January 25th, 2009, 07:01 AM  #1 
Newbie Joined: Nov 2008 Posts: 12 Thanks: 0  Riemann Sum help please
I'm stuck on this Riemann sum and I have no idea what I'm doing wrong: y= 3x^2 + 2 on [1,2] Ive gotten Delta X = 1/n xi = 1 + 1/n the equation i used was  3[(1+(1/n)i)^2 + 2] [1/n] i simplified that down into this by distributing [3 + 3/n^2 i^2 +2] [1/n] that turns into  (3/n) + (3/n^3)i^2 + (2/n) that then becomes (3/n) and (2/n) can be taken out of the equation to become 5... and I replace i^2 with (n(n+1)(2n+1))/6 and i multiply it with 3/n^3 I take the limit of that as it approaches infinity and i get 1. 3+2+1 is 6, but the when I take the anti derivative of this it is 9?!?!?!? I need to know where I went wrong!!! and would someone write this out for me step by step from the beginning if you have the time? Thanks! 
January 25th, 2009, 12:54 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,709 Thanks: 675  Re: Riemann Sum help please
(1+i/n)^2 = 1 + 2i/n + (i/n)^2 In you calculation you left out the middle term. 
January 31st, 2009, 02:47 PM  #3 
Member Joined: Dec 2008 From: Washington state, of the United States Posts: 37 Thanks: 0  Re: Riemann Sum help please
Definite Integral is defined with the infinite Riemann Sum: Your problem is this: f(x) = 3x^2 + 2; a=1; b=2 That there is your problem... Simplify... Now you need to know how to interpret summations... rewrite them in a simplified algebraic form... Evaluate the limit... 

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