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Date: 5/23/96 at 16:17:3
From: Anonymous
Subject: Integration

Could you please integrate the following function:	

x^2 * e^(-2x^2)

Any help would be appreciated!

Date: 6/1/96 at 11:20:19
From: Doctor Anthony
Subject: Re: Integration

Use integration by parts INT[x(xe^(-2x^2).dx]. Here we differentiate x 
and integrate xe^(-2x^2).  On integration by substituting 2x^2=u, this 
last expression gives (-1/4)e^(-2x^2)

So the parts integration gives:

   I = x(-1/4)e^(-2x^2) + (1/4)INT[e^(-2x^2).dx]

From here we have to integrate e^(-2x^2), and this is done as follows:

For simplicity I shall save some ugly sqrt(2) terms by integrating 
e^(-x^2). The integral cannot be done by ordinary elementary methods, 
and one way of proceeding is to write:

   I^2 = {INT[e^(-x^2).dx]}^2 = INT[e^(-x^2)]INT[e^(-y^2).dy]

       = double integral  INT{INT[e^-(x^2+y^2).dx.dy]}

where the integration is over the appropriate region of the xy plane.

We change to polar coordinates and note that in making the change, 
each element of area dx.dy in cartesian coordinates is represented by 
elements r.d(theta).dr in polar coordinates. If you want a more 
thorough explanation of change of variable with multiple integrals you 
should look up 'Jacobians' in the textbooks.

We have:  I^2 = INT{INT[e^(-r^2).r.d(theta).dr]}

              = INT{d(theta)INT[e^(-r^2).r.dr]}

Substitute r^2 = u  so 2r.dr = du
                        r.dr = (1/2)du

So INT[e^(-r^2).dr] = (1/2)INT[e^(-u).du]
                    = -(1/2)e^(-u) + const
                    = -(1/2)e^(-r^2) + const
                    = f(r)

If we have limits for r or some initial conditions this can be 
evaluated.  In any case, we can treat f(r) as constant for the last 
part of the integration with respect to theta.

   I^2 = f(r)INT[d(theta)]
       = f(r){theta + const}

Finally take the square root to find I:

   I = sqrt{f(r)[theta+const]}  

If you want to find INT(e^(-2x^2).dx), make the substitution 

sqrt(2).x = u
       dx = (1/sqrt(2)).du and we have to integrate: 

       (1/sqrt(2))INT[e^(-u^2).du] using the method outlined above.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
High School Calculus

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