lim _{n rightarrow infty}left[frac{sqrt{n^2-1 | vedclass

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(A) આપેલ પદાવલિ $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{\sqrt{n^2-r^2}}{n^2}$ છે.આને આપણે $S = \lim _{n \rightarrow \infty} \sum_{r=1}^

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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(A) આપેલ પદાવલિ $S = \lim _{n ightarrow \infty} \sum_{r=1}^{n} \frac{\sqrt{n^2-r^2}}{n^2}$ છે.આને આપણે $S = \lim _{n ightarrow \infty} \sum_{r=1}^{n} \frac{n \sqrt{1-(r/n)^2}}{n^2} = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sqrt{1-(\frac{r}{n})^2}$ તરીકે લખી શકીએ.નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ,$\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.અહીં,$f(x) = \sqrt{1-x^2}$.તેથી,$S = \int_{0}^{1} \sqrt{1-x^2} dx$.પ્રમાણિત સંકલન સૂત્ર $\int \sqrt{a^2-x^2} dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$ નો ઉપયોગ કરતા:$S = [\frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1}(x)]_{0}^{1}$.સીમાઓ મૂકતા: $S = (0 + \frac{1}{2} \sin^{-1}(1)) - (0 + \frac{1}{2} \sin^{-1}(0)) = \frac{1}{2} 	imes \frac{\pi}{2} = \frac{\pi}{4}$.

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

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જો $\lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n \cdot n!}\right)=\int_1^2 f(x) d x$ હોય, તો $f(x)=$

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