Conceptual doubt in Definite Integration regarding limits of $\int_{0}^{2}x\cdot\sqrt{x+2}\ dx$ Question:

Evaluate $I=\int_{0}^{2}x\cdot\sqrt{x+2}\ dx$.

My Approach:

Put $x+2=t^{2}$.
Now changing limits: Lower Limit $\left(x=0\right)\ :\ t^{2}=0+2=2$ $\to$ $t=\pm\sqrt{2}$

How do I know if i have to take the t with + or - sign? Same confusion is with the upper limit. Sorry if this might be too bad of a question to ask but I have only began with definite integration. I know This doubt is really specific because taking some other substitution would not have landed me in this situation. But how do I proceed this way?
Edit:
I think everyone is right, It is just a matter of preference.
What alex is saying is:
Let me take $-\sqrt{2}$ as the lower limit and $-2$ as the upper limit, since I have substituted $t^{2}=x+2$
We have, $dx=2tdt$ and the integral becomes:
$I=2\int_{-\sqrt{2}}^{-2}\left|t\right|\cdot t\cdot\left(t^{2}-2\right)dt$ which gives the same answer. Here we will open |t| with - sign because we know it is varying in negative number.
We can even take lower limit as $-\sqrt{2}$ and upper limit as $+2$ and evaluate the integral which gives the same thing but this time I have to split the integral because of the mod, specifically:
$$I=2\int_{-\sqrt{2}}^{2}\left|t\right|\cdot t\cdot\left(t^{2}-2\right)dt$$
$$I=2\int_{-\sqrt{2}}^{0}\left(-t\right)\cdot t\cdot\left(t^{2}-2\right)dt+2\int_{0}^{2}\left(+t\right)\cdot t\cdot\left(t^{2}-2\right)dt$$ .
Both give the same answer. Obviously integrating from $+\sqrt{2}\to2$ is easy but it's nice to get the concepts right.
 A: Instead of starting with the equation $t^{2}=x+2$ you should start with $t =\sqrt {x+2}$ the positive square root of $x+2$.
(The notation $\sqrt {x+2}$ in the question stands for the positive square root of $x+2$). So you  take $t >0$.  The intergal w.r.t. $t$ is from $\sqrt 2$ to $2$.
A: In my humble opinion the correct answer is: it doesn't matter, since $\sqrt{t^2}=|t|$, so if you choose $\sqrt 2$ you have to choose $|t|=t$, if you choose $-\sqrt 2$ you have to split the cases, namely $|t|=-t$ between $-\sqrt 2$ and $0$, and $|t|=t$ between $0$ and $2$. I admit of course that the choice $\sqrt 2$ is way more rational than the other one, but both are correct.
A: I want to add a hopefully last perspective on this. You can in fact choose any combination of the limits you want: With $f(t) =(t^2-2)\cdot |t|\cdot 2t$ we have
$$
\int_{\sqrt 2}^2f(t)\,dt
= \int_{-\sqrt 2}^2f(t)\,dt
= \int_{\sqrt 2}^{-2}f(t)\,dt
= \int_{-\sqrt 2}^{-2}f(t)\,dt.
$$
This is because all of them are equal to $\int_0^2x\sqrt{x+2}\,dx$ through the substitution $x:=t^2-2$. The first and last version correspond to the standard substitutions $t=\pm\sqrt{x+2}$. But all of them work if we read the theorem of substitution in reverse, which amounts to letting $x$ depend on $t$ and not the other way round.

Let's look at it formally. The precise statement of substitution is, if $\varphi$ is $C^1$ and $f$ is continuous on the relevant regions, then
$$
\int_{\varphi(a)}^{\varphi(b)}f(x)\, dx
= \int_a^b f(\varphi(t))\varphi'(t)\,dt.
$$
I've written it in the unusual order, because we want to use the above from left to right. Then with $f(x) = x\sqrt{x+2}$, $\varphi(t)=t^2-2$, $\varphi(a)=0$ and $\varphi(b)=2$, we get the result. Note that $a$ and $b$ are any numbers that satisfy $\varphi(a)=0$ and $\varphi(b)=2$. We see that we always have complete freedom of $a$ and $b$ in cases like this.
A: Both of the substitutions $t=\sqrt{x+2}$ and $t=-\sqrt{x-2}$ are valid transformations of the integral, as the functions $x\mapsto\sqrt{x+2}$ and $x\mapsto-\sqrt{x+2}$ are both invertible on the interval $[0,2]$. This condition is necessary, as the term $\left(\sqrt{x+2}\right)'=\frac{1}{2\sqrt{x+2}}$ does not appear in the integrand.
For further discussion of the theory behind integration by substitution, see Michael Spivak's Calculus, in particular, chapter 19: integration in elementary terms.
