Aspire Faculty ID #18109 · Topic: JEE Main 2026 (28 January Morning Shift) · Just now
JEE Main 2026 (28 January Morning Shift)

The value of $ \sum_{n=1}^{20} \left[ \sqrt{\pi} \int_0^\pi [x]\sin(nx),dx \right] $ is ______

Solution

Let $ I_n = \int_0^\pi [x]\sin(nx),dx \quad ...(1)$

Apply King Property

$ = \int_0^\pi (\pi - x)\sin(nx),dx \quad ...(2)$

By (1) + (2)

$ 2I_n = \int_0^\pi \pi\sin(nx),dx = \pi \int_0^\pi \sin(nx),dx $

$ I_n = \frac{\pi}{2} \int_0^\pi \sin(nx),dx $

$ I_n = \frac{\pi}{2} \cdot \frac{2}{n}(1 - (-1)^n) $

$ = \frac{\pi}{n}(1 - (-1)^n) $

Thus only odd $ n $ contribute

$ S = 1 + 2 + 3 + \cdots + 20 = \frac{20 \cdot 21}{2} = 210 $
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