Improper integrals question: why does this work? So, I'm solving 
$$
\int_0^3{  \frac{dx}{ \sqrt{ 9 - x^2 } }  }
$$
The catch here is $f(x)$ is defined on ( -3, 3 ) only.
And I know the way to solve this is to integrate (yielding $ \arcsin{ \frac{x}{3} } + C $), then evaluate $ \lim_{c \to 3^-}{ \arcsin{\frac{c}{3}} - \arcsin{ 0 } } $, which is simply $ \frac{ \pi }{2} $
My question is why does this work?  I don't intuitively understand how, although $\sqrt{ 9 - x^2 }$ blows up very slowly to infinity as $ x \to 3 $, it still goes there, so how is the sum of the area under that curve not infinity?
 A: How can an "infinite region" have only finite area? For much the same reason that an "infinite sum" can still add up to a finite number: even though each of $\frac{1}{2^n}$ is positive, and we add an infinite number of them when we do
$$\sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+\cdots$$
yet this still adds up to a finite value, namely $1$. 
But your confusion is exactly what made Zeno's paradoxes paradoxes: it contradicts the intuitive expectation (at least until you get so used to these kinds of things that your intuition starts pointing in the "right" direction). 
Why does it add up only to $1$, even though you have an infinite number of positive numbers getting added? Because the things you are adding "get very small fast enough" to not amount to much, even though there are infinitely many of them.
Or: imagine taking a sequence of rectangles which are getting thinner but higher. If you make them higher by the same proportion as you are making them thinner, the total area will remain the same: if you start with a $1\times 1$ rectangle, with area one, and in the next step you have a rectangle half as wide but twice as tall, $\frac{1}{2}\times 2$, then the area is still $1$; then a rectangle half as wide but twice as tall as the latter one, $\frac{1}{4}\times 4$, then the area is still $1$; etc. But what if you make them thinner a lot faster than you are making them taller? If you halve the width, but you only add $25$% to the height? You start with a $1\times 1$ rectangle; then a $\frac{1}{2}\times \frac{5}{4}$, which only has an area of $\frac{5}{8}$; then $\frac{1}{4}\times 1.5625$, with an area of only $0.390625$; etc. If you can make them thinner a lot faster than you make them taller, you can ensure that the total area doesn't get too large (that is, it's not infinite, even though you have an infinite number of rectangles, each with positive area).
The same is happening with the portions of the graph under, say, $\frac{1}{\sqrt{x}}$ near $0$ on the right. The integrals
$$\int_{1/2}^1 \frac{dx}{\sqrt{x}},\quad \int_{1/4}^{1/2}\frac{dx}{\sqrt{x}},\quad \int_{1/8}^{1/2}\frac{dx}{\sqrt{x}},\quad\ldots, \int_{1/2^{n+1}}^{1/2^n}\frac{dx}{\sqrt{x}},\ldots$$
are all positive, and if you "add" all of them, you'll get
$$\int_0^1\frac{dx}{\sqrt{x}}$$
So how can this integral be finite, if you are adding an infinite number of positive numbers? Because the integrals are getting sufficiently small sufficiently fast that they don't add up to much:
$$\begin{align*}
\int_{1/2^{n+1}}^{1/2^n}\frac{dx}{\sqrt{x}} &= \int_{1/2^{n+1}}^{1/2^n}x^{-1/2}\,dx\\
&= 2x^{1/2}\Biggm|_{1/2^{n+1}}^{1/2^n} = 2\sqrt{\frac{1}{2^{n}}}  - 2\sqrt{\frac{1}{2^{n+1}}}\\
&= \frac{2}{2^{n/2}} - \frac{2}{2^{(n+1)/2}} = \frac{\sqrt{2}-1}{\sqrt{2^{n-1}}}.
\end{align*}$$
As $n\to\infty$, the contributions of these portions get so small, so fast, that even taking infinitely many of them doesn't add up to much. So the area is finite, even though the region is infinite.  Intuitively, you are adding thinner-but-higher rectangles. But the rectangles get thinner a lot faster than they are getting high, so the total area contribution is actually diminishing very fast. 
(The fact that an infinite shape may have finite area is, when you stop to think about it, not that surprising: a line is an infinite shape, but it has zero area, after all...) 
The case of 
$$\int_0^3\frac{dx}{\sqrt{9-x^2}}$$
is pretty much the same thing. As you get closer and closer to $3$ from the left, the thin rectangles of area you are adding are just not getting tall fast enough to counteract how fast they are getting thinner, so the total area doesn't add up to much (in fact, it's bounded above, which is why the integral is finite). 
A: A related question: How can you have infinitely many positive numbers that "sum" to a finite number?  For example, $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots = 1$. This can be made intuitive by repeatedly bisecting a unit interval, and seeing that $\frac{1}{2}+\cdots+\frac{1}{2^n}$ gets as close as you want to $1$ by taking $n$ big enough.  The definition of the infinite "sum", i.e., the value of the series, is the limit of the finite sums, so the value is $1$.
When assigning a meaning to the "area" under a curve that has unbounded height, the same idea can be applied.  You can take a limit of areas of curves with bounded height, for which the area can be visualized.  If it turns out, as it does here, that the areas approach a finite value, then that value can be thought of as the "area" under the unbounded curve.
The relationship between the two notions can be seen for the case of $\displaystyle{\int_0^1\frac{1}{\sqrt x}}$ as follows.  On the interval $\left[\frac{1}{(n+1)^2},\frac{1}{n^2}\right]$, $\frac{1}{\sqrt{x}}\leq n+1$.  Therefore the area under the curve on that interval is less than $(n+1)\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)=\frac{2n+1}{n^2(n+1)}<\frac{3n}{n^3}=3\frac{1}{n^2}$.  The finiteness of the "area" under the curve then follows from the finiteness of the "sum" $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots$.
The convergence problem in your example is essentially the same as that for $\frac{1}{\sqrt x}$, because $\sqrt{9-x^2}=\sqrt{3-x}\sqrt{3+x}$, and the $\sqrt{3+x}$ part approaches $\sqrt 6$.
