My Math Forum Riemann Sums

 Calculus Calculus Math Forum

 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: 2,403 Thanks: 1306 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top