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 $\Delta = \dfrac{ba}{n} = \dfrac{3}{n}$ as observed in the rightmost factor 

Tags 
riemann, sums 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Riemann Sums  slabbxo  Calculus  1  February 25th, 2014 04:52 AM 
Riemann Sums...  n3rdwannab3  Calculus  1  January 17th, 2014 03:41 PM 
Help please on Riemann sums  mctiger  Real Analysis  0  May 5th, 2013 07:03 AM 
Limit of Riemann sums  nubshat  Calculus  1  November 17th, 2012 07:55 AM 
Riemann sums  nubshat  Calculus  2  November 13th, 2012 05:04 PM 