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 April 28th, 2017, 11:13 AM #1 Member   Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1 Riemann Sums $\lim_{n \to \infty} \sum_{i=0}^n sin \left[ \left( 2 + \frac{3}{2n} \right) + \frac{3}{n}i \right] \cdot \frac{3}{n} ~=~ \int_a^b \sin(x)$ dx What will be the value of b and a? April 28th, 2017, 01:58 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 ignoring the limit for a moment our leftmost limit occurs at $i=0$ $a_n = 2 + \dfrac {3}{2n}$ our rightmost limit occurs at $i=n$ $b_n = 2 + \dfrac{3}{2n} + 3$ pretty clearly $a= \lim \limits_{n\to \infty} a_n = 2$ $b = \lim \limits_{n\to \infty} b_n = 5$ and this of course agrees with $\Delta = \dfrac{b-a}{n} = \dfrac{3}{n}$ as observed in the rightmost factor Tags riemann, sums 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 slabbxo Calculus 1 February 25th, 2014 04:52 AM n3rdwannab3 Calculus 1 January 17th, 2014 03:41 PM mctiger Real Analysis 0 May 5th, 2013 07:03 AM nubshat Calculus 1 November 17th, 2012 07:55 AM nubshat Calculus 2 November 13th, 2012 05:04 PM

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