Scribe Post # 3

Hello again, I am today's scribe. Today was the first lesson of the unit "The Definite Integral". So there was not a lot of information that was covered in class today except the basics of the unit The Definite Integral. If you take a look at the second slide that Mr. K posted just below this scribe post you will see a question like; Give a lower and upper estimate of the distance traveled in 5 seconds.

In order to find the lower estimated distance you must add all the values on the table excluding the last value of velocity at 5 seconds which is 5.7 ft/s. The reason why you must exclude the 5.7 ft/s is because the lower estimate should not have the highest velocity which was recorded. As even with the high estimated distance you do not count the lowest value on the table, because you want to find the highest estimated distance, in this case 1.4 ft/s is not used. So the lower estimated value is 1.4 + 2.7 + 3.5 + 4.5 + 5 = 17.1 the reason why we are adding is because the values in change of time is 1 so 1.4 divided by 1 is still 1.4. The upper estimated value is 2.7 + 3.5 + 4.5 + 5.0 + 5.7 = 21.4.

If you move to the third slide it asks for the actual distance travel. Well you are never going to find the actual distance, only the approximate distance travel. So you begin by adding the upper and lower estimated values and divide by two ( like when trying to get the average of two numbers). Now in this case the value is 19.25 Although there must be an error because more than likely the actual distance is not the average. So we say that there is an error of + or - 2.15.
As there is a distance of 2.15 from 19.25 to either 17.1 or 21.4.

If you move to the fourth slide it asks for representation of the lower estimate, well it is the blue graph, as it is lower then the red graph, which means it is the lower estimate.

If you move to the fifth graph it asks to fill out the rest of the table and pick values for the 0.5 second interval ( which cannot be the average of the whole seconds) so intern once done picking values find the lower and upper estimates in my case I had gotten lower: 18.25 upper: 20.4 . (When finding the estimates you have to divide by 2 because of the 0.5 second intervals) Then find the actual distance in this case after adding and dividing by 2 and you received 19.325. In this case also we received the same error which is + or - 1.075. So in other words the shorter the interval of time, the smaller the error for actual distance.

So that was the class, not much information, but a lot of knowledge on Definite Integrals.
Next class scribe will be Mark.