Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Area and estimating with finite sums. The integral has many applications in statistics, economic science and engineering integration. We can effectively compute many quantities by breaking them into small pieces and then summing the contributions from each small part. The upper sum, the area is greater than or equal to the area of the given region. A lower sum, The area is less or equal to the area of the region. Midpoint role. We're not really sure how the area relates to the area of the region, but it's usually more accurate than the lower or the upper sum left endpoint area. On the rectangle, we always use the Y value on the left or the right endpoint area. We always use the Y value on the right of a given interval. Total distance traveled is going to be our speed times the interval of time plus our speed times the interval of time for each of these increments. Recall that displacement is just our final location minus our initial location. It is not the total distance traveled. Average value is the area under a curve divided by the interval b -, a. So if we look at some examples, we're going to start with the F of X equaling X ^3, and we want to figure out what's happening between zero and one. We're going to use a lower sum with two rectangles. So if we think of F of X equaling X ^3 from zero to 1, if we thought about our Y value here, we'd have one one, and our Y value here would be 00. So what we're going to do is we're going to split it up into rectangles. And if we have two rectangles, we know that each of the bases as a half 1/2 and 1/2 as one hole, and we're going to do the lower sums. So when we look here, we're going to have this point being 1/2, one eighth. If we look at this first interval, 0 to one half, where would the lower Y value be? Well, the lower of the two Y values is right here. So the area for the first rectangle would be a base of 1/2 times AY value or a height of 0. And we're going to add to it. When we look at the next rectangle from 1/2 to one, we have AY value of 1/8 and AY value of one, and we want the lower, so we're going to use the 1/8. So we're going to have the base of 1/2 times the height of an eighth. So using the lower sum, we're going to get a total of 116th. And what we've really done is we've looked at just this area and we can see clearly that the area that's shaded is definitely less than the total area under that curve. Well, what if we took this exact same thing and we split it up into four rectangles instead of two? So if we have 4 rectangles, we're going to have each base being 1/4. So we're going to have one one here, we're going to have 00. If we have 1/4, we're going to have one. 4th, 164th, we're going to have the one half 1/8 that we did a minute ago and we're going to have 3 fourths, 2760 fourths. So if we're talking about the lower sum, we're going to look at each interval, the lower sum on this first quarter base times the height, which is 0. On the next quarter, we want the lower sum. So we're going to have 1/4 times 164th. On the next quarter, we're going to have a fourth for the base times the height of 1/8. And then on the last quarter, we're going to have 1/4 * 2760 fourths. And if we do our computation on this, we should get 960 fourths. OK. So pictorially, what we're doing is we're taking this rectangle here and adding it with this rectangle here and adding it with that rectangle there. So we can see once again that the area with these technically 3 rectangles, because the first one was just a line segment, but those rectangles are definitely less than the area under the curve. We're going to do the same thing now, but we're going to look at the upper sums. So if we look at the upper sum of two rectangles coming back to that first picture, we know that each base was 1/2 and the upper sum now is 1/8 for the first interval and the upper sum for the 2nd interval would be 1. So if we thought about drawing our picture here, we have this being our one and this being our half. Now we're talking about the rectangles that would go from zero to 1/2 and would go from 1/2 out to one. If we looked at this area, we can see that this area is more than the area under the curve. So the upper values are bigger than the area under the curve. If we computed this together, we'd have 9 sixteenths 116th plus 1/2 half. Is 8 sixteenths doing the same thing with the four for the upper sum? If we thought about dividing this into four pieces and this time we're going to do the upper sum. So we're going to look at the higher Y value each time, the higher Y value each time, the higher Y value each time, and the higher Y value each time. Now, in this particular incident, incidents, the higher values on the right and the lowers on the left, that doesn't always happen to be the case. So we're going to have 1/4 for our base. Our first height was going to be 164 base times height. So 1/4 * 1/8 + 1/4 * 2760 fourths plus 1/4 * 1. And that should give us, I believe, 2560 fourths. Let's look at another example. The next one, we're going to use the midpoints. So we're going to look at in between -2 and two. But we're going to do midpoints doing 2 rectangles, and then we're going to do midpoints doing 4 rectangles. So the first thing is, if we thought about our curve 4 -, X ^2, we know that that's a parabola going down with the Y intercept of 04. And if we're going to split it up into two equal rectangles in between -2 and two, we can see that our base is going to be two and another two. So we're going to get 2. But now we want to use instead of one of the endpoints, we want to use the midpoint. The midpoint in between -2 and 0 is going to be at -1. So what happens here is we have some of the area that's above the curve and some of the area that's below the curve in this rectangle. Then we're going to do the same thing over here with one. So we're going to have remember my pictures are never the scale. OK, so base of two times the height. Well, the height is going to occur at -1 plus the next rectangle, which is going to have a base of two times the height that occurs at 1:00. So 2 * 4 - -1 ^2 2 * 3 + 2 * 3. So the height here is going to be approximately or. The area, I'm sorry, is approximately 12 square units. All of the previous examples should have been square units. If we did 4 rectangles instead of two rectangles. Now each of our bases is going to be one, and we're going to want to do the midpoints again. So if we know that each of our bases -2 negative 1012, then we're going to use a height of -3 halves and a height of negative 1/2 and another height of 1/2, and then a height of three halves for our four rectangles. So base of one 4 - -3 halves squared plus base of 1 * 4 minus negative 1/2 ^2 plus base of one 4 - 1/2 ^2 plus plus base of 1 * 4 - 3 halves squared. 4 - -3 halves squared is going to give us 7 fourths and a 15 fourths and another 15 fourths and a 7 fourths. So 11 square units. So as we use more rectangles, we should be getting a more accurate area underneath the curve. And eventually we're going to look at what happens if we do infinitely mini rectangles underneath. The next example we're going to look at is going to be a bottle floating down a river. So you're sitting on the Bank of a tidal river watching the incoming tide. Carry a bottle upstream. You record the velocity of the flow every 5 minutes for an hour with the results shown. About how far upstream did the bottle travel during that hour? Find an estimate using the intervals. And we're going to do it with the left end point values and a right end point value. So first of all, we want to pay attention to the fact that our units aren't the same. We have minutes here and we have meters per second over here. So if we changed our minutes into seconds, we realize that there is zero to 5 minutes. So that's 300 seconds. So each of these are going to be an interval time of 300 seconds. So we're going to take 300. And if we're going to do the left endpoints, if we thought about these being intervals from zero to five, the interval on the left endpoint would be my 1.0. The interval from 5:00 to 10:00 for the left endpoint would be 1.2. The interval from 10 to 15 left endpoint would be 1.7. So we're going to take all these values and add up the left hand side as we're going 1.21 point O 1.8, 1.5 and 1.2. So this has gone about 5220 meters. If we were going to do the right end point, each interval time was still 5 minutes or 300 seconds. But now if we're doing the right end point, we're going to do the 1.2 + 1.7 +2 point O + 1.8 + 1.6 + 1.4 + 1.2 plus one point O + 1.8 + 1.5 + 1.2 + 0. And we would get 4920 meters just taking the left side of each interval versus the right side. If you're not seeing what I'm doing, you could think about plotting this. At zero time we were at one and at five we were at 1.2. So the left side would be 0, the right side would be 1.2. From 5:00 to 10:00, we were at 1.7. So the left side of the interval 5 to 10 would be 1.2. The right side would be 1.7, etcetera, if we wanted to find the average value of a function on an interval. So we're going to look at the average value of F of X of 1 / X on the interval 1 to 9. And we're going to do it in four some intervals evaluated at midpoints. So if we thought about our graph, 1 / 1 would be one to 1 / 9, which would be 1/9. So we would be doing something that looked like this. If we want to evaluate it in four intervals, we can see that the distance here, 1 to 9 is 8. So we're going to have an interval base of every two. So one to three is 2/3 to five is 2/5 to seven is 2, seven to nine is 2. So we're going to have two times. Now we're wanting the midpoints. So the midpoint in between 1:00 and 3:00, the one midpoint in between 3:00 and 5:00, the midpoint in between 5:00 and 7:00, the midpoint in between 7:00 and 9:00. So we're going to draw up and we're going to put in our rectangle here. We're going to draw up on four to where it intersects the curve, and we're going to put in our new rectangle. We're going to go up at six to where it intersects the curve, put in a rectangle up on 8 where it intersects the curve, put in a new rectangle. So our base is 2, our height the two is going to be 1 / 2, our next interval base is going to be of two, and our height is going to be 1 / 4. We're literally just finding our Y values at those midpoints. Our next base is 2 and our height is going to be 1 / 6, and our last one is 2 with our height of 1 / 8. So if we evaluate this, we'd get 1 + 1/2 + 1/3 + 1/4 or 25 twelfths. Now that's the total area using the midpoints, but we want the average. So the average says we're going to take that total area and we're going to divide it by the interval. And the interval in this case was 9 -, 1 or a distance of eight. So 25 twelfths divided by 8 is going to give us 2596. So each interval is about 2596 or the average of the whole thing. The last type of problem we're going to talk about is the area of a triangle. And what we're going to do is we're going to look at a circle and we're going to come up with formulas no matter how many Poly polynomial sided Polygon we put inside. So if we look at the area of a triangle within a circle, we know that the area of a triangle is 1/2 base height. So if we drew a circle with our radius and our Theta and we dropped an altitude so it's perpendicular to our R side, we would know that our height is. Then the Theta sine of Theta would be H / r sine Theta H / r so H would be R sine Theta. So in 1/2 base height our 1/2 our base in this case would be our R and our height would be R sine Theta. So the area of a triangle 1/2 R-squared sine Theta. We know from our trig days and our precalc days that the area of a triangle is 1/2 AB sine C. This is the same formula, but this time our A and our B are both the radius of the triangle and our C is the angle, the central angle, so sine Theta. So if we wanted to inscribe a regular 8 sided Polygon within a circle of radius one, we're going to use this formula. So we're going to have area equaling 1/2 the radius of 1 ^2 sine. Well, if we're putting 8 triangles in there, we realized that 360° is a whole circle, and now we're subdividing that, so we have 8 triangles. So 360 / 8 is really just going to be 45. And we know that the sine of 45 is root 2 / 2. So root 2 / 4 will be the area of one of those triangles. If we had eight triangles, we're going to take that area and we're going to multiply it by 8. So eight times. Root 2 / 4, which is 2, root 2, which is approximately 2.8284. What if we did 16 sides instead of eight sides? So if we wanted a 16 sided Polygon, we'd have area equaling one half 1 ^2 sine 360 / 16. So we'd have 1/2 sine 22 1/2 degrees. And if we took sixteen of those, we'd really end up with eight sine 22 1/2° and that would be about 3.06147. What do you think would happen if we did 32 sides? So we'd have 32 * 1/2 sign 360 / 32. What happens is we get more and more sides. So this is supposed to be a thought kind of question. What happens when we had 64 sides or 128 sides etcetera? What happens as our N gets larger and larger and larger? I'm going to leave you to ponder that. Thank you and have a wonderful day.