Solve ∫e^(-x^2)dx

  • Thread starter Jamin2112
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  • #1
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Homework Statement



On the next exam I'm supposed to show that ∫e^(-x^2)dx = √(π/2).

Homework Equations



?????

The Attempt at a Solution



When the professor was showing us one way to do it, I remember him doing a step that was like

∫e^(-x^2)dx ∫e^(-x^2)dx = ∫∫e^(-x^2 - y^2) dx dy.

Is that legal? I never knew ∫f(x)dx ∫g(y)dy = ∫∫f(x)g(y) dx dy.
 

Answers and Replies

  • #3
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Yes, it's called Fubini's theorem:
http://en.wikipedia.org/wiki/Fubini's_theorem
(See the corollary where f(x,y)=g(x)h(y))
There are some conditions on when it can be used, you can read about it on the wiki article.

Why was never told this before? Many of my previous classes would've been easier if I could solve a double integral by just splitting it into two easier single integrals.
 
  • #4
LeonhardEuler
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Why was never told this before? Many of my previous classes would've been easier if I could solve a double integral by just splitting it into two easier single integrals.

The normal way of evaluating integrals in dimensions 2 and higher basically is this method, though it isn't usually presented using this theorem at first from what I remember.
 
  • #5
dextercioby
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Homework Statement



On the next exam I'm supposed to show that ∫e^(-x^2)dx = √(π/2).

Pay attention, you need to have a definite integral, with integration limits. The common integration limits for the Gauss bell are 0 and +infinity.
 
  • #6
HallsofIvy
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Why was never told this before? Many of my previous classes would've been easier if I could solve a double integral by just splitting it into two easier single integrals.
That certainly is in any Calculus text I have ever seen. I can't speak for your class but every introduction to multiple integration class I have ever seen (and I have seen many) starts from the fact that if f(x,y)= g(x)h(y) and the area of integration is the rectangle [itex]a\le x\le b[/itex], [itex]c\le y\le d[/itex], then
[tex]\int f(x,y)dxdy= \left(\int_a^b g(x)dx\right)\left(\int_c^d h(y)dy\right)[/tex].

Of course, NOT every integral can be separated like that.
 

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