2.7.3 AP calculus Exam Riemann sum

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Discussion Overview

The discussion revolves around the application of Riemann sums in the context of an AP Calculus exam problem. Participants explore how to calculate the total area under a curve represented by a function R(t) over a specified interval, considering the nuances of interval lengths and heights.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant notes that the time intervals are 3 hours apart except between 7 and 12, questioning whether to interpret this differently.
  • Another participant calculates an approximate total volume using Riemann sums, presenting the formula and resulting value of 114.9 L.
  • A participant expresses confusion about whether the intervals must be equal, suggesting that they should follow the given data instead.
  • There is a question regarding the graphical representation of the data, specifically about displaying only relevant ticks on a graph.
  • Another participant clarifies that the intervals do not need to be equal and describes how to determine the bases and heights for the Riemann sum calculation.

Areas of Agreement / Disagreement

Participants express differing views on whether the intervals must be equal, with some asserting that they do not need to be, while others seem uncertain. The discussion remains unresolved regarding the interpretation of the intervals and the graphical representation.

Contextual Notes

There is ambiguity regarding the assumptions about interval equality and the specific definitions of the terms used in the discussion, such as "ticks that apply." These aspects remain unresolved.

karush
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ok basically t is 3 hours appart except between 7 and 12 of which I didn't know if we should intemperate.
other wise it is just adding up the 4 $(t)\cdot(R(t))$s.
 

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$\displaystyle 50 + \int_4^{15} R(t) \, dt \approx 50 + (3 \cdot 6.2 + 5 \cdot 5.9 + 3 \cdot 5.6) = 114.9 \text{ L}$
 

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skeeter said:
$\displaystyle 50 + \int_4^{15} R(t) \, dt \approx 50 + (3 \cdot 6.2 + 5 \cdot 5.9 + 3 \cdot 5.6) = 114.9 \text{ L}$

ok, well I thot the intervals had to be equal but just take what is given here!

on your graph program is it possible just to show the ticks that apply?
 
karush said:
ok, well I thot the intervals had to be equal but just take what is given here!

on your graph program is it possible just to show the ticks that apply?

the intervals do not have to be equal ... what do you mean by "ticks that apply" ?

first base [4,7], height is R(7)

second base [7,12], height is R(12)

last base [12,15], height is R(15)
 

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