Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Definite integrals. The limit as the partition goes to 0. Remember the partition is the biggest of the rectangular intervals of the summation K equal 1 to N of function of each C sub K. That C sub K is going to be the height the Y value, and then the delta X sub K is going to be the width of each of those rectangles. That's going to equal some I and that I by definition is going to be an integral from the lower limit of A to the upper limit of B of the function F of XDX being the variable of integration, the F of X being the integrand. The S sign sort of is called the integral sign. Now that's really the equivalent of. If we have N being infinitely mini rectangles, we'd have the limit as N goes to Infinity of K equal 1 to north of the height times the width of each of those individual rectangles equaling I. So the integral of A to B of F of XDXA continuous function is integrable. That is, if a function F is continuous on an interval AB, then the definite integral over AB exists. So by definition A to B of little F of XDX equals the capital F of B minus capital F of A. F of X is our anti derivative. F of B is the original function. So if we were trying to figure out the derivative of the original function would give us out the little F of X. We have some properties here. If we have the integral of A to B of F of XDX, that's the same thing as the opposite of the integral B to A of F of XDX. Usually our A&B are left to right, so if we're thinking about going from right to left, we're changing the order, so we're changing the orientation and hence the negative. If we go from one location to the same location, A to A, we're going to get F of A -, F of A, which is just zero. If we go A to B of KF of XDX, that K is a constant, so it can come in and out of the integral. So the K * F of A minus the K * F of B factoring out the constant K, we get F of A -, F of B. And then the definition of that capital F of A -, F of B is just the integral A to BF of XDX. The next one we can add or subtract 2 functions and it's just the integral of each of those functions done separately. If we have two different functions that we're adding, and the upper limit of one is the same as the lower limit of the other, and they're continuous, we can rewrite that as the integral of A to C of F of XDX. We could also split it up so if we have A to B of F of XDX, we could think of it as A to 0 + 0 to B, or we could think of it as A to C + C to B. We can split those pieces up so F of F of 0 -, F of A + F of B -, F of 0. The F of zeros would cancel whatever they are and you'd end up with the FB minus FA, which is what the original integral was. Max min inequalities. If F has a maximum value, we call it Max F, and a minimum value min F on the integral AB or on the interval AB, then the minimum of F times the interval B -. A has to be less than or equal to the integrin A to BF of XDX, which has got to be less than or equal to the maximum of F on the same integral. So if we're looking at this interval here, we know that we have some value that would be lowest if it's a closed and bounded interval. We have some value that's going to be the highest on the closed bounded interval. And what we're trying to figure out is that the area underneath the curve from A to B with the minimum height would be too little for the area under the curve, but the maximum height from on the interval A to B would be too big. S somewhere in between has got to be a constant that actually gives us the right value. So currently we just know that the minimum height times the interval has got to be less than or equal to the actual area under the curve, which has got to be less than or equal to the maximum area times the interval. The domination theorem says that if F of X is greater than or equal to G of X on some interval, then we know that the interval the integration from A to B of F of XDX has got to be greater than or equal to the integration of A to B of G of XDX. Remember we're talking about areas under a curve average value of a continuous function 1 / b -, A of the integral A to BF of XDX. Looking at a few examples, if we had. First of all, we're going to ask to be writing a partition as an integral. So if we have the limit as the partition goes to zero of the norm of the partition goes to 0K, equal 1 to north of two CK cubed times delta XK where the partition is going on the closed and bounded interval -1 to 0. So negative ones are lower bound, zeros are upper bound. This 2C sub K ^3 is really our Y value, our height. So it'd be two X ^3 for each and every one of those rectangles. Our delta X sub K is just our DX, it's our small change in the X direction delta. The triangle means change. So if we thought about this taking the antiderivative, we're going to add 1 to the exponent and divide by the new exponent. So we're going to get 1/2 X to the 4th from -1 to 0, so one half 0 to the 4th -1 half -1 to the 4th. So we'd get 0 -, 1/2 or negative 1/2. Now that negative really just tells me that it's the area underneath the X axis. The actual area is the absolute value of that, but the negative tells me where it's positioned. It's under the X axis. If we thought about doing our Max and our min, if we looked at the interval 0 to one, the smallest height would be a height of 0. At 0, the biggest height would be a height of one at -1. If we thought about just the heights, not taking into account the direction. So the base distance of one times the height of 0 would have to be less than or equal to the actual absolute value of the area underneath the curve, which would have to be less than or equal to the base distance of one times the height of one. And we can see that zero less than or equal to 1/2 less than or equal to one is a true statement. If we look at the next one, we have the integral of 1/2 to three halves of negative two X + 4 DX. Negative two X + 4 is a line Y interceptive 04 slope of -2 down 2 and over one down 2 and over one, etcetera. If we drew a picture of this, this would actually be a figure of a trapezoid, and we know that the area for a trapezoid is 1/2 the height. In this case, the height is along the X axis, the base one plus the base two. So 1/2 height base one, one half height times the quantity of base one plus base two is the area of a trapezoid. So if we're looking over here on the right to start, 1/2 height, base one plus base two, 1/2 the height, the height being along the X axis. So the furthest to the right minus the furthest to the left three halves -1 half. And then we actually need the height there. So we're going to use that negative two X + 4 at 1/2 to get the distance of this base and the -2 X +4 at three halves to get the height or the distance of the space. So -2 * 1/2 + 4 + -2 * 3 halves +4. When we compute that, we get the areas 2 by using a geometric formula for the trapezoid. Now we can do the integration of this. If we added one to the exponent and divided by the new exponent, we get negative X ^2 if we added one to the exponent divided by the new exponent. Remember, if there is no X, it's like X to the zero. So we'd have 4X putting in our upper bound and subtracting putting in our lower bound. Doing the math there, we get X is going to be two or the area under the curve is also two. So both directions, whether we did the integral or whether we actually used a known geometric formula. Now what's going to happen is as the problems get more complex, we're not going to have geometric formulas that we can just plug them into. Thank you and have a wonderful day.