41-44. {Use of Tech} Nonuniform grids
Use the indicated ...

Question

41-44. {Use of Tech} Nonuniform grids

Use the indicated methods to solve the following problems with nonuniform grids.

fig

41. A curling iron is plugged into an outlet at time t = 0. Its temperature T in degrees Fahrenheit, assumed to be a continuous function that is strictly increasing and concave down on 0 ≤ t ≤ 120, is given at various times (in seconds) in the table.

a. Approximate (1/120)∫(0 to 120)T(t)dt in three ways using a left Riemann sum, using a right Riemann sum and using the Trapezoid Rule

Interpret the value of (1/120)∫(0 to 120)T(t)dt in the context of this problem.

Accepted Answer

Hello. In this video, we are told that a pot of soup is placed on a stove at a time T is equal to 0. Its temperature, capital T, in degrees Celsius for time between 0 and 100, is measured at the following times in seconds. The function is assumed to be continuous and strictly increasing on the given interval. We want to approximate 1 divided by 100, multiplied by the definite integral from 0 to 100 of T of little TDT, using the trapezoid rule for non-uniform grids, and then we want to determine what the value represents. So, this problem is asking us to approximate the value of the given integral by using the trapezoidal rule for non-uniform grids. And so what this means is that we need to determine the area for each of the intervals from 20 to 30, 30 to 45, 45 to 62, 62 to 78, 78 to 85, and 85 to 95. Those areas are going to be defined as the following. Every area is defined as delta XI divided by 2 multiplied by TI minus 1 plus TI. Now, what we need to do in this case is we need to understand that all these areas are non-uniform, and that means that every area has an independent width that is different from each other. So the first thing we're going to need to do is we're going to need to determine the sub-intervals for each of the seconds. So, for the sub-intervals, We are going to start with delta X1. Now, Delta X1 is going to be the subinterval from 10 seconds to 0 seconds. So that is going to be defined as 10 minus 0, which is 10. Then we will repeat this process for the subinterval between 30 and 10. So 30 minus 10, which gives us 20. Then we need the sub interval for 50 minus 30. Which is going to give us 20. We need this sub interval. For 70 to 50. Which again gives us 20. Then we need one more subinterval. X of 5, which is going to be from 85 to 75, or it's 85 to 70. And that is going to give us 15. And finally, we have 1/6 subinterval, which is defined from 100 to 85, and that is going to leave us with 15. So these are going to be the independent widths for each of the sub-intervals we are working with for the trapezoid rule. Now what we need to do is we just need to determine the independent areas for each subinterval. So, starting with the independent area. From 20 to 30. That area is going to be defined as 10 divided by 2 multiplied. By 20 plus 30 And that is going to simplify to leave us with 250. So what this means is that the area of the first sub-interval from 20 to 30 is going to be 250. Then we need to repeat this process. For the sub-interval from 30 to 45. So that area is going to be defined as 20 divided by 2 multiplied by 30 plus 45, and this is going to simplify to be 750. Again, with the repeating this process from 45 to 62, that area, a sub 3, is to be defined as 20 divided by 2. Multiplied by 45 plus 62, and this is going to simplify to be 1070. For a sub 4, which is the area from 62 to 78, that area is going to be defined as 20 divided by 2. Multiplied by 62 plus 78 and this is going to simplify. To be 1400. Repeating this process for the 5th interval. This is going to be the integral from 78 to 85, and this is going to be defined as 15 divided by 2. Multiplied by 78 plus 85, and this is going to simplify to be 2,445 divided by 2. And for the final interval, a 6, this is the interval from 85 to 95 for the temperature, and that is going to be 15 divided by 2, multiplied by 85 plus 95, and this will simplify to be 1350. So, we have all the independent areas for each of the sub-intervals we are working with for the trapezoidal rule. And so what this means is that the given definite integral, which was 1 divided by 100, multiplied by the definite integral from 0 to 100 of the function capital T of little TDT can be approximated by taking the sum of all the subareas. So that is going to be the sum of all of our areas we calculated, which is 250 plus 750 plus 1070, plus 1400. Plus 2,445 divided by 2, plus 1350. So we will need to take the sum of all of these areas together, and that's going to be the approximate value of the given definite integral. And by using the help of a calculator, this is going to simplify to be 60.4. So this is going to be the value of the given definite integral. Now, what does this value mean exactly? Well, keep in mind that this definite integral represents the temperature over the amount of seconds we are working with. And so what this means is that this is going to be the average temperature of the soup over the 1st 100 seconds, and that is going to be 60.4 °C. So, 60.4 °C is the average temperature for the 1st 100 seconds. And that is going to be the solution to this problem. And with that being said, the overall solution is going to be a. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Riemann Sums

Trapezoid Rule

Average Value of a Function

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