यदि [ x ] महत्तम पूर्णांक leq x है, तो pi^{2} | vedclass

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Question

$\pi^{2}\left[\int_{0}^{1} \sin \frac{\pi \mathrm{x}}{2}\, \mathrm{~d} \mathrm{x}+\int_{1}^{2} \sin \frac{\pi \mathrm{x}}{2}(\mathrm{x}-1)\, \mathrm{d

Vedclass · Accepted Answer

$4(\pi-1)$

Vedclass · Answer

$\pi^{2}\left[\int_{0}^{1} \sin \frac{\pi \mathrm{x}}{2}\, \mathrm{~d} \mathrm{x}+\int_{1}^{2} \sin \frac{\pi \mathrm{x}}{2}(\mathrm{x}-1)\, \mathrm{d} \mathrm{x}\right]$

$=\pi^{2}\left[-\frac{2}{\pi}\left(\cos \frac{\pi \mathrm{x}}{2}\right)+\left((\mathrm{x}-1)\left(-\frac{2}{\pi} \cos \frac{\pi \mathrm{x}}{2}\right)\right)_{1}^{2}-\int_{1}^{2}-\frac{2}{\pi} \cos \frac{\pi \mathrm{x}}{2}\, \mathrm{dx}\right]$

$=\pi^{2}\left[0+\frac{2}{\pi}+\frac{2}{\pi}+\frac{2}{\pi} \cdot \frac{2}{\pi}\left(\sin \frac{\pi \mathrm{x}}{2}\right)_{1}^{2}\right]$

$=4 \pi-4=4(\pi-1)$

यदि $[ x ]$ महत्तम पूर्णांक $\leq x$ है, तो $\pi^{2} \int \limits_{0}^{2}\left(\sin \frac{\pi x }{2}\right)( x -[ x ])^{[ x ]} dx$ बराबर है

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