Solving Integration Problem: Finding Int. of x^2e^(-x^2)

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Homework Help Overview

The discussion revolves around evaluating the integral of the function x^2e^(-x^2) over the interval from negative to positive infinity. Participants reference a known integral involving e^(-x^2) and seek methods to approach the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts substitution and integration by parts but finds them unhelpful. Some participants suggest generalizing the known integral and differentiating with respect to a parameter. Others propose using differentiation under the integral sign or integration by parts with specific variable assignments.

Discussion Status

Participants are exploring various methods to approach the integral, with some providing alternative strategies and insights. There is no explicit consensus on a single method, but several productive lines of reasoning are being discussed.

Contextual Notes

The original poster expresses uncertainty and frustration, indicating a struggle with the problem despite it being perceived as simple. There is an emphasis on the need for careful handling of terms when applying integration techniques.

qspeechc
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I feel really dumb for asking this, because I know it's something simple I'm just not seeing. Ok, given that

[tex]\int _{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi }[/tex]

how to I find

[tex]\int _{-\infty}^{\infty} x^2e^{-x^2}dx = ?[/tex]

I have tried the substitution u=x^2, and integration by parts, but nothing is working. Any help? Thanks
 
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The easy way to do this problem is to generalize your first integral. Can you show the integral of exp(-ax^2) is sqrt(pi/a) for a>0? (Use a substitution x=sqrt(a)*u). Now differentiate that with respect to a. Finally put a=1 again.
 
This might help:

d/dx (x*[e^-x^2]) = ...
Solve it and then integrate!
 
Integration by parts should work. You have

[tex] \int_{-\infty}^\infty x^2 e^{-x^2} \, dx = \sqrt \pi[/tex]

Set

[tex] u = x, \quad dv =x e^{-x^2} dx[/tex]

Then

[tex] \int u \, dv = uv - \int v \, du[/tex]

should, with careful attention to the [tex]uv [tex]term at the infinities, work fine.[/tex][/tex]
 
Thanks everyone.
 

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