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

$\lim _{n}$ ${\rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{n}{n^2+(2n)^2}\right)=$

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

જો $k \in N$ હોય,તો $\lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots+\frac{1}{k n}\right]=$

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$ નું મૂલ્ય શોધો.

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