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Hi there. In this example, we're going to approximate the area under the graph of the function f of x. The problem states, let's approximate the area under the graph of f of x equals x, from x equal 1 x equal 2 by partitioning the interval 1, 2 into in part a, 2 subintervals of equal length and choosing t to be the left endpoint. In part b, 4 subintervals of equal length and choosing t to be the left endpoint. In part c, 6 subintervals of equal length and choosing t to be the left endpoint. And in part d, compare the approximations found in parts a through c with the actual area. We let A be the area of a region below the graph of y equals x, above the x-axis, and between the vertical lines x equals 1 and x equals 2. In part a, we're going to find 2 subintervals of equal length and choose t to be the left endpoint. Here, the graph shows that the interval from 1 to 2 is split into 2 intervals of equal length. We have one interval from 1 to 3 over 2, and another interval from 3 over 2 to 2. They're both of equal length, 1/2. So we can write that down that we take the endpoints of our interval from 1 to 2. We have 2 minus 1 divided by 2 equals 1/2. That is the width of each of these subintervals. Therefore, the subintervals are 1, 1 plus 1/2 and 1 plus 1/2, 2. We can simplify both of these subintervals. We have 1, 3/2 and 3/2, 2, with left endpoints for both of these subintervals of 1 and 3/2. So now, we can find the approximate area under the curve of f of x equals x on that interval from 1 to 2 as follows. The area is approximately f of 1. That is the value of f at x equals 1 times the width of the first subinterval. So that's 1/2. We're going to go ahead and add to that f of x evaluated at the left endpoint for the second subinterval. That is 3/2, again, times the width of the second subinterval, so 1/2. So this is equal to f of 1 is 1. We're going to multiply that by 1/2. f of 3/2 is 3/2. And we multiply that by 1/2. We go ahead and simplify. We have 1/2 plus 3/4 is equal to 5/4, or 1.25. So the approximate area under the graph of f of x equals x on the interval from 1 to 2 based on 2 subintervals with width 1/2 is 1.25. So now, we're going to do the same thing in part b, but this time, we have 4 subintervals instead of 2. Again, our interval is from 1 to 2. And we've divided that interval into 4 subintervals of equal width. So to find the width of the subintervals, we're going to take the difference of the endpoints. So we have 2 minus 1 divided by 4, the number of subintervals. And we have that the width of the subintervals are 1/4. Therefore, we have 4 subintervals where we start with the left endpoint of the first subinterval as 1. The subintervals are 1, 1 plus 1/4; 1 plus 1/4, 1 plus 1/4 plus 1/4. Then we have 1 plus 1/4 plus 1/4 to 1 plus 1/4 plus 1/4. And we add to that another 1/4 for the width of the next subinterval. And finally, we have a fourth subinterval with this left-most endpoint so it's 1 plus 1/4, plus 1/4, plus 1/4. And the right-most endpoint to this last subinterval is 2, adding 1/4 to this end point. We can simplify all these subintervals down. So we have 1, 1 plus 1/4 is 5/4, 5/4, 6/4 or 3/2, for the third subinterval 3/2. Adding a fourth to that 3/2 gives us 7/4. And finally, we have for the fourth subinterval, 7/4, 2. So now, we can find the approximate area under the graph of f of x equals x on that interval from 1 to 2 using these 4 subintervals, based on the fact that the width of each of these subintervals is 1/4. The approximate area is the evaluation of f of x, at x equals 1, times the width of the first subinterval, which is 1/4, plus f of 5/4, the left-most endpoint of the second subinterval, times the width of that second subinterval. We add to that f of 3/2, again, the left-most end-point of the third subinterval. We multiply that by the width of the subinterval. And finally, we add to that the term f of 7/4, where again, 7/4 is the left-most endpoint of the last subinterval, and multiply that by the width of the subinterval. We go ahead and evaluate f of x at each of these left-most endpoints to the 4 subintervals. And we have that f of 1 is 1, and that's times 1/4. f of 5/4 is 5/4, and that's times 1/4. f of 3/2 is 3/2, and that's times 1/4. And f of 7/4 is 7/4, and that's times 1/4. All of this simplifies to 22/16, or 11/8. And that as a decimal is 1.375. Therefore, the approximate area under the graph of f of x equals x on the interval from 1 to 2 based on 4 subintervals of equal length, 1/4, is 1.375. So now, we move on to part c. We're going to evaluate the area under the curve of f of x equals x, on the interval from 1 to 2, using 6 subintervals of equal length. And we're going to choose 1 to be the left-most endpoint. So first of all, we want to find the width of each of these 6 subintervals. So we're going to take the difference in the two endpoints. We have 2 minus 1. We're going to divide that up by 6, and so we have 1/6 as the width of these 6 subintervals. Therefore, we can find the left and right endpoints to each of the 6 subintervals as follows. We have the left-most endpoint for the first subinterval is 1. We can add 1/6 to that to get 7/6. 7/6 will be the left-most endpoint for the next subinterval. We add 1/6 to that, and we have 4/3. 4/3 will be the left-most endpoint for the third subinterval. We add 1/6 to that, and we get 3/2. 3/2 is the left-most endpoint for the fourth subinterval. We add 1/6 to that, we have 5/3. Next is 5/3 and we add 1/6 to that to get 11/6. And finally, for the sixth subinterval, we have 11/6, 2. So now, we can approximate the area under the curve as follows. A is approximately f of 1 times the width of the subinterval, 1/6, plus f of 7/6 times 1/6, plus f of 4/3 times 1/6, plus f of 3/2 times 1/6, plus f of 5/3 times 1/6. And finally, f of 11/6 times 1/6. Evaluating f of x at each of these left-most endpoints to the 6 subintervals, we have 1 times 1/6, plus 7/6 times 1/6, plus 4/3 times 1/6, plus 3/2 times 1/6, plus 5/3 times 1/6, plus 11/6 times 1/6. All this simplifies to 17/12, and that as a decimal is approximately 1.417. So we have that area under the curve of f of x equals x on the interval from 1 to 2 based on 6 subintervals of equal length, 1/6, is approximately 1.417. So now in part d, we want to compare the approximations found in parts a through c to the exact area under the curve of f of x on the interval from 1 to 2. We can do that using two approaches. In the first approach, we can actually calculate the area of the square that's shaded here, and then the area of the triangle above that square. So we have that the area under the graph of f of x equals x is equal to the addition of the area of the triangle and the area of the square. So that the area is equal to, for the box we have a side equal to 1. So that's 1 times 1. For the area of the triangle, we have 1/2 base times height. So that is 1/2 times 1, the base, and the height 1. So we can simplify that as 1 plus 1/2, or 3/2, which as a decimal is 1.5. So that's the first approach. In the second approach, we can take the area of the trapezoid formed by the area under the curve on that interval from 1 to 2. That area is equal to 1/2 times 1, which is b1, the height of the trapezoid, times, in parentheses, the addition of the two bases, b1 and b2 for the trapezoid. That's 1 plus 2. We simplify this and we have 3/2, which again as a decimal is 1.5. So based on these two approaches, we have the exact area under the curve of f of x equals x on the interval from 1 to 2 as A equals 1.5. So comparing our answer in part d to our answer is in parts a through c for the approximations of the area under the curve on the interval from 1 to 2, we see that as these subintervals increase in number from 2 to 6, that our answer gets closer and closer to the actual area of 1.5. More specifically, we have in part a, the area was 1.25 for 2 subintervals. In part b, we had 4 subintervals and the area was 1.375. In part c, we had 6 subintervals, and the area got closer to 1.5. It was 1.417. And so the exact area of 1.5 is approached the higher number of subintervals that we have.

Video Details

Duration: 14 minutes and 38 seconds
Country: United States
Language: English
License: Dotsub - Standard License
Genre: None
Views: 62
Posted by: 3play on Jul 24, 2017

Please translate to spa_la. Account ID: 585. Notes on format and other things are here: http://s3.amazonaws.com/originp3/app/translation-profiles/profiles/c728d56a6e3afc44c0a63b925c143995.html

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