TITLE: Is there ever a requirement to change the limits of integration?
QUESTION [5 upvotes]: I don't have issues with doing integration problems, but occasionally I see the solution changing the limits of integration whenever a $u$-substitution is done.
I obviously don't have a problem doing this, and I just recently noticed my book doing this under the chapter involving "area of surface of revolution." 
My question is did I develop a bad habit by never changing the limits of integration, or is it a best practice to always change limits of integration?  
My teacher said on a test, and in general, if we do not change the limits of integration then we should be signifying this by labeling our limits of integration $x=$ lower-limit and $x=$upper-limit.
EDIT EXAMPLE INCLUDED
After further investigation, my confusion is because of the two below equations:
Find the exact area of the surface obtained by rotating the curve about the x-axis
$$y=\sqrt{1+4x}, 1\le x\le 5$$
The limits of integration were changed in the solution to this problem.  
The given curve is rotated about the y-axis.  Find the area of the resulting surface.
$$y=x^\frac{1}{3}, 1\le y \le 2$$
The limits of integration were NOT changed in the solution to this problem.

REPLY [3 votes]: Suppose you have
$$
\int_0^{\pi/4} \left(\tan x\right)^3 \Big(\sec^2 x\,dx\Big)
$$
(where you should take the $\Big($Big parentheses$\Big)$  as a hint about what $u$ should be), then you can write
\begin{align}
u & = \tan x \\
du & = \sec^2 x\,dx
\end{align}
and then the integral is
$$
\int_{x=0}^{x=\pi/4} u^3 \,du = \left[\frac{u^4}4\right]_{x=0}^{x=\pi/4} = \left[\frac{(\tan x)^4}{4}\right]_0^{\pi/4} = \frac14.
$$
In this case, you could say changing the bounds is optional, although it speeds things up and simplifies things:
$$
\int_0^1 u^3\,du = \left[ \frac{u^4}{4}\right]_0^1 = \frac14.
$$
But there are situations in which you'll get nowhere if you don't change the bounds.  For example, the Beta function is defined by
$$
B(\alpha,\beta) = \int_0^1 x^{\alpha-1} (1-x)^{\beta-1}\,dx.
$$
So suppose we have
$$
\int_0^5 x^{\alpha-1}(5-x)^{\beta-1} \,dx.
$$
(I've seen integrals like this arising in statistics problems.)  Then let
\begin{align}
u & = x/5, \\
du & = dx/5.
\end{align}
The integral then becomes
$$
\int_0^1 (5u)^{\alpha-1} (5-5u)^{\beta-1}\,\Big(5\,du\Big) = 5^3 \int_0^1 u^{\alpha-1}(1-u)^{\beta-1}\, du = 125 B(\alpha,\beta). 
$$
Without changing the bounds, how could one recognize this as being the Beta function?