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

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