Why the integral $\int _0^4\:\frac{\sqrt{x}}{\left(4-x\right)^{\alpha }}dx $ is equivalent to this two expressions? The integral $\int _0^4\:\frac{\sqrt{x}}{\left(4-x\right)^{\alpha }}dx $ in the neighborhood of $0$ it's equivalent to $$ \frac {\sqrt{x}}{4^{\alpha}}  $$ and in the neighborhood of 4 it is equivalent $$ \frac{2}{(4-x)^{\alpha}} $$  Why in the first expression we replaced in the denominator  $x$ with $0$ and then in the second expression we replaced in the numerator x with 4? What property of the integrals allow us to do that? Plus we didn't even solve the integral we just replaced directly. So I don't understand how this property works.
 A: First of all: the integral $\int_0^4 (\cdots)dx$ is a number, not a function of $x$. So you can't say that it is approximated by one thing near $0$ and another thing near $4$. It doesn't have any concept of "near"ness. So I'm pretty sure these approximate expressions are approximations for the integrand, not the integral as a whole.
Now then: in the neighborhood of a number $x_0$, you can approximate a reasonably "normal" function $f$ as being constant, and equal to the value of the function at that number, $f(x_0)$. In this case, the function is the integrand,
$$f(x) = \frac{\sqrt{x}}{(4 - x)^{\alpha}}$$
and the number $x_0$ is either $0$ (for the first approximation) or $4$ (for the second approximation). So you could plug in $0$ and $4$ respectively, and find that
$$\begin{align}
f(0) &\approx \frac{\sqrt{0}}{(4 - 0)^{\alpha}} &
f(4) &\approx \frac{\sqrt{4}}{(4 - 4)^{\alpha}}
\end{align}$$
Well, but hold on a minute: this just means $f(0) \approx 0$, which is not terribly useful, and $f(4)$ is undefined, which is not at all useful. Here's where you can use a little trick: don't plug in the numerical value for $x$ where it would make the function zero or undefined. For $f(0)$, this means you leave the $\sqrt{x}$ in the numerator, which gives you a slightly more useful function than $f(0) = 0$, and for $f(4)$, that means you leave the $x$ in the denominator, which gives you something with an actual value.
$$\begin{align}
f(x\approx 0) &\approx \frac{\sqrt{x}}{(4 - 0)^{\alpha}} &
f(x\approx 4) &\approx \frac{\sqrt{4}}{(4 - x)^{\alpha}}
\end{align}$$
If you want to be more technically precise, you would consider series expansions of the function $f(x)$ around the points $x=0$ and $x=4$. I won't get into the details, but there is a fairly procedural way to calculate a series of progressively better and better approximations to a function in the neighborhood of a point. The approximations you're asking about are the leading-order approximations: the first entries in their respective series.
