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Note by Rajath Krishna R
3 years, 11 months ago

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\( \int_1^{3}[x]cos(\frac{\pi}{2}(x - [x]) dx \\ = \int_1^{2}[x]cos(\frac{\pi}{2}(x - [x]))dx + \int_2^{3}[x]cos(\frac{\pi}{2}(x - [x]))dx \\ = \int_1^{2} 1cos(\frac{\pi}{2}(x - 1) dx + \int_2^{3} 2cos(\frac{\pi}{2}(x - 2) dx \\ = \frac{1}{\frac{\pi}{2}} + 2 \frac{1}{\frac{\pi}{2}} \\ = \frac{6}{\pi} \) Jatin Yadav · 3 years, 11 months ago

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