Trapezoidal Rule Homework: Equations and Solutions

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Homework Help Overview

The discussion revolves around the application of the trapezoidal rule for numerical integration, specifically focusing on the formulation and interpretation of the equation used to approximate the area under a curve. Participants are examining the components of the trapezoidal rule and how they relate to the problem at hand.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to clarify the formula for the trapezoidal rule and its application, questioning the presence of specific coefficients in front of the areas of segments. There is also an exploration of how to calculate the area of segments based on average heights and widths.

Discussion Status

The discussion is ongoing, with participants raising questions about the formula and the reasoning behind the coefficients used in the calculations. Some guidance has been offered regarding the interpretation of segment areas, but no consensus has been reached on the overall approach.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is a noted confusion regarding the specific formula and its application to the problem.

iRaid
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Homework Statement


2427rbk.png

Part b.


Homework Equations





The Attempt at a Solution


Well I thought the trapezoidal rule was:
[tex]\int_a^b f(x)dx = \frac{b-a}{2n}[f(1)+2f(2)+2f(3)+...+f(n)][/tex]

But when I did that I got the wrong answer and the work shows a completely different formula that I didn't know:
2dw9ekw.png


The part I'm confused about is the (2((66+60)/2)+...)
 
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(66+60)/2 is the avg height and 2 is thw dith of the segment so then 2*(66+60)/2 = area of that segment. add them up to get the approx area under the curve.
 
Why is there an 2, then 3, then 4, then1 infront of each of the areas of the segments :|
 
iRaid said:
Why is there an 2, then 3, then 4, then1 infront of each of the areas of the segments :|
The first subinterval is of width 2 (sec); the 2nd's width is 3 sec; the 3rd's width is 4 sec; the last subinterval's width is 1.
 

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