What does infinity as an upper/lower limit mean in a definite integral? Eg, $\int_0^\infty e^{-x} dx$ I was finding
$$\int_0^\infty e^{-x} dx$$
So the integral becomes
$$\left.(-e)^{-x}\right\rvert_0^\infty$$

Now here's the confusing thing for me.  What exactly does infinity mean in upper/lower limit?
How do I actually calculate definite integrals when one of the limit is infinity?

 A: Naively, you can do this, and it gets the right answer:
$$\int_0^\infty e^{-x}dx = \left[-e^{-x}\right]_0^\infty=e^0-e^{-\infty}=1-0=1$$
More formally, you can define
$$\int_0^\infty e^{-x}dx \,\,\,\underset{\text{def}}{=}\,\,\,\lim_{t\to\infty} \int_0^t e^{-x}dx = \lim_{t\to\infty} \left[-e^{-x}\right]_0^t=\lim_{t\to\infty}(e^0-e^{-t})=1-\lim_{t\to\infty}e^{-t}=1$$
A: This type of integral is called an 'improper integral'. I have always assumed that the term 'improper' refers to the fact that an improper integral is not really an integral, but rather the limit of one:
$$
\int_a^\infty f(x) \, dx := \lim_{n \to \infty} \int_a^n f(x) \, dx \, .
$$
Thus, '$\int_{0}^{\infty}e^{-x}$' is just a shorthand for $\lim_{n \to \infty} \int_{0}^{n}e^{-x}$, in the same way that the 'infinite sum'
$$
\sum_{n=1}^{\infty} \frac{1}{n}
$$
is really just a shorthand for
$$
\lim_{N \to \infty}\sum_{n=1}^{N} \frac{1}{n} \, .
$$
And just like series can be either convergent or divergent, the same is true of improper integrals. The example you gave is convergent:
\begin{align}
\int_{0}^{\infty}e^{-x} &= \lim_{n \to \infty} \int_{0}^{n}e^{-x} \\
&=  \lim_{n \to \infty} \left[-e^{-x}\right]_{0}^{n} \\
&= \lim_{n \to \infty} (-e^{-n}+1) \\
&= 1 \, .
\end{align}
Other improper integrals are not so well-behaved. Consider
$$
\int_{1}^{\infty}\frac{1}{x} \, dx \, .
$$
There's another class of improper integrals that are a little different. This is when the bounds of the integral are finite, but the function itself is unbounded. Try computing
$$
\int_{0}^{1}\frac{1}{\sqrt{x}} \, dx \,
$$
using the fact that
$$
\int_{a}^{b} f(x) \, dx := \lim_{n \to a^+} \int_{n}^{b} f(x) \, dx
$$
if $f(x)$ is undefined at $x=a$.

Informally, the improper integral
$$
\int_{a}^{\infty} f(x) \, dx
$$
represents the area under the curve between $x=a$ and $x=\infty$. I feel a little uncomfortable making such a statement, since it is hard to see what 'area' means in the infinite case. However, this gives us a great intuitive picture of what improper integrals are all about, something which David G. Stork's answer (+1) alludes to. Here is another exercise. How would you define
$$
\int_{-\infty}^{\infty} x \, dx \, ?
$$
There isn't a correct answer to this question, since definitions can't be 'true' or 'false', but notice that different definitions yield different values. If you are interested in the typical way such an improper integral would be defined, feel free to ask.
A: Look at the area here, and imagine extending to the right "forever":

A: Just like
$$\sum_{k=0}^\infty t_k$$ is a shorthand for
$$\lim_{n\to\infty}\sum_{k=0}^n t_k\ ,$$
$$\int_0^\infty f(x)\,dx$$ is short for
$$\lim_{u\to\infty}\int_0^u f(x)\,dx.$$
