any point within any type of partition that you wanted to make because as soon as you make the number of rectangles go to Infinity those all no matter what the size they all go close to zero does that make sense even though some are closer than others doesn't matter they're all going to zero anyway so it doesn't even matter they were equal we did that for the sake of making it easy on us uh but this is what this means it means adding up some all the little rectangles no matter what their size is as we go the number of rectangles to Infinity so basically this is a limit of a sum and I told you where that comes from that's like the Delta X so this is the same thing we had before which was called net signed area this is net signed area so basically this is this and this is what you've been doing your whole like a lot right aren't you glad I started you at like whatever I started you at you did it my way right the way I showed you hope so did it my way a a lot of effort not a lot of results now before I tell you how to actually compute these with very little problems because you basically already know I'm going to show you geometrically what this means okay I give you just three examples on on what exactly we are doing here so when we talk about geometrically what this represents let's talk about the integral from 1 to 4 of 2 DX now firstly can you tell me what the bounds of this integration is okay so X should start at one and end at four that's where our area starts and stops what's our function can you graph two what what is it straight line very good straight line or straight line straight so basically horizontal lines what you you probably mean by that and it's not this way it's this way yeah so this is our our two our yal 2 or f ofx = 2 and we're going from 1 to 4 this folks is what this integral represents it's this area right here that's what that is now for constants that's pretty easy right it just says oh we have this this horizontal line we're going from one spot to another let's actually calculate the area geometrically what's our our base good 1 to four gives you a three what's our height so what's the area here's what I know right now this integral is equal to six square units whatever the units are but it's an area that's what this talks about it says look you can even partition of it no matter what you do you add up the areas of those rectangles it's going to equal six you got me that's what we're talking about let's talk about a couple more let's do uh let's do this one how about x+2 could you graph X+ 2 hope so so X+ 2 says I'm going to start at 2 my slope is one so this is going to give me that line and I'm going to go from1 to two and this is the area that I would like to represent is the area of this region how do you find area of that region could you do it geometrically could you do it how might you how much you do it it up a triangle probably a triangle and a rectangle is how I would do that so maybe break this one off right here you okay with that let's find the area what's the area of this rectangle down here how are you getting three so a AE base of three a height of one so the purple area is three okay very good how about the area of this this region let's do it in blue what's the base of our our triangle what's the height sure and we divide by two so the area is also What's the total area by the way total area does always equal six just so you know just so you know the area everything is six just put six no we did something wrong here folks the height is not that three times sorry my it's not the scale but we did something wrong uh this is actually Four so 3 * 3 What Now 3 * 3 * 3 over two N9 Hales if we want to add those together we'd get 9es plus three Hales is going to be 15es you okay with the 15 Hales sorry about that one yet it is 3 * 3 15 hves we just you're just proving that not all areas equal six yes that's what I was doing that's what I was doing I knew something didn't add up right but uh my DOT was uh not exactly in the right spot how many people feel okay with our area So Far So geometrically speaking this is not so bad rectangle rectangle with a triangle as long as we calculate our height right we can do that base times height over two what about something like this one this is going to look nasty until you really think about what's going on how about that one what what's it mean what's it mean what does an integral stand for you guys over here asleep come on integral integral come on integral what's a definite integral what are we doing here areas that's what we're talking about we're find the area of whatever this function is whatever that function is whatever this function is now I think Michael might have said it but what is this function oh it's not quite a circle it's a half circle is the top or the bottom bottom negative would be the bottom there go this would be the that's the top of the circle a circle is plus and minus sare root of radius minus radius s- X so this is the top half of a circle that's a radius of one a radius of one very good radius of one that right there you okay with this so far you sure so this right here is that hey tell me someone the left hand side of the room what are my bounds of integration where am I starting and where am I stopping no zero read my integral read my integral see my integral doesn't always say start where my function starts and end where my function ends this actually says start here and there just like we did over here start here and here so basically quarter we have a quarter of a circle we have a quarter of a circle with what type of a radius how much is the radius here can you find the area of a circle area Circle what's the formula for area circle come on now R 2 what y * 1/4 this would be the whole circle we're taking a four of it so this is basically a quarter circle with a radius of one which is interesting right without even having to calculate these integrals we're able to find some of those areas we're able to do them we didn't do anything about those integrals but the idea is what you're doing is finding areas there's a geometrical approach to it if you can do the basic geometry you can do some of these integrals now of course do we always have basic geometry to do yes you would like to hope so have my fingers crossed no but if I give you something like uh X cubed there's no basic geometry with that X squ there's no basic geometry with that so we do have to have a better way and we had the the remon sums or those those limits of summations we did those that was one way what we're going to do now is think about maybe a better way to do that a faster way to do that there is no better way it's the same way it's just a short cuts about that just like how with derivatives I gave you limits first and then I gave you a quick and dirty way to do it quick and clean it's kind of nice pretty cool do you feel okay with this uh area approach to begin with can you always do it no but you you're going to practice some of this on your homework be able to break up a rectangle rectangular triangle some circle ideas now before we do go further I have to give you some properties things we can and can't do with these integrals so properties number one property okay think through this than think about this yeah exactly what's the are remember an integral is an area an integral is an area what's the area why what's the area under a single point there is zero the width would be zero the height it doesn't matter 0 times anything is zero so whenever your bound goes from one number to another it's the same number that's going to be zero no matter what the function is that's zero this is the area under a point basically which is why that doesn't work okay second property this is kind of an interesting one I I get evil looks from people on this one the interval is backwards that's exactly right which means this in terms of integrals if your integral is written backwards like what I mean the bounds are backwards see how we start at B and go to a here that would be like starting at one and going to zero what that's basically saying is the if the interal is written backwards it's talking about the area that's under the x- axis uh this is this is this definition it says that in order to get this written properly so you can do the integral this is negative A to B well it's negative integral from A to B and yeah they they really should be a lot simp but that's the definition that do you feel okay with that so far now typically our integrals are given to us in in this fashion where we start smaller number we n bigger number so this we don't use it a whole lot but it is there for you get something that we can do you all right with that one okay okay third property just like before even with definite integrals you can always pull a constant outside of your integral so if we have C * FX where C is a constant this is the same thing as C * that integral also just like before this also works with definite rles if you have any sort of functions in terms of X being added or subtracted any functions whatsoever you can split those things up by addition and subtraction as long as you don't change your bounds of integration basically just says Hey integral from A to B of f ofx plus G ofx or F ofx minus G ofx split it up we've already done this several times with indefinite I'm just telling you now this also works with definite integrals which is what we're talking about here also don't forget your your DX that's got to go with both of those integrals okay but show hands how many of you feel okay with the properties we've talked about so far there's one more I got to give you I just want to make sure you're okay on this as we as we are here so area under a point P has got to be zero this says basic well my integral is written backwards it represents a function that's actually below the x-axis we're just writing it backwards uh so we can make it negative and reverse our our bounds of integration that's great we can still pull constants outside of our integral that's still appropriate we can still separate integrals by addition and subtraction not multiplication division unless we have that constant but it is subtraction absolutely just like derivatives did now the last one is interesting is just for definite integrals this doesn't work for indefinite and you're going to see why in a second but here's what we'll say now I'll give it to you with the picture first and then write it out let's suppose that we had some function really so easy and there's some number between A and B basically like some some number c would you agree that the total area of this figure from A to B is written like that if this is f ofx then this is the total area of that whole figure would you also agree that the area of this whole figure could be written as the area of this plus the area of this that's also true statement it says if you find some number c that's between your bounds of integration then this is the same thing as a to c and then from from where to where signic very good that's another way that you can split up an integral find the area of just piece by piece you can do that as well as long as this number is between a and b and they match right up to it so you start it you end at C and then you start right again at C also I don't know if I'll call this a property but I I'll list it anyway yeah you actually could do do in Reverse these are not one way I'm not going to write it twice you need to know that if this is the case you could write one integral out of that for sure absolutely that if you have integral from a to c and then C to B and it's the same exact function matched together that works just fine okay with that all right now last up this one it's kind of a common sense thought but but think about this if f ofx is greater than or equal to zero what's that mean in terms of a graph what's that mean positive that's above or below the x-axis so if the function is always above the xaxis the function's always above the x-axis for every x on a certain interval that right there says X is an element of ab it just says X is within that okay so if you're confused on the notation says for all X's that happen to be in my interval in English that's what it says for all X's that happen to be or or an element of the interval you okay with the notation on that if this is true for everything you mean you plug in any X and it's always positive it's always above the x-axis tell me something about the area under the curve on AB it's got to be positive that's what this says if f ofx is greater than equal to zero for all these X's then the integral from A to B of f ofx DX must also be positive or equal to zero does that make sense to you can you think through the the uh the logic on that the critical thinking on that one if your function is always up here and you can actually take an area of it then well the are is going to be positive because it's always above the xaxis you don't have that net sign area which this represents that's actually below you don't have anything that takes away from it it's always above you follow same thing as in Reverse I'll just state it out for you if f ofx is less than or equal to Z for all X and A B then the integral from A to B of FX DX is less than equal to Z so it's kind of same statement basically those are really all the six properties we have area could be zero if the numbers are exactly the same our bounds are the same it's are in point no problem we can flip bounds of integration by making our integral negative that's possible we can fill out constants we can separate by addition subtraction definitely not multiplication we can also break up an integral if it is to our benefit or put one together if it's to our benefit provided our functions are the same and there's a number between there that we can split up that's okay also this is just kind of a statement but it is truth if the function's always above the x-axis areas going be positive if the function is always below the x-axis the area is going to be negative that's the idea how many people feel okay with our properties is just fine cool let's try one more geometrically and then I will go on into how to actually calculate these definite integrals trust me it's it's not hard all right if you know how to do indefinite integrals you're almost set just one more thing so last little example here let's use some of these properties and what we know about the uh the area approach to these integrals and find the area of this thing there we go now that looks kind of nasty in fact you probably wouldn't even be able to take the integral of that if you didn't think about the area here because when you look at that you're thinking well substitution can I even do that with the definite integral first of all because we haven't talked about it does that even possible will be but secondly even if I did a substitution the substitution is doesn't really work out that well for us at least not exactly so that might not even work so we have to kind of think about the geometry in this particular part of it using some of these properties tell me something I could do definitely sep SE seate SE so from 0 to one of 4 DX minus tell me something else I could do on the second part of it after I write this way I'll write every step for you yeah absolutely I can pull that constant 2 out in front of my integral you okay with that so far now this point well we got some things we've talked about already today tell me can you find that that's pretty easy actually that's going to be I'll draw it in a second that's just going to be a horizontal line at four from 0 to one find that area this one what is that that's a half circle and I'm going from zero to one that's a we've done that problem already today so what this says is 4 0 to 1 what this one says is a quarter of a circle so basically what are intergral comes down to our area is what's this whole area minus 2 * don't forget the two this was pi over what was it pi 4 we got this area minus 2 * this area so 4 - 2 * 4 if we simplify just a little bit area is 4 - piun / 2 if you really want to find a common denominator and do that I think you'd get a 8 - piun over 2 if you really want it doesn't do you any good to do that but you could if you want to do you feel okay with breaking these things up and using some some geometry so in this section that's what you're going to do I haven't taught you how to directly attack a definite integral yet I'm going to do that right now this is kind of a roundabout approach we we say let's let's think about the geometry let's let's draw a picture of it let's use our geometrical approach to find area and that's what this represents would you like to see how to hammer at a definite interal just right on