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April 28th, 2017, 11:13 AM   #1
ZMD
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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?
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April 28th, 2017, 01:58 PM   #2
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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
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