lim _{n rightarrow infty}left{frac{1}{sqrt{4 | vedclass

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(D) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{\sqrt{4n^2 - r^2}}$.We can rewrite the term inside the summation

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(D) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{\sqrt{4n^2 - r^2}}$.We can rewrite the term inside the summation as $\frac{1}{\sqrt{n^2(4 - (r/n)^2)}} = \frac{1}{n \sqrt{4 - (r/n)^2}}$.Thus,$S = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{4 - (r/n)^2}}$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.Here,$f(x) = \frac{1}{\sqrt{4 - x^2}}$.So,$S = \int_{0}^{1} \frac{1}{\sqrt{2^2 - x^2}} dx$.Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \arcsin(\frac{x}{a}) + C$,we get:$S = [\arcsin(\frac{x}{2})]_{0}^{1} = \arcsin(\frac{1}{2}) - \arcsin(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6}$.

$\lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{4 n^2-1^2}}+\frac{1}{\sqrt{4 n^2-2^2}}+\frac{1}{\sqrt{4 n^2-3^2}}+\dots+\frac{1}{\sqrt{4 n^2-n^2}}\right\}=$

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