Solved: Integral Question - Finding Average Area Under Curve

  • Thread starter Thread starter rjs123
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The discussion revolves around finding the average area under the curve of the function \( e^{-2x^2} \) over the interval [0, 2]. The original poster expresses an intention to calculate this using an integral with a prefactor of 1/2, indicating a focus on average value calculation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of substitution in evaluating the integral and question the correctness of the original poster's approach. There are mentions of numerical methods like Simpson's Rule and the Trapezium Rule as alternatives for evaluation.

Discussion Status

The discussion is ongoing, with participants providing insights into the challenges of finding an elementary solution for the integral. Some guidance has been offered regarding numerical methods, and there is an exploration of the error function as a potential avenue for further understanding.

Contextual Notes

There is a recognition that the integral does not have an elementary solution, which may affect the approach taken by the original poster. Additionally, there are indications of missing factors in the substitution attempt that could impact the evaluation process.

rjs123
Messages
90
Reaction score
0

Homework Statement


[tex]1/2\int_0^2e^{-2x^2}dx[/tex]

I'm basically trying to find the average area under the curve for a rectangle from [0, 2] inclusive...which is why i put the 1/2 in front.




The Attempt at a Solution



I let u = to everything above e
du = -4xdx

[tex]-1/8x[e^{-2x^2}]dx[/tex]

then plug in 2 ...0...subtract...am i on the right path?
 
Last edited:
Physics news on Phys.org
There really is no elementary answer for that integral. You will have to use an approximate method in order to evaluate the integral, such as Simpson's Rule or the Trapezium Rule.
 
rjs123 said:

Homework Statement


[tex]1/2\int_0^2e^{-2x^2}dx[/tex]

I'm basically trying to find the average area under the curve for a rectangle from [0, 2] inclusive...which is why i put the 1/2 in front.




The Attempt at a Solution



I let u = to everything above e
du = -4xdx

[tex]-1/8x[e^{-2x^2}]dx[/tex]

then plug in 2 ...0...subtract...am i on the right path?
In addition to what rock.freak667 said, you attempted to use substitution on this integral, but didn't do it correctly.

In your substitution u = -2x2, and du = -4xdx, but you are missing an x factor in your integrand, and there's no way to get it.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K