My Math Forum Riemann Sum help please

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 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: $\int_{a}^{b} f(x) dx= \lim_{n\to\infty} \left(\frac{b-a}{n}\right) \sum_{i=0}^n f(a + \frac{b-a}{n}i)$ Your problem is this: $\int_1^2 3x^2 + 2 dx$ f(x) = 3x^2 + 2; a=1; b=2 $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{2-1}{n}\right) \sum_{i=0}^n f(1 + \frac{2-1}{n}i)$ $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{2-1}{n}\right) \sum_{i=0}^n 3(1 + \frac{2-1}{n}i)^2 + 2$ That there is your problem... Simplify... $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{1}{n}\right) \sum_{i=0}^n 3(1 + \frac{1}{n}i)^2 + 2$ $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{1}{n}\right) \sum_{i=0}^n 3(1 + \frac{2}{n}i + \frac{1}{n^2}i^2) + 2$ $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{1}{n}\right) \sum_{i=0}^n 5 + \frac{6}{n}i + \frac{3}{n^2}i^2$ $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{1}{n}\right) \cdot \left[ \sum_{i=0}^n 5 + \sum_{i=0}^n \frac{6}{n}i + \sum_{i=0}^n \frac{3}{n^2}i^2 \right]$ $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{1}{n}\right) \cdot \left[ 5\sum_{i=0}^n 1 + \frac{6}{n}\sum_{i=0}^n i + \frac{3}{n^2}\sum_{i=0}^n i^2 \right]$ Now you need to know how to interpret summations... rewrite them in a simplified algebraic form... $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{1}{n}\right) \cdot \left[ 5n + \frac{6}{n}\frac{n(n+1)}{2} + \frac{3}{n^2}\frac{n(n+1)(2n+1)}{6} \right]$ $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left(\frac{1}{n}\right) \left[ 5n + 3(n+1) + \frac{(n+1)(2n+1)}{2n} \right]$ $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left[ 5 + 3\frac{n+1}{n} + \frac{(n+1)(2n+1)}{2n^2} \right]$ Evaluate the limit... $\int_{1}^{2} f(x) dx= \lim_{n\to\infty} \left[ 5 + 3(1+{1\over n}) + \frac{(1+{1\over n})(2+{1\over n})}{2} \right]$ $\int_{1}^{2} f(x) dx= 5 + 3(1+0) + \frac{(1+0)(2+0)}{2}$ $\int_{1}^{2} f(x) dx= 5 + 3 + 1 = 9$

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