in this video we will introduce and discuss the definite integral we're going to start with a generalization of the left and right sums that we've already discussed definition suppose f is a function defined over the interval from A to B divide this interval from A to B into capital N Sub intervals just like we've been doing and just like we've been doing we want these sub intervals to be of equal length of equal length and we'll call that Delta X just like we usually do so note as usual we can actually compute Delta X because it's the whole interval divided into n pieces and the whole interval has length B minus a and we divide it into n equal pieces so that gives us that Delta X is B minus a quantity ided n now what we want to do is for each interval select a point that is from the first interval pick some point and call it X1 from the second interval pick a point and call it X2 and so on and so forth from the nth interval pick a point and call it X subn now these could be the left end points or the right end points or the midpoints or any points at all but we have some point from each interval now we're ready to make our definition we call the sum Sigma K = 1 to n of F at xub K time Delta X we call this the remon sum or the nth remon sum now this sum is the same kind of sums that we've been adding together we're adding together things that look like F at XK * Delta X where the K goes from 1 to capital N so it's going to be F at X1 Delta X Plus F at X2 Delta X and so on until we get to the end F at xn Delta X this is a sum like the left sum and the right sum that we've been looking at but these points can come from anywhere in their respective intervals the left sum is an example of a remon sum where all of the points are the left end point and the right sums are also remon sums where all of the points are right end points this is a more General concept where in each interval you can pick any point that you want to make things a little bit easier instead of writing down this whole thing I'm just going to call it R subn now in this case I don't mean r as in right or right end point I mean r as in remon this is the nth remon sum the motivation for these remon sums comes from the calculations of area that we've been doing so if your function is positive you want this remon sum to be an approximation of the area and you want the approx imation to get better and better as n gets larger and for that to make precise we invoke the notion of a limit if the limit as n goes to Infinity of R subn exists then we have a special name for the result and we have a special symbol too and what is it well we denot not the limit by the following symbol so this is called an integral symbol and we're going to write a little a at the bottom and a b at the top the integral symbol looks a bit like an elongated s and that's because it is it stands for sum because it should remind us of these remon sums that Define it so this is an integral symbol and we write a little a at the bottom and a little B at the top to represent the interval from A to B that we've been working with now what goes next is f ofx DX this is equal to the Limit as capital N goes to Infinity of R subn if such a thing exists now let's remind ourselves that this limit as n goes to Infinity of RN is actually the limit of a sum and that sum is Sigma K = 1 to capital n f at XK * Delta X so you can see that our integral looks similar in form to the sum that gave rise to it and that's not an accident here in the sum we have a sigma Sigma stands for sum and similarly this integral symbol also stands for sum integral symbol because this is what we call an integral the object that we've just defined integral from A to B F f x DX is called a definite integral or in particular it's called the definite integral of f over the interval from A to B the definite integral of f over A to B that's what this is called it's called a definite integral and the symbols are supposed to indicate what it is it's a limit of these remon sums so not the notation is this elongated s for sum from A to B F ofx DX so as I mentioned this elongated s is called the integral symbol the A and B are what we call limits of integration so a is the lower limit of integration and B is the upper limit of integration the E the function f here is what we're integrating this is called the integrand the integrand is the thing that we're integrating the DX here is called a differential and it's supposed to evoke in our minds the Delta X which was a small change in X so we think think of this differential DX as being a very small change in X as well if this limit exists if the limit as capital N goes to Infinity of this remon sum R subn exists we say the integral exists the integral that we just wrote down here the integral from A to B F ofx DX exists and we say that the function f is integrable over this interval from A to B integrable as in we are able to integrate it it's not true that this limit will always exist sometimes it won't and in a case like that the integral doesn't exist that is the function is not integrable but for many functions this integral will exist and we have a theorem to that effect and the theorem says that if your function f is continuous on your interval from A to B then f is integraal on that that interval from A to B so if your function f is continuous then you can integrate it that's actually stricter than we need to be or a stronger claim than it needs to be because it's true that F can be discontinuous sometimes as long as it's only finitely many times and the discontinuities are jump discontinuities so let's write that down as part of our theorem too so in fact f is integrable if F has only finite finitely many jump discontinuities on this interval so if f is continuous or if it's continuous except for finitely many Jump discontinuities Then f is integrable on that interval we saw that these limits could lead to area under a curve if your function was non- negative at least we saw that for left sums and write sums but in fact it's true for this General remon sum so we can write that down as a theorem as well the theorem states that if f is continuous or has only finitely many Jump discontinuities Then if F at X is greater than or equal to zero for all X in the interval from A to B then this definite integral from A to B of f ofx DX equals the area under the curve now when I say area under the curve I mean something a little more specific than that I mean that if I have my XY plane and I have some function and it's integrable between A and B then the area in question is the area under the curve above the xaxis and between xal A and xals B so the integral from A to B of f ofx DX is the area of the Shaded region this tells us what the definite integral is in the specific example when f is non- negative but a natural question to ask what about when the function is negative or less than or equal to zero I can say let's draw the picture let's draw our XY axis here now I want this function to be negative so I'd better have my x- Axis up here like this so this is X and this is y and let's have some function down below the x-axis yal F ofx this is a continuous function and our theorem tells us that since this is continuous the integral exists but what does it represent well let's again mark off a spot for a and a spot for B now if I'm Computing a remon sum what does that look like in terms of this picture well I'm going to divide this interval up into a bunch of sub intervals like so and let's move our B over here so that way our intervals are of the right size there we go now suppose I'm going to compute a remon sum so I've divided my interval into N Sub intervals which in this case is 1 2 3 4 5 6 7 so I have seven sub intervals but the actual number doesn't matter it's just N N Sub intervals and in these intervals I'm going to choose a point and I can choose end points if I want but they don't have to be end points they can be any points that I want to choose like so and each one of these is going to give me the height of a rectangle angle well it's not exactly the height anymore but I'm going to construct it the same way that we did before I'm going to start on the point that I've selected and I'm going to draw a straight line until I reach the curve and as soon as I reach the curve that's going to determine in this case the base of the rectangle like so and I can do that for as many of these as I want in fact I want to do it for all of them so there we are I'm not going to actually draw it in but we know that we can we can do this for all of them now notice that these rectangles are underneath the xaxis so what about the area of these things let's draw one more here so we can look at this one well I guess we might as well if we're going to draw this one in we might as well draw them all but actually let's just leave it like that so let's look at this region right here now suppose I wanted to compute the area of this thing well I know what the length is or the width rather it's Delta X just like it always is but what about the height what is what is the height of this thing going to be well it's not going to be F of of X so these points remember that's X1 X2 X3 X4 this one's X5 then we have an X6 and an X7 now if I look at F evaluated at X5 that's going to give us this this yalue right here but notice this yvalue is below the x-axis so this is negative this is a negative number F X5 is negative in particular that means it can't be the height of the rectangle it's the opposite of the height so if you wanted to compute the height of the rectangle you'd have to put a minus sign in front of this and make it a positive number or you could also say that the height of the rectangle is the absolute value of f of X5 okay well that's fine but what does that mean for us well if we look at the remon sum R subn where in this particular case n is equal to 7 this is equal to the sum K = 1 to n f XK * Delta X but all of these quantities here are negative Delta X is still positive so the result here is now a negative number RN is a negative number but what does it represent well each of these quantities F at XK * Delta X is still the width times the height except it's not exactly the height it's the opposite of the height so this is the opposite of the height times the width so that means it's the opposite of height times width so it's the opposite of area so it's not giving you the area of these rectangles it's giving you the opposite of the area of these rectangles so that means that r subn is equal to minus an estimate of the area now if I compute the limit as n goes to Infinity that estimate of the area gets better and better and better but it's always going to have this minus sign in front of it so the integral from A to B of f ofx DX which is equal to the Limit as n goes to Infinity of R subn it isn't going to be the area it's going to be the opposite of the area so if your function is positive it gives you area but if it's negative it gives you the opposite of area and in fact we can say something a little bit more because some functions are both positive and negative and what do we do in a case like that well if f is continuous or the usual caveat now if F has finitely many jump discontinuities then we know that this integral will exist but more to the point we can actually interpret it in terms of area so what's it going to look like well the integral from A to B F ofx DX is going to equal well if the function is positive then it's going to give you area and if the function is negative it's going to give you the opposite of the area let's look at a picture before we do anything else let's suppose that this is our x- axis and let's suppose that we have a function that's positive sometimes negative sometimes and maybe positive again so this is our function yals f ofx and let's say that this is a over here and this is B over here if I were to look at my remon sums then in this first bit where the function is positive the remon sums will approximate the area same in this last bit over here but in the middle it will give us not an approximation of area but the opposite of the area so if this first part has area let's say equal to a and the second piece here has area equal to B and let's say this third piece has area equal to C then this integral from A to B of f ofx DX is going to equal the area where it's positive so a and then opposite of the area where it's negative and then where it's positive again it's going to give you the area so it's the area above the x-axis minus the area below the x-axis so if we look at our theorem here the integral from A to B of f ofx DX is going to be equal to the area above the xaxis minus the area below the x-axis let's take a look at an example that I prepared earlier so here we have an example of a function this is a function which is sometimes positive sometimes negative it is continuous over the entire interval from 0 to 7 now what we would like to do here is we would like to compute the definite integral from 0 to 7 F ofx DX what is this equal to well we're going to compute this by looking at the area some of this area is above the xaxis and some is below so let's mark the area that's above the x-axis blue just for fun and let's mark the area below the x-axis red again just for fun just for the visual aid now this integral is continuous so we know that or rather this function is continuous so we know that this integral exists the only question is how to find it and what we know is that this integral from 0 to 7 F ofx DX is equal to the area above the axis minus the area below the x-axis so what that means the Blue Area minus the red area all right well we can figure out what all of this area is because these are all triangles so the first one here which is a red triangle that has area 12 * the base time the height which is of course2 now what about this first Blue Area this is a triangle two so it's 1/2 the base which is two times the height which is also two so this is equal to two we have another triangle here which is 1 12 the base which is two and the height here is two as well so that's also two and finally over here we have 1/2 the base which is 2 * the height which is 1 so it's one so that means we have the area above the x-axis the area above the x-axis is going to be 2 + 1 minus the area below the xaxis which is going to be 12 + 2 uh so that two cancels with that two and we end up with 1 -2 is 1 12 so the definite integral of this function from 0 to 7 is 12 1 12 does not tell us anything about the area that is bounded by this curve we know that the difference between the area above the x-axis and the area below the x-axis is 1/2 but the number 1/2 doesn't itself tell us anything about the area the definite integral is related to area but it's not the same as the area if I wanted to compute the area of all of the Shaded region so let's write this as a further example compute the area bounded by yal F ofx the curve and the x-axis in that case we want to add up all of the area so that's going to be 12 + 2 + 2 + 1 which is five and a half or 11 halves now if we want to write this as a definite integral we can but it's not the integral of F this is the integral from 0 to 7 of the absolute value of f the absolute value of f is always a positive function so instead of the graph that we see here what we would have is that all of these things would be reflected so if I was just to copy and paste this whole thing and paste it down here all of these red areas would become blue areas they'd look like this so there would be that and then there would be this one so we reflect everything so that it's above the xaxis so this is the graph of y equals the absolute value of f ofx so in this case everything is positive now I lost my my leg there there we are so this is why equals the absolute value of f so that makes everything positive so all of the negative things are reflected across the x-axis and they become positive areas so now we'd want to add these areas together so this one is 12 this one is two and so we'd add them all together 12 + 2 + 2 + 1 as we said is 11 over2