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

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(B) दी गई सीमा $l = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} \sin \left( \frac{\pi}{4} + \frac{k \pi}{12n} \right)$ है।यह $\int_{0}^{1}

Vedclass · Accepted Answer

$\frac{6(\sqrt{2}-1)}{\pi}$

Vedclass · Answer

(B) दी गई सीमा $l = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} \sin \left( \frac{\pi}{4} + \frac{k \pi}{12n} \right)$ है।यह $\int_{0}^{1} f(x) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} f(\frac{k}{n})$ के रूप का रीमान योग है।यहाँ,$f(x) = \sin(\frac{\pi}{4} + \frac{\pi}{12}x)$ है।अतः,$l = \int_{0}^{1} \sin(\frac{\pi}{4} + \frac{\pi}{12}x) dx$ है।समाकलन का मूल्यांकन करने पर:$l = \left[ -\frac{12}{\pi} \cos(\frac{\pi}{4} + \frac{\pi}{12}x) \right]_{0}^{1}$$l = -\frac{12}{\pi} \left[ \cos(\frac{\pi}{4} + \frac{\pi}{12}) - \cos(\frac{\pi}{4}) \right]$$l = -\frac{12}{\pi} \left[ \cos(\frac{\pi}{3}) - \cos(\frac{\pi}{4}) \right]$$l = \frac{12}{\pi} \left[ \frac{1}{\sqrt{2}} - \frac{1}{2} \right] = \frac{6(\sqrt{2}-1)}{\pi}$.

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

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