Limn-r-1n1-nsinr-2n is

The correct option is C 2/π #160;lim _ n→∞ #160;∑ _ r=1 ^ n 1 / n sin #160; rπ #160;/ 2n =lim _ n→∞ #160; 1 / n ∑ _ r=1 ^ n sin #160; ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

limn→∞∑r=1n1/...

Question

$lim n \to \infty n \sum r = 1 \frac{1}{n} sin \frac{r π}{2 n}$ is

π / 2

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2

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2 / π

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none of these

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Solution

The correct option is C $2 / π$
$lim n \to \infty n \sum r = 1 \frac{1}{n} sin \frac{r π}{2 n}$

$= lim n \to \infty \frac{1}{n} n \sum r = 1 sin \frac{r π}{2 n}$

$= \int_{0}^{1} sin \frac{π x}{2} d x = - {[\frac{2}{π} cos \frac{π x}{2}]}_{0}^{1} = \frac{2}{π}$

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limn→∞n∑r=11nsinrπ2n is

The correct option is C 2/πlimn→∞n∑r=11nsinrπ2n=limn→∞1nn∑r=1sinrπ2n=∫10sinπx2dx=−[2πcosπx2]10=2π

$lim n \to \infty n \sum r = 1 \frac{1}{n} sin \frac{r π}{2 n}$ is

The correct option is C $2 / π$
$lim n \to \infty n \sum r = 1 \frac{1}{n} sin \frac{r π}{2 n}$

$= lim n \to \infty \frac{1}{n} n \sum r = 1 sin \frac{r π}{2 n}$

$= \int_{0}^{1} sin \frac{π x}{2} d x = - {[\frac{2}{π} cos \frac{π x}{2}]}_{0}^{1} = \frac{2}{π}$