Riemann Sums in Sigma Notation

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  1. Ken Schwartz
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Комментарии
Автор

Thank you! Best explanation I have ever found.

themathcorner
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This is very helpful! I followed the steps and managed to solve the following with much clarity and (no gaps)!!
What I probably missed was, however, can I replace r = 1 to n with r = 0 to n-1?
log y = {n to ∞} Σ (log 1/k - r/kn) (1/n) {from r = 0 to n-1}
Let's say 1/k = a.
⇒ log y = 1/(0-a) {n to ∞} Σ log [a + (0-a)r/n] (0-a)/n {from r = 0 to n-1}
⇒ log y = -1/a ∫ log (x) dx {from a to 0}
And I got the answer as 1/ke (which was an answer option given, but I don't know if its right).
Thank you for an exact and crisp explanation.

mithileshprajapati
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great explanation, thank you for helping me finally understand this

JoelNUmana