IntegrationDate: 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/ |
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