Strange use of u-substitution I'm struggling with the following problem (I'm in a Calc I class) - we've never covered this particular kind of integration question before, so any advice would be most welcome.
The problem asks:

$F\left(x\right)=\int _0^{3x}\:\sqrt{t^2+2t}dt$ 
Find $F'\left(1\right)$.

Is F(x) distinct from f(x)? What would F'(x) be then?
The answer sheet I have proceeds with solving the problem using u-substitution. u = 3x, which makes u' (or du) = 3. I'm already a little lost, because I'm not accustomed to setting u to the boundaries (though I have rewritten the boundaries in terms of u for other problems).
The answer sheet ends up with:
$\int _0^u\:\sqrt{t^2+2t}\:dt\:$
From there it ends up with:
$u'\sqrt{u^2+2u}$
From there it substitutes back the original value of u, from which we can solve F'(1). This step is maybe the most confusing, I'm not sure why u' is being multiplied against the function, and why u is being substituted for t in general. Any advice/insight?
 A: Yes, $F(x)$ is distinct from $f(x)$  They are two different functions.  If both are used in a problem they are often related in some way.  Here we do not have any $f(x)$.  You are given a definition of $F(x)$ as a particular integral. 
If you define $u=3x$ you then can substitute it anywhere $3x$ appears, which happens here to be one of the limits.  They then are using the fundamental theorem of calculus and the chain rule to say 
$$\begin {align}F'(x)&=\frac d{dx}\int_0^u \sqrt{t^2+2t} \; dt\\&=\frac {du}{dx}\frac d{du}\int_0^u \sqrt{t^2+2t} \; dt\\
&=u'\sqrt{u^2+2u} \end{align}$$
Now you can back substitute $x=\frac 13u$ and evaluate the result at $x=1$
A: The fundamental theorem of calculus says that if $f(x)$ is continuous then $F(x)=\int_0^xf(t)dt$ is differentiable and $F'(x)=f(x).$ Now, have a look at the upper limit of your integral. It is $3x$ and not $x.$ That's the reason to make the change $u=3x.$ 
So, $F(u)=\int_0^u f(t)dt$ is differentiable and $\frac{dF}{du}=F'(u)=f(u).$ Since $u=3x$ we have that $$\frac{dF}{dx}=\frac{df}{du}\frac{du}{dx}=3f(3x).$$
