TITLE: Integration by substitution, why do we change the limits?
QUESTION [6 upvotes]: 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

REPLY [5 votes]: 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.)