Re: Positive/Negative after taking the square root

On Nov 28, 2:36 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}

[That should read: du = 2x dx, dx = +- sqrt(u) du]

Imagine you were doing a definite integral. u is
always positive, but x can take on either positive
or negative values.

The substitution x = -sqrt(u) is correct for x<0,
and the substitution x = sqrt(u) is correct for x>0.
There is nowhere (except u = x = 0) where both
signs are correct.

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}

Then EITHER y = sqrt(x) or y = -sqrt(x). If there are no
other requirements on y, both are valid solutions. If
there are other requirements, both may not be valid.
A common example from physics is in trajectory problems,
which involve finding the intersection of a parabola
with a straight line, and the unknown might be a
forward distance. So there is an additional requirement
that y>0, in addition to y solving the quadratic equation.

In your case, if you are doing a definite integral,
there is a requirement that x>0 on some parts of
the integration region and x<0 on other parts.

- Randy
.

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