Here's a first year calculus student's (my) guide to integration.
Integration is a way of finding the area between the xaxis and a curve (or line) on a graph. According to my maths teacher, it works by dividing the area up into an infinite number of infinitesimally small rectangles.
Firstly, let's choose a polynomial to integrate:
x^{3} + 6x^{2} + 3x + 8
We also choose a limit, which is how much we want to integrate, we'll choose to integrate the area under the curve between x = 0 and x = 4.
For each term in the equation, we raise the exponent by one and divide that term by the new exponent.
x^{3} becomes (x^{4}) / 4
6x^{2} becomes (6x^{3}) / 3, which is the same as 2x^{3}.
3x becomes (3x^{2}) / 2, because 3x is the same as 3x^{1}.
8 becomes 8x, because 8 is the same as 8x^{0} so it becomes (8x^{1}) / 1 which is 8x.
We use a funky looking ∫ thing to show that it's an integral, apparently it's an Old English S and stands for 'sum'. Our integral is:
/\4

 x^{3} + 6x^{2} + 3x + 8 dx
0 
\/
x^{4} 3x^{2}
=  + 2x^{3} +  + 8x
4 2
We then plug the upper and lower limits into our integral.
4: 4^{4} / 4 + 2 * 4^{3} + 3 * 4^{2} / 2 + 8 * 4 = 248
0: 0^{4} / 4 + 2 * 0^{3} + 3 * 0^{2} / 2 + 8 * 0 = 0
We then take the difference between these two values, so (248  0) = 248. The area under the graph where x is between 0 and 4 is 248.
This was a simple polynomial example. You should know that there are some things that cannot be integrated using this technique. See integrating sin(ln(x)) and impossible integral.
A simple mechanics example:
A train starts from rest and accelerates for 5 seconds. While it is accelerating, its velocity in ms^{1} is given by the equation v = t^{2}. Find out how far it's travelled after the first 5 seconds.
(Okay, this is pretty unrealistic, a train wouldn't accelerate like that, but just humour me.)
In a velocitytime graph, the area under the graph is always displacement (how far it's travelled), so we integrate to find the area under the graph, which will give us the distance travelled.
Our graph would look something like this...
v
50
 '

 .


 '
 .//
 ///
 '/////
.__'////////

0 5 t
The shaded area is the area that we're trying to find. Alright, our equation is:
v = t^{2}
t^{2} becomes (t^{3})/3
So our integral is:
/\5

 t^{2} dx
0 
\/
t^{3}
= 
3
So we plug in our values.
5: 5^{3} / 3 = 41.666
0: 0^{3} / 3 = 0
Then find the difference to get the area. (41.666  0) = 41.666.
ms^{1} is just the same as "meters per second", so our answer will be in meters. So the train has travelled 41.7m.
And now you know.