Re: Positive/Negative after taking the square root

On Nov 28, 7:36 am, Taras_96 <taras...@xxxxxxxxx> wrote:
I'm not totally mathematically inept, but I find I'm getting confused
by the following:

I'm confused as to what happens to the +/- in the following examples:

As a simple example, integrate x^2

int{x^2 dx}

Let x^2 = u, then dx = du/(2x) and du = 2x, dx = +/- 2sqrt{u}

Then the integral becomes
=int{u/(+/-)2sqrt{u} du}
= +/- int{1/2sqrt{u} du}
= +/- 1/3u^1.5
= +/ 1/3x^3 (substitute back in for u)

This is obviously not correct, and the correct answer is in taking
only the positive square root when representing dx interms of u and
du. However, is there a mathematical/logical reason for discounting
the negative root? After all, if y^2 = x, then y = +/-sqrt{x}

When you substitute u = x^2 you're losing information about which of
the two possible values of x you're dealing with. (This always happens
when the inverse function is multi-valued.) Strictly speaking you need
to keep track of the separate cases, so that everything is uniquely
defined, and then it all should come right:

If x > 0 then u = x^2 implies sqrt(u) = x, dx = du/(2*sqrt(u)) and we
end up with Int x^2 dx = 1/3*u^(3/2). Since what we know is that
sqrt(u) = x (not just that u = x^2), when re-substituting we should
write 1/3*u^(3/2) in terms of sqrt(u), giving 1/3*u^(3/2) =
1/3*sqrt(u)^3 = 1/3*x^3.

If x < 0 then u = x^2 implies sqrt(u) = -x, dx = -du/(2*sqrt(u)), and
we end up with Int x^2 dx = -1/3*u^(3/2). Now we have -1/3*u^(3/2) =
-1/3*sqrt(u)^3 = -1/3*(-x)^3 = +1/3*x^3, as before.

.

Relevant Pages

  • Re: Positive/Negative after taking the square root
    ... the two possible values of x you're dealing with. ... when the inverse function is multi-valued.) ... multivalued relationships, one needs to be more careful than ... when playing around with functions. ...
    (sci.math)
  • Re: Positive/Negative after taking the square root
    ... the two possible values of x you're dealing with. ... when the inverse function is multi-valued.) ... With the substitution u = x^2 this appears to become an integral ...
    (sci.math)