Given the definite integral:


We can make the Power Substitution: $$\begin{align} u^2=&&x+3 \\ 2u\text{d}u=&&\text {d}x \end{align}$$

We get the following: (without the limits)

$$\int{\left(\left(u^2-3\right)\times u\times 2u\right)\text{d}u}$$

(Yes I know, there is a much simpler way to go about this.)
However, when we change the bounds of integration we are left with this upper limit for example: $$\begin{align} u&&=\pm\sqrt{x+3} \\ &&=\pm\sqrt{2+3} \\ &&= \pm\sqrt{5} \end{align}$$

Should we use $+\sqrt{5}$ or $-\sqrt{5}$, How do we know which value to use?

I assume it depends on the integrand function's domain. But what about the case where both values satisfy the domain?

  • $\begingroup$ Why don't you make the substitution $u = x+3$ as well and compare it against the substitution you have tried? That should help you identify the proper bounds on your integral. $\endgroup$ – Mattos May 8 '16 at 2:50
  • $\begingroup$ @Mattos I've actually done that before. However I haven't been able to crack out any pattern. I don't want to have to verify whether I've chosen the correct boundaries with a linear substitution $\endgroup$ – David Yue May 8 '16 at 2:54

The problem you have is that the map $u \mapsto x$ is not one-to-one (injective). This is discussed with some complexity here. There are two intervals in $u$ that are mapped to the same interval in $x$: $[2,\sqrt{5}]$ and $[-\sqrt{5}, -2]$. You have two choices:

  • Pick one interval, either $\int_2^\sqrt{5} \dots \,\mathrm{d}u$ or $\int_{-\sqrt{5}}^{-2} \dots \,\mathrm{d}u$, or
  • Use both intervals $\frac{1}{2} \left( \int_2^\sqrt{5} \dots \,\mathrm{d}u + \int_{-\sqrt{5}}^{-2} \dots \,\mathrm{d}u \right)$ and realize you're getting two copies of the result, so you divide by two.

You know your $x$ interval, $[1,2]$, is connected and the two intervals we got are not connected to each other, so either is fine. This can be much more complicated...

  • $\begingroup$ If i get a complex number after trying to change the intervals. Should I pick the other interval then? $\endgroup$ – David Yue May 8 '16 at 3:01
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    $\begingroup$ If you get a complex number, you have two choices: figure out what path in $\Bbb{C}$ you need for your interval, or pick a different substitution. Odds are, your class is not dealing with complex numbers and complex integration, so I'd go with the second option. $\endgroup$ – Eric Towers May 8 '16 at 3:06

It does not matter, as long as we are consistent. We can either integrate from $u=2$ to $u=\sqrt{5}$ or from $u=-2$ to $u=-\sqrt{5}$.

  • $\begingroup$ Wait. So would you have to choose one over the other. If a specific limit of integration converted to $u$ is complex? $\endgroup$ – David Yue May 8 '16 at 2:56
  • $\begingroup$ Why it is don't matter? I cannot follow the logic behind it. and then again is it's so called, intuitive. $\endgroup$ – Salech Rubenstein May 8 '16 at 2:58
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    $\begingroup$ In this case there was no issue, for each "limit" we used the same branch of the inverse function. If for some choice we get non-reals, then from the real variable point of view it is not an appropriate substitution. However, remarkably often formal manipulation leads to the right answer. As Euler once apparently wrote, sometimes my pencil is smarter than I am. $\endgroup$ – André Nicolas May 8 '16 at 3:01
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    $\begingroup$ @SalechAlhasov: The two choices just lead to different parametrizations of the curve. $\endgroup$ – André Nicolas May 8 '16 at 3:02

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