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

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(D) ધારો કે $I = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{\sqrt{4n^2-r^2}}$.આ પદાવલિને નીચે મુજબ લખી શકાય:$I = \lim _{n \rightarrow \infty}

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(D) ધારો કે $I = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{\sqrt{4n^2-r^2}}$.આ પદાવલિને નીચે મુજબ લખી શકાય:$I = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{\sqrt{4 - (\frac{r}{n})^2}}$.આ નિશ્ચિત સંકલન માટે રીમાન સરવાળો છે:$I = \int_{0}^{1} \frac{1}{\sqrt{4 - x^2}} dx$.પ્રમાણિત સંકલન સૂત્ર $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$ નો ઉપયોગ કરતા:$I = \left[ \sin^{-1}(\frac{x}{2}) \right]_{0}^{1}$.$I = \sin^{-1}(\frac{1}{2}) - \sin^{-1}(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6}$.

$\lim _{n \rightarrow \infty} \left( \frac{1}{\sqrt{4n^2-1}} + \frac{1}{\sqrt{4n^2-4}} + \dots + \frac{1}{\sqrt{4n^2-n^2}} \right)$ ની કિંમત શોધો.

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