Integration by substitution, why do we change the limits? I've highlighted the part I don't understand in red. Why do we change the limits of integration here? What difference does it make?

Source of Quotation: Calculus: Early Transcendentals, 7th Edition, James Stewart
 A: By the substitution $x=\varphi(t)$ we have
$$\int_a^b f(x)dx=\int_{\varphi^{-1}(a)}^{\varphi^{-1}(b)}f(\varphi(t))\varphi'(t)dt$$
A: You asked two questions: “Why do we change the limits of integration here?” and “What difference does it make?” I wonder whether you are also asking “why do we substitute $u=1+\frac9 4 x$ ?” so I’ll cover that briefly too.
Why do we change the limits of integration?
The limits of integration are not actually being changed – just expressed in the language of the new variable $u$. Ironically, this is to stop the value of the answer getting changed!
If you were somehow able to perform the integral in your question directly (without substituting in $u=1+ \frac94 x$), you would naturally then evaluate the answer from $x=1$ to $x=4$.  So when someone writes $$L=∫_1^4√(1+9/4 x) dx$$ it's assumed (by them and you) that $4$ and $1$ are $x$-values. That is, they really meant $$L=∫_{x=1}^{x=4}√(1+\frac94x)dx$$
(Make sense? If you performed $∫_1^42xdx$ you'd evaluate the answer, $x^2$, by substituting in $4$ and $1$ for $x$, giving $4^2-1^2=15$.  i.e. you just evaluated the answer at $x=4$ and $x=1$, but it was so obvious that $4$ and $1$ referred to $x$-values, that no-one needed to write $∫_{x=1}^{x=4}2xdx$ even though that's what the question really meant. It’s important because if the question was, say, $∫_{y=1}^{y=4}2xdx$ then the answer would probably be different. After you got your general answer of $x^2$, you couldn’t then just substitute $4^2-1^2$; you would need to find out what $x$ is worth at $y=4$ and $y=1$.)
So the question could be more precisely written  $$L=∫_{x=1}^{x=4}√(1+\frac94x)dx$$
By the way, we are going to substitute $u=1+\frac94 x$ because it looks like it’s going to be so much easier to integrate:
$$L=∫_{x=1}^{x=4}√u dx$$
But we have a problem: integrating $u$ with respect to $x$, we may as well just be back with the original question. So instead we figure out what $dx$ means in the world of $u$: 
$$\frac{du}{dx}=\frac{d}{dx}(1+\frac94x)=\frac94⇒du=\frac94dx$$ or $$dx=\frac49du$$
So
$$L=∫_{x=1}^{x=4}√u⋅\frac49du=\frac49∫_{x=1}^{x=4}u^{\frac12}du=\frac49⋅\frac32 u^{\frac32}\Bigg|_{x=1}^{x=4}=\frac23u^{\frac32}\Bigg|_{x=1}^{x=4}$$
Oh dear. Problem. All very well figuring out what $dx$ meant in the world of $u$, and all very well substituting $u$ for $1+\frac94x$, but we also need to know what $u$ is worth at $x=4$ and $x=1$ or we can’t evaluate the integral. That is why we "change" or "substitute for" the limits of integration:
$$u=1+\frac94x⇒$$
$$at\, x=4,\,u=1+\frac94×4=10$$
$$at\, x=1,\,u=1+\frac94×1=3.25$$
Now we can evaluate the integral because the following two statements mean exactly the same thing:
$$\frac23u^{\frac32}\Bigg|_{x=1}^{x=4} \,and \, \frac23u^{\frac32}\Bigg|_{u=3.25}^{u=10}$$
the difference being that we can work with the second, but not with the first.
Answer:
$$L=\bigg(\frac23×10^{\frac32}\bigg)-\bigg(\frac23×3.25^{\frac32}\bigg)=21.08-3.91=17.17 (2 d.p.)$$
What difference does it make?
If we hadn’t changed the limits of integration (i.e. recognised that $4$ and $1$ meant $x=4$ and $x=1$ not $u=4$ and $u=1$) then we’d get a different answer by mistake:
$$L=\frac23u^{\frac32}\Bigg|_1^4=\bigg(\frac23×4^{\frac32}\bigg)-\bigg(\frac23×1^{\frac32}\bigg)=16/3-2/3=14/3\,\,i.e.\,4.33 \,(2 d.p.)$$
because we wrongly assumed that 4 and 1 referred to u rather than to x…
By the way, it also makes sense to change the limits earlier i.e. at the same time as the other substitutions (although doing it earlier won’t affect your answer.)
A: Because the function has changed. Let's do an example:

$$\int_{-1}^1 x\,dx =0$$

because the integrand is odd and the interval is symmetric (you can also check directly).
Let's do a simple $u=x+1$ so that $du=dx$ then the right way we have:

$$\int_0^2 u-1\,du = {1\over 2}(u-1)^2\bigg|_0^2={1\over 2}-{1\over 2}=0$$

but if we fail to change the limits:

$$\int_{-1}^1(u-1)\,du = {1\over 2}(u-1)^2\bigg|_{-1}^1=-2$$

The underlying reason is that integration comes from Riemann sums, the function values depend on the interval of integration. When you change the interval, the heights of the rectangles you use in the definition change (remember the heights are the function values) so that you end up adding up different things if you don't change the function to compensate.
A: The original limits is for variable $x$, and the new limits is for the new variable $u$. If you can get a primitive function 
$$\frac{8}{27}(1+\frac{9}{4}x)^\frac{3}{2}$$
by observation, it is unnecessary.
A: No, we don't change the limits. Who said that. They stay the same. Only their representation changes.
$$
L = \int\limits^{x=4}_{x=1}\sqrt{1 + \dfrac{9}{4}x}dx
$$
Now we substitute $u=1+\dfrac{9}{4}x$:
$$
L = \int\limits^{x=4}_{x=1} \sqrt{u} dx
$$
See? We didn't change anything. We only substituted $u$ for the $x$ expression in the integrand. But now, it contains mixed variables and not solvable as is. We need to substitute $u$ expressions for other $x$ expressions too.
$$
L = \int\limits^{x=4}_{x=1} \sqrt{u} dx = \int\limits^{\dfrac{4}{9}(u-1)=4}_{\dfrac{4}{9}(u-1)=1} \sqrt{u} d\left( \dfrac{4}{9}(u-1) \right)
$$
Now, all the variables are in $u$, and the integral is solvable.
$$
\dfrac{4}{9}(u-1)=4 \implies u=10, \quad \dfrac{4}{9}(u-1)=1 \implies \dfrac{13}{4} \\
d\left( \dfrac{4}{9}(u-1) \right) = \dfrac{4}{9}du
$$
We rewrite the integral expression:
$$
L = \int\limits^{\dfrac{4}{9}(u-1)=4}_{\dfrac{4}{9}(u-1)=1} \sqrt{u} d\left( \dfrac{4}{9}(u-1) \right) = \dfrac{4}{9} \int\limits^{u=10}_{u=\dfrac{13}{4}} \sqrt{u} du
$$
Have we ever changed the limits in any step? No! They are the exact same limits. There only represented by $u$ instead of $x$ in the final expression.
