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.

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.
