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.
Applications of Integration
a cross section of an airplane wing is shown. Measurements of the thickness of the wing in centimeters at 20 centimeter intervals are 5.82 point 20.3 26.7 29 27.6 27.3 23.8 20.5 15.1 point seven and 2.8 used the midpoint rule to estimate the area of the wings cross section. Okay, since we know measurements at every 22nd centimeter intervals and we want to use the midpoint rule, what we want to do is we want to break up this wing into various, um, smaller areas and then approximate the area of those, uh, areas. So we want to make it these the size of these intervals so that we can use the mid points. Since we have measurements at every 20 centimeters, we want to break it up into areas of with of 40 centimeters. So that's going to give us 12 three for in five. So this happens at 40 centimeters 81 20 1 60 and this last one at 200. Of course, this 1st 1 is at zero. Okay, So now we want us Look at what are the mid points. So that's going to be a 20. This is going to be It's 60. This one's at 100 1 30 and one 80. Okay, so we're actually not interested in all of our measurements since we're using the midpoint rule S O, we have measurements at 20 centimeter intervals. So we're interested in the first value, which happens at 20. We're not interested in the value that happens at 40. So we skipped this number. We're interested in the number that happens at 60 centimetres. 100 centimetres come. So I'm in a mistake. This is 1 40 over here. So this is 1 40 centimeters and this is 1 80 centimeters. So we're actually not going to be using all of our measurements to approximate the area area is approximately. So What we do is we take the midpoint. So looking at this first region over here between zero and 40 are we take the measurements at the midpoint at 20. So we have 5.8 centimeters and then we multiply it by the width of this region, which is 40 centimeters. Since we're going from 0 to 40. So that's how approximation of the first region. And we're going to do the same thing for all the other four regions. So the midpoint of the second region is has a width of 26.7 centimeters and the width of this region is 40 centimeters. So we just carry on 27.6. I'm going to drop all the centimeters since everything here is our in centimeters for 40 centimeters plus 23.8 times 40 plus 15.1. I'm multiplied by 40. So after we add up all of these, well, we multiply first. Then we add up. Ah, we get the final answer of 4232 and our units are centimeters squared since we are multiplying centimeters, time, centimeters in each of these terms. So our approximation for the area of this wing is 4232 centimeters squared