Evaluate Definite Integral Using Trapezium Rule: 4, 8 & 16 Subs

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Homework Help Overview

The discussion revolves around evaluating the definite integral of the function cos x + sin x from 0 to π using the trapezium rule with different numbers of subintervals. Participants are attempting to understand their calculations and the correct application of the trapezium rule.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants share their calculations for the integral and express confusion about the number of subintervals used. There are questions regarding the correct endpoints for the subintervals and the values obtained from the function.

Discussion Status

Some participants have pointed out potential errors in the original poster's calculations, specifically regarding the number of subintervals and the values derived from the function. There is a recognition of the need to ensure calculators are set to the correct mode (radians) for accurate results.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of guidance they can receive. There is an ongoing exploration of the assumptions made in the calculations and the definitions of the intervals.

escobar147
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∫cos x + sin x

lower limit: 0 upper limit: pi

correct answer: 4 intervals: 1.8962, 8 intervals: 1.9742, 16 intervals: 1.9936

i cannot seem to get the correct answer, here is my attempt:

x values cosx + sinx values
0 1
1.042 0.998
2.0944 0.993
3.1415 0.9984

2(0.998 +0.993) + 1 + 0.9984 = 5.9966

h= b-a/n n=4, b-a = pi

h= 0.7853

h/2(5.9966) = 2.3548

this is for the 4 sub interval part of the question and is incorrect... any help would be massively appreciated!
 
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escobar147 said:
∫cos x + sin x

lower limit: 0 upper limit: pi

correct answer: 4 intervals: 1.8962, 8 intervals: 1.9742, 16 intervals: 1.9936

i cannot seem to get the correct answer, here is my attempt:

x values cosx + sinx values
0 1
1.042 0.998
2.0944 0.993
3.1415 0.9984
You have three subintervals, not four. The endpoints of your subintervals should be at 0, [itex]\pi[/itex]/4, [itex]\pi[/itex]/2, [itex]3\pi[/itex]/4, and [itex]\pi[/itex].
escobar147 said:
2(0.998 +0.993) + 1 + 0.9984 = 5.9966

h= b-a/n n=4, b-a = pi

h= 0.7853

h/2(5.9966) = 2.3548

this is for the 4 sub interval part of the question and is incorrect... any help would be massively appreciated!
 


Mark44 said:
You have three subintervals, not four. The endpoints of your subintervals should be at 0, [itex]\pi[/itex]/4, [itex]\pi[/itex]/2, [itex]3\pi[/itex]/4, and [itex]\pi[/itex].

i believe these are the correct values:
0 1
.785 1.414
1.570 1
2.356 0
3.1415 -1

how are they found? when i plug the x values into cos x + sin x, my answers are different?
 


Your calculator should be in radian mode, not degrees!
 


HallsofIvy said:
Your calculator should be in radian mode, not degrees!

ah... i see... there's 3 hours i will never get back :)
 

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