Integral calculus (pretty sure it's definite but correct me if wrong...)
Anyway: Calculate the area under the following functions in the indicated region:
f(x) = 3sin(2x) dx from x=0 to x=3pi
I have two answers but I'm confused so I would like some help (the answer isn't needed but some guidance in the right direction)
1) I antidiff. the function to get: -1/6 cos (6x) + c (am I even allowed to change the thing inside the bracket to begin with, left me thinking this solution isn't correct)
2) [-1/6 cos(6x)]0-3 - [-1/6 cos(6x)]0-3 (I don't know how to use the equation writer but this is from 0 to 3pi)
3) I then got confused, do I sub in 3pi? and 0 for the other one which will result in 0? or do I just sub in 3 alone?
The second solution I got was:
1) Antidiff. to get -3/2 cos(2x) + c (according to the formula sheet this is what I do)
2) [-3/2 cos(2x)]0-3 - [-3/2 cos(2x)]0-3
3) I'm still not sure whether to sub in 3pi or something else
Help will be +1 and appreciated. And if someone can help with the question above thank you 
What's the difference between definite and indefinite (is it that one is specific i.e. it's defined to a specific area?)
Not able to answer the whole lot right now but can offer this on the very last part:
 dx)
An indefinite integral is an answer to the question 'what do I differentiate (w.r.t. x) if I want to end up with this thing between the ∫ and the dx: f(x)?'
 dx)
A definite integral is an answer to the question 'what do I get if i slice the x axis (or at least the part between the two terminals) into an infinite number of tiny chunks, multiply each chunk's tiny width by f(x), and add together all the results'
It turns out that this definite integral thing represents the same thing as the net area 'under a graph' because you can imagine splitting the area into tiny rectangles with width dx and height f(x) and adding these tiny rectangles gives you the total net area ('net' because if f(x) is negative you have negative 'heights' and so negative-area 'rectangles' actually take away from the sum total)
It also turns out that (and this is my guess as to why the notations are so similar?) that if you want to calculate a definite integral, you don't have to go and split up the x axis into infinite tiny pieces, you can just solve an indefinite integral instead, and then sub-in the terminals like:
 dx = \left [ \int f(x) dx \right ]_a^b)
*
where those square brackets mean substitute in b, substitute in a, and then find the difference between those two results (you are probably familiar with this bit):
]_a^b = g(b) - g(a))
That's a huge help because antidifferentiation (indefinite integrals) is usually a lot easier than adding an infinite number of things together (definite integrals) and this fact is so profound it earned the name 'The Fundamental Theorem of Integral Calculus'
Hope that perspective helps a bit
May have time to tackle your other questions a night or two from now if nobody else gets around to them first!
*this way of writing it isn't perfect. I specifically mean to antidifferentiate inside the square brackets THEN substitute into the answer. But I think it's a little more natural than the more common form something like:
 dx = [F(x)]_a^b)
where
 = \int f(x) dx)
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