Indefinite IntegralDate: 6/1/96 at 0:48:9 From: Anonymous Subject: Integrals I just want to find the indefinite integral: ( | e^(x*x) dx "e to the x squared" ) AND: ( | e^(-x*x) dx "negative x squared" ) Thank you. Date: 6/1/96 at 8:11:43 From: Doctor Anthony Subject: Re: Integrals The integral of e^(x^2) cannot be done by elementary methods. One way of proceeding is to write I^2 = [INT(e^(x^2)).dx]^2 = [INT(e^(x^2)).dx][INT(e^(y^2)).dy] which can be written as a double integral over a region of the xy plane. I^2 = INT.INT[e^(x^2+y^2).dx.dy] Now change to polar coordinates, remembering that an elementary unit of area dx.dy becomes an elementary unit of area r.d(theta).dr This is known as the Jacobian of the transformation if you want to look up change of variables in multiple integrals. Also x^2+y^2 = r^2 I^2 = INT.INT[e^(r^2).r.d(theta).dr] From here we can integrate first with respect to r. I^2 = INT[d(theta)INT[e^(r^2).r.dr]] Put r^2 = u, so 2r.dr = du and r.dr = (1/2)du INT[e^(r^2).r.dr] = (1/2)INT[e^(u).du] = (1/2)e^(u) = (1/2)e^(r^2) + const. = f(r) Now if we had limits for r we could evaluate this expression, but in any case treat r as a constant for the integration with respect to theta. Returning to the main integral I^2 = f(r).INT[d(theta)] = f(r)[theta + const.] And finally take the square root I = sqrt{f(r)[theta+const.]} In the case of integrating e^(-x^2) we proceed in exactly the same manner. The only difference is that f(r) would be (-1/2)e^(-r^2) + const. In this case if we are integrating x from 0 to infinity, the region of integration is the positive quadrant of the xy plane, and r is from 0 to infinity while theta is from 0 to pi/2. We would have f(r) = (-1/2)[0-1] = 1/2 and I^2 = (1/2)[pi/2 - 0] = pi/4 Giving I = (1/2)sqrt(pi) This last integral is, of course, used in evaluating areas under the Normal distribution curve. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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