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

$\frac{\pi}{4}$

Vedclass · Answer

(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: $S = (0 + \frac{1}{2} \sin^{-1}(1)) - (0 + \frac{1}{2} \sin^{-1}(0)) = \frac{1}{2} 	imes \frac{\pi}{2} = \frac{\pi}{4}$.

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

Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}}\left[ {1\cos \frac{1}{{{n^2}}} + 2\cos \frac{4}{{{n^2}}} + 3\cos \frac{9}{{{n^2}}} + .... + 2n\cos 4} \right]$ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ ની કિંમત શોધો.

જો $a$ અને $b$ ધન પૂર્ણાંકો હોય કે જેથી $b > a$,તો $\lim_{n \to \infty} \left[ \frac{1}{na} + \frac{1}{na + 1} + \frac{1}{na + 2} + \dots + \frac{1}{nb} \right] = $

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $ ની કિંમત શોધો.

$\lim _{n}$ ${\rightarrow \infty} \frac{1}{n}\left[\sin \frac{\pi}{4}+\sin \frac{\pi}{12}\left(3+\frac{1}{n}\right)+\sin \frac{\pi}{12}\left(3+\frac{2}{n}\right)+\ldots+\sin \frac{\pi}{3}\right]=$

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