# 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.

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$\sqrt 1 \ne \pm 1$, but $\sqrt 1 = 1$ – Kaster May 29 '13 at 17:19

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.

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Thanks! You were chosen as the correct answer... by coin flip. – d0rmLife May 29 '13 at 18:24
Saying $\sqrt x=+\sqrt x$ is saying nothing; it wouldn't exclude taking $\sqrt 9$ to be $-3$. What you mean is that it is a convention that $\sqrt x\geq0$. – Marc van Leeuwen May 31 '13 at 12:16

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.

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Thanks! I literally flipped a coin to determine the correct answer, as I found both provided similar information. The other poster was "tails", which won. +1 anyways, of course! – d0rmLife May 29 '13 at 18:22

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

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