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Find the anti-derivative

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Find the anti-derivative

Post  funion987 on Mon Jan 12, 2009 11:10 pm

Find the anti-derivative of:

v(t) = ln(t2+1)

Or, find the area of the curve between t=0 and t=3 for the above function.
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Re: Find the anti-derivative

Post  bfrsoccer on Mon Jan 12, 2009 11:45 pm

Using my Voyage 200, its antiderivative is: t*ln(t^2+1) + 2(arctan(t)-t). It seems kind of complicated if you actually are expected to find this, but something like partial fractions might possibly help, I don't know. You might also be able to take the inverse of the function and find the area using that, if that somehow would make it easier (the limits of integration would be from 0 to where the inverse x coordinate where y=3, and you'd subtract that from the rectangle of area formed by that x coordinate where y=3 times 3, or something like that). Then there are Riemann sums (which you could take probably an infinite amount of, using some sort of limit), I guess, as you probably know.

I'm really not sure what's the best way.
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Re: Find the anti-derivative

Post  funion987 on Mon Jan 12, 2009 11:48 pm

I didn't think i was supposed to use the fundamental theorem of calculus, but since I don't have a graphing calculator, I can't really find left and right-hand sums.

Thanks anyway.
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online graphing calculator

Post  karooomph on Tue Jan 13, 2009 4:32 pm

I have no idea what you guys are talking about, but funion said he doesn't have a graphing calculator or something.

here's the link to an online graphing calculator you can get online for free (TI-89, I believe)

http://mathematics.mc.maricopa.edu/seims/TI-Emulator.htm

just download and unzip, instructions are included.

Very Happy
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Re: Find the anti-derivative

Post  funion987 on Tue Jan 13, 2009 9:48 pm

bfrsoccer wrote:Then there are Riemann sums (which you could take probably an infinite amount of, using some sort of limit), I guess, as you probably know.
I totally thought of coolv's "question" when Ms. Flaws told us about Reimann sums.
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Term Papers

Post  orten999 on Thu Apr 22, 2010 1:06 am

Keep up this good work of helping others. Two thumbs for you.


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