Q49 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
The value of $\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r} x|\sin \pi x| d x\right)}\right|\right)$ is $\_\_\_\_$
Correct answer: 210
Solution
Let $I_r = \int_{0}^{r} x|\sin \pi x| \, dx$ .....(1)
Apply King Property
$= \int_{0}^{r} (r - x)|\sin \pi x| \, dx$ .....(2)
By (1) + (2)
$2I_r = \int_{0}^{r} r|\sin \pi x| \, dx \Rightarrow I_r = \dfrac{r}{2} \int_{0}^{r} |\sin \pi x| \, dx$
$I_1 = \dfrac{1}{2} \int_{0}^{1} |\sin \pi x| \, dx = \dfrac{1}{2\pi} \int_{0}^{\pi} |\sin t| \, dt = \dfrac{1}{2\pi}(2)$
$I_2 = \dfrac{2}{2} \int_{0}^{2} |\sin \pi x| \, dx = \dfrac{2}{2\pi} \int_{0}^{2\pi} |\sin t| \, dt = \dfrac{2}{2\pi}(4)$
$S = \sqrt{\pi \cdot \dfrac{1}{2\pi} \cdot 2} + \sqrt{\pi \cdot \dfrac{2}{2\pi} \cdot 4} + \sqrt{\pi \cdot \dfrac{3}{2\pi} \cdot 6} + \ldots + \sqrt{\pi \cdot \dfrac{20}{2\pi}(2 \cdot 20)}$
$= 1 + 2 + 3 + \ldots + 20$
$= \dfrac{20 \times 21}{2} = 210$
Apply King Property
$= \int_{0}^{r} (r - x)|\sin \pi x| \, dx$ .....(2)
By (1) + (2)
$2I_r = \int_{0}^{r} r|\sin \pi x| \, dx \Rightarrow I_r = \dfrac{r}{2} \int_{0}^{r} |\sin \pi x| \, dx$
$I_1 = \dfrac{1}{2} \int_{0}^{1} |\sin \pi x| \, dx = \dfrac{1}{2\pi} \int_{0}^{\pi} |\sin t| \, dt = \dfrac{1}{2\pi}(2)$
$I_2 = \dfrac{2}{2} \int_{0}^{2} |\sin \pi x| \, dx = \dfrac{2}{2\pi} \int_{0}^{2\pi} |\sin t| \, dt = \dfrac{2}{2\pi}(4)$
$S = \sqrt{\pi \cdot \dfrac{1}{2\pi} \cdot 2} + \sqrt{\pi \cdot \dfrac{2}{2\pi} \cdot 4} + \sqrt{\pi \cdot \dfrac{3}{2\pi} \cdot 6} + \ldots + \sqrt{\pi \cdot \dfrac{20}{2\pi}(2 \cdot 20)}$
$= 1 + 2 + 3 + \ldots + 20$
$= \dfrac{20 \times 21}{2} = 210$
Hence, the correct answer is 210.