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A cross-section of an airplane wing is shown. Measurements of the thickness of the wing, in centimeters, at 20 -centimeter intervals are $5.8,20.3,26.7,29.0,27.6,27.3,$ $23.8,20.5,15.1,8.7,$ and $2.8 .$ Use the Midpoint Rule to estimate the area of the wing's cross-section.

4232 $\mathrm{cm}^{2}$

Calculus 2 / BC

Chapter 6

Applications of Integrals

Section 1

Areas Between Curves

Applications of Integration

Campbell University

Harvey Mudd College

Baylor University

Boston College

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Hello. My name is Lucas, and today we're gonna be taking a look at this calculus. Question is shown right here, and hopefully we all will be arriving at this answer shown right here. If you're following along the textbook, this is chapter six, Section one, question 19. And if you want to pause the video and write this down, feel free, but to summarize, we have this shape right here. It is 200 centimeters wide and were given the height of the shape or the context of the problem. The thickness of the wing at each 20 centimeter interval. And those numbers are given right here. With this information, we have to use the midpoint rule to estimate the area of this ship. We're going to begin with some conceptual things here. Um, a big part of calculus is finding the area under the curve because, as we will learn, uh, area under the curve has a lot of applications and calculus. Um, in many cases, such as this one, we're gonna be given very irregular looking shape, and we're not exactly gonna know how to find the exact area off this shape to get around this. We're gonna use approximation and so they were gonna be learning. One of these methods is called Raymond Some. And this is essentially using a bunch of rectangles whose areas we do know how to find. It's just a simple base, times height. And to visualize this, we're gonna be essentially putting a bunch of rectangles next to each other spanning the width of this shape, adding all of their areas together toe hopefully find a pretty good approximation of what the true area of this shape is. So translate what we're given in the textbook onto paper here, Um, this is a very poorly drawn. I'm sorry about that, but in black we have the approximate shape of what we're given. It's 200 centimeters wide and at each 20 centimeter partition, we have, ah than the height, the heights that were given in the problem. And in green. Here are the rectangles we're going to be using to add together to ultimately find approximation off our shape. And to do this, we have to use the midpoint rule formula, which is shown here. It does look like a lot to look at it first, but it's essentially it's essentially a fancy way of saying we're gonna be calculating the areas of a bunch of rectangles and adding them together. Ah, in this case, um, another way to look at this is we can view Delta X as the width of the rectangles wreck triangles. As I wrote there, that's a mistake. Sorry. Um, and in this case, Delta X er, as we can look at the problem, we can have five, uh, rectangles of equal length and each vehicle with I'm sorry, and each winth will be 40 centimeters 20 plus 20. So Delta X, in this case in our problem will be 40 centimeters and n in this case will be five. The number of rectangles that we will be using and f off What's going on here is essentially finding the height of the function between two given points. And what I mean by that is we have fo zero. The function at zero is 5.8 and the function at X equals 40. Right here is 26.7. We want to find the mid point of that. So the average of zero and 40 is 20 and the function at 20 in this case is 20.3. So in our first rectangle that we will be calculating. Here, Um, we have 20.3 here and in on this page here. Green represents the heights of the triangles. Rectangles. Sorry that we will be calculating, and Rand represents Delta X or the width of the rectangles. So for our first rectangle, our height will be f of 20. Wish, as we know, is 20.3. We'll be multiplying that by Delta X, which is 40 and once again shown right here are next rectangle shown here is 29 times 40 29 is shown right here, which is F of 60 the midpoint between 40 and 80. And we keep on going, you know, f of 100 half of 140 half of 180 each multiplied by Delta X, which is always going to be 40 centimeters in this case in this problem. So if we, uh, grab our calculators, we can find that each of these rectangles have areas shown here, and if we add all of these rectangles together, we should hopefully get 4232. And in this case, it's always. Well, it's always important to include units in this case. That will be centimeters squared if we take a look at what our rectangles look like in relation to the curve, Um, we would hope, and we should see that this is actually a pretty good approximation of what it actually is. Because, as we see here, we have some area of the rectangle that isn't included in the curve, but at the same time, in the same rectangle. We have this area here which is being excluded, and these should hopefully cancel each other out pretty well. And same thing goes for each of these rectangles shown here. And as we calculated, um, it all cancels out to a pretty good approximation. Well, if once again, little over 4000 centimeter square. Thank you.

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