Finding area under cosx using midpoint rule

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Homework Help Overview

The discussion revolves around using the midpoint rule to approximate the integral of the sine function, specifically ∫sin(x) dx, while there is a mention of finding the area under cos(x) in the thread title. Participants are attempting to clarify the correct application of the midpoint rule and the appropriate limits of integration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are discussing the setup of the midpoint rule, including the calculation of Δx and the evaluation of function values at specific points. There are questions about the correctness of the limits of integration and whether certain values fall within the specified interval.

Discussion Status

The discussion includes attempts to clarify the limits of integration and the number of points in the partition. Some participants have provided guidance regarding potential issues with calculator settings, while others are exploring the implications of using different values in the calculations. There is no explicit consensus on the correct approach yet.

Contextual Notes

There is confusion regarding the limits of integration, with mentions of both 0 to 1 and the inclusion of 9/8, which may not be within the specified interval. The original poster's mention of cosine in the title adds to the ambiguity of the problem context.

pvpkillerx
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Use the midpoint rule to approximate the following integral:
∫sin(x) dx

This is what I did:
Δx = (1-0)/4 = 1/4

1/4(f(1/8) + f(3/8) + f(5/8) + f(7/8) + f(9/8))

But the answer i get is wrong. Is that the correct midpoint rule formula, and are the values I plugged in right? Any help is appreciated, thanks.
 
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pvpkillerx said:
Use the midpoint rule to approximate the following integral:
∫sin(x) dx

This is what I did:
Δx = (1-0)/4 = 1/4

1/4(f(1/8) + f(3/8) + f(5/8) + f(7/8) + f(9/8))

But the answer i get is wrong. Is that the correct midpoint rule formula, and are the values I plugged in right? Any help is appreciated, thanks.

Are the limits on your integral 0 and 1?. How many points in your partition? Is 9/8 in your interval?
 
Ye, its from 1 to 0, and n = 4.
 
LCKurtz said:
Are the limits on your integral 0 and 1?. How many points in your partition? Is 9/8 in your interval?[/QUOTE]

pvpkillerx said:
Ye, its from 1 to 0, and n = 4.

What about my last question?
 
I tried with and without 9/8, both don't give me the right answer.
 
pvpkillerx said:
I tried with and without 9/8, both don't give me the right answer.

Why would you try it with 9/8 in the first place? Unless you show us your work how can we help you find what you are doing wrong? Maybe something simple like having your calculator in degree mode instead radians? Show us your calculations.
 
ohh, i got the answer, u were right, the calculator was suppose to be in radians mode -_- oops. Thanks.
 
pvpkillerx said:
Use the midpoint rule to approximate the following integral:
¢
 
pvpkillerx said:
Use the midpoint rule to approximate the following integral: \intsin(x) dx

But title says: Finding area under cosx using midpoint rule

Better establish whether sine or cosine :confused:
 

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