What is the mathematical/physical meaning of the second integral? Next semester I will be starting Calculus 2, and I opened my textbook to get a hint of the course material, and on one of the pages I saw a function being integrated twice. I've been racking my brains trying to figure out the meaning of the second integral of a function but have been unsuccessful - I really don't want to wait two months to find out, either.
The meaning of the first integral is the area bounded between the function and the x-axis; what is the meaning of the second integral?
 A: If $a(t)$ is the acceleration at time $t$, then the first integral gives the velocity $v(t)$, and the second integral gives the displacement $s(t)$. That is the most important application, perhaps the only application, of integrating twice that you will meet this coming term.  
A: Andre is correct for the physics interpretation. Mathematically, one can describe a double integral as the volume under a surface within a particular region in the plane. For instance,
$$\int_{-1}^1\int_{-1}^1{x^2+y^2}dydx$$
would represent the area under the paraboloid $z=x^2+y^2$ in the square whose vertices are at (-1,-1),(-1,1),(1,1),(1,-1)
A: As one of the comments mentioned, you can have many. 
From physics, they have defined:
$\bullet$ $r(t)$ -  position
$\bullet$ $v(t)$ - velocity (1st derivative)
$\bullet$ $a(t)$ - acceleration (2nd derivative)
$\bullet$ $w(t)$ - jerk (3rd derivative)
$\bullet$ $s(t)$ - snap (4th derivative)
$\bullet$ $c(t)$ - crackle (5th derivative)
$\bullet$ $p(t)$ - pop (6th derivative)
You can integrate from the 6th derivative all the way back up to position.
Here is a paper on the matter.
Regards
A: Some Advice: Almost every calculation in higher level calculus is solved by reducing the problem to an elementary Calculus 101 calculation. This is one example. 
You are probably asking about a multiple integral- such as a volume or surface integral like the one mentioned in a previous answer. 
If so, don't worry because Fubini's Theorem allows you to reduce such integrals to a a set of ordinary integrals you are already familiar with. 
You never have to actually integrate over a volume or surface directly (by adding up the contributions from each volume or surface element) because you will learn how to evaluate volume and surface integrals by calculating multiple ordinary integrals- which you already know how to do.
But you will have to think carefully about the region of integration. That is new.      
