IntegrationDate: 03/30/99 at 01:19:58 From: Jason Spuller Subject: Integration Integrate (2-x^2)^4xdx I have tried to go through the steps by letting u = 2-x^2 and du = -2x but I really do not know where to go after that. I hope you can give me some directions. Date: 03/30/99 at 08:29:04 From: Doctor Jerry Subject: Re: Integration The theorem is int(f(g(x))*g'(x)*dx) = int(f(u)*du), where u = g(x) and it is understood that after you finish the easier integration of int(f(u)*du), you must replace u by g(x). For the problem you gave, f(u) = u^4 and g(x) = 2 - x^2, so that g'(x) = -2x. Now the given integral does not quite have this form, but it is possible to force it a little: int((2-x^2)^4*x*dx) = (-1/2)int((2-x^2)^4*(-2x)*dx) We were able to adjust since what we were missing was a constant - and constants can be taken across the integral sign. Now, from the theorem, int((2 - x^2)^4*x*dx) = (-1/2)int((2 - x^2)^4*(-2x)*dx) = (-1/2)int(u^4*du). This integral is just (1/5)u^5, etc. Do not forget to replace u by 2-x^2 and, of course, add the constant. I think it is a bit easier to do this: The 2-x^2 inside and x outside suggests that u = 2 - x^2 might be useful, since the x will be part of the du. After thinking of this substition, just follow the symbols: u = 2-x^2 means du = -2x*dx. So, by substitution, int((2-x^2)^4*x*dx) = int(u^4*du/(-2)) = (1/(-2))int(u^4*du) and now just do as above. It is almost the same, but from du = -2x*dx you can solve for what is in the original integrand, namely x*dx. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/