Quote:
Originally posted by KnifeMissle
...so trivial that you proceed to not state the antiderivative...
Oh, and sqrt(x) = x^(1/2). It's a square root and not an a'th root, right?
The fact of the matter is that almost everyone here knows enough calculus to find the area given the antiderivative. The hard part is finding that antiderivative. I can tell that you're new to calculus because the rather simple looking sqrt(r^2 - (r - x)^2) is actually quite hard to find an antiderivative for.
For example, here's the antiderivative of a very similar expression:
∫ sqrt(r^2 - x^2) dx = sqrt(r^2 - x^2)/2 + r^2*arcsin( x/r )/2 + C
I got this by looking it up in an integral table (there happens to be a physics text book in the office). How would you "trivially" do it, tehhappyboy--integration by parts?
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Ok, somebody here has to have a TI-89 floating around...