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

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Question

(C) આપેલ પદ $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r}{n} \sin ^{-1} \left( \frac{r}{n} \right)$ છે.આ રીમાન સરવાળો છે,જેને નિશ્ચિ

Vedclass · Accepted Answer

$\frac{\pi}{8}$

Vedclass · Answer

(C) આપેલ પદ $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r}{n} \sin ^{-1} \left( \frac{r}{n} \right)$ છે.આ રીમાન સરવાળો છે,જેને નિશ્ચિત સંકલન $\int_0^1 x \sin ^{-1} x \, dx$ તરીકે લખી શકાય.ખંડશઃ સંકલનનો ઉપયોગ કરતા,$u = \sin ^{-1} x$ અને $dv = x \, dx$ લેતા,$du = \frac{1}{\sqrt{1-x^2}} \, dx$ અને $v = \frac{x^2}{2}$ મળે.$\int_0^1 x \sin ^{-1} x \, dx = \left[ \frac{x^2}{2} \sin ^{-1} x \right]_0^1 - \int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} \, dx$.પ્રથમ પદની કિંમત: $\left[ \frac{1^2}{2} \sin ^{-1}(1) - 0 \right] = \frac{\pi}{4}$.બીજા પદ માટે: $-\frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int_0^1 \frac{x^2 - 1 + 1}{\sqrt{1-x^2}} \, dx = \frac{1}{2} \int_0^1 \sqrt{1-x^2} \, dx - \frac{1}{2} \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx$.પ્રમાણિત સંકલનનો ઉપયોગ કરતા: $\frac{1}{2} \left[ \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin ^{-1} x \right]_0^1 - \frac{1}{2} \left[ \sin ^{-1} x \right]_0^1 = \frac{\pi}{8} - \frac{\pi}{4} = -\frac{\pi}{8}$.કુલ સરવાળો: $\frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8}$.

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

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