Calculus Calculus Math Forum

 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: 6,770 Thanks: 700 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... Tags riemann, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post alphabeta89 Real Analysis 2 February 11th, 2012 03:44 AM Pixel Calculus 2 July 15th, 2010 08:01 AM selenne431 Complex Analysis 0 March 24th, 2010 12:23 PM chrisf Calculus 9 January 31st, 2009 02:19 PM cheloniophile Real Analysis 1 November 23rd, 2008 05:30 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      