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Integration


Date: 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/   
    
Associated Topics:
High School Calculus

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