Verifying Int. of f(x) Using Trapezoid Rule

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Discussion Overview

The discussion revolves around verifying the calculation of the integral of a function $$f(x)$$ using the trapezoid rule. Participants are examining the correctness of the calculations and the application of the trapezoid rule to estimate the integral from 0 to 1.25.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents their calculation for estimating the integral using the trapezoid rule, providing specific function values at given points.
  • Another participant points out an error in the constant factor used in the trapezoid rule calculation, suggesting that the result is twice as large as it should be.
  • A later reply confirms the correction and proposes a revised estimate for the integral, questioning if the new value appears more accurate.
  • Another participant agrees with the revised estimate, indicating it looks good.

Areas of Agreement / Disagreement

Participants generally agree on the correction of the initial calculation, but there is no explicit consensus on the final value of the integral, as it is still being discussed.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations and the specific application of the trapezoid rule, particularly concerning the number of intervals used.

Who May Find This Useful

Students or individuals interested in numerical methods for integration, particularly those learning about the trapezoid rule and its application in estimating integrals.

shamieh
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Just need someone to verify my solution and that I have done my calculations correctly. Thank you again for all of your help in advance, everyone on this forum has helped me SO much with my classes this semester.

Using the trapezoid rule.

The following data was collected about the function $$f(x)$$

Estimate $$\int^{1.25}_0 f(x) \, dx$$

$$x | f(x)$$

$$0 | 3.000$$

$$.25 | 2.540$$

$$.50 | 1.583$$

$$.75 | 1.010$$

$$1.00 | 1.346$$

$$1.25 | 2.284$$

I ended up with $$.25[3.000 + 2(2.540) + 2(1.583) + 2(1.010) + 2(1.346) + 2.284] \approx 4.5605$$
 
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The constant in front should be:

$$\frac{b-a}{2n}=\frac{1.25-0}{2\cdot5}=0.125$$

Your result is twice as large as it should be. :D
 
MarkFL said:
The constant in front should be:

$$\frac{b-a}{2n}=\frac{1.25-0}{2\cdot5}=0.125$$

Your result is twice as large as it should be. :D

I knew something didn't look right...lol
Thanks Mark!

- - - Updated - - -

so does $\approx 2.28025$ look more accurate?
 
Yes, that looks good. :D
 

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