Choosing a bound when it can be plus or minus? I.e. $\sqrt{4}$ My textbook glossed over how to choose integral bounds when using substitution and the value is sign-agnostic. Or I missed it!
Consider the definite integral: $$ \int_1^4\! \frac{6^{-\sqrt{x}}}{\sqrt x}  dx $$
Let $ u = -\sqrt{x} $ such that $$ du = - \frac{1}{2\sqrt{x}} dx $$
Now, if one wishes to alter the bounds of the integral so as to avoid substituting $ - \sqrt{x} $ back in for $ u $, how is the sign of the integral's bounds determined?
Because: $ u(1) = -\sqrt 1 = -(\pm 1) = \pm 1 $ and $ u(4) = -\sqrt{4} = -(\pm2) = \pm2 $
How does one determine the correct bound? My textbook selected $ -1 $ and $-2 $ without explaining the choices.
 A: When $x$ varies between $1$ and $4$ (as in this integral), $\sqrt{x}$ varies between $1$ and $2$, and $-\sqrt{x}$ varies between $-1$ and $-2$.
$\sqrt{x}$ is not a multi-valued function on the reals.  Its input is a nonnegative number, and its output is a nonnegative number.  This is different from solving $x^2=y$, which DOES typically have two solutions in the reals.
A: It is convention that $\sqrt{x} = + \sqrt{x}$. Thus, you set $u(1) = -\sqrt{1}=-1$ and $u(4) = -\sqrt{4}=-2$.
The only situation where you introduce the $\pm$ signs is when you are finding the root of a quadratic such as $y^2=x$ in which case both $y=+\sqrt{x}$ and $y=-\sqrt{x}$ satisfy the original equation.
A: The correct description of the convention is that √a ≥ 0.
If you want the negative square root of the equation  x^2 = a, the you write  -√a.
The sign is NEVER ambiguous.
Now, with √x appearing at two places in the integral  --  one with a negative sign  --
we can either let u = √x  (x = u^2 with u ≥ 0)
or let u = -√x  (x = u^2 with u ≤ 0)
In the former case, when x = 1  then u = 1; and when x = 4, u = 2.  
                                                   (This is the bit you need.)

In the latter case, the u values have opposite sign.
Either way is a good first step.
Try the former:
Let  I = [ ∫ 6^(-√x) / √x dx  for 1 ≤ x ≤ 4 ]
   = [ ∫ 6^(-u) / u * 2u du for 1 ≤ x ≤ 4 ]

   = [ 2 ∫ 6^(-u) du for 1  1 ≤ u ≤ 2 ]

   = [ -2 * 6^(-u) / ln 6 : 1 ≤ u ≤ 2 ]

   = 2 * (6^(-1) - 6^(-2)) / ln 6   

   = 5 / (18 ln6)

   = 0.1550

