JEE Main 2026 — 28 January, Morning Shift — Question 23

Let $\mathrm{I}_{\mathrm{r}}=\int_{0}^{\mathrm{r}} \mathrm{x}|\sin \pi \mathrm{x}| \mathrm{dx}$

Apply King Property $\begin{gathered} =\int_{0}^{\mathrm{r}}(\mathrm{r}-\mathrm{x})|\sin \pi \mathrm{x}| \mathrm{dx} \end{gathered}$ By + (2) $2 \mathrm{I}_{\mathrm{r}}=\int_{0}^{\mathrm{r}} \mathrm{r}|\sin \pi \mathrm{x}| \mathrm{dx} \Rightarrow \mathrm{I}_{\mathrm{r}}=\frac{\mathrm{r}}{2} \int_{0}^{\mathrm{r}}|\sin \pi \mathrm{x}| \mathrm{dx}$

$\mathrm{I}_{1}=\frac{1}{2} \int_{0}^{1}|\sin \pi \mathrm{x}| \mathrm{dx}=\frac{1}{2 \pi} \int_{0}^{\pi}|\sin \mathrm{t}| \mathrm{dt}=\frac{1}{2 \pi}(2)$

$$\begin{aligned} & \mathrm{I}_{2}=\frac{2}{2} \int_{0}^{2}|\sin \pi \mathrm{x}| \mathrm{dx}=\frac{2}{2 \pi} \int_{0}^{2 \pi}|\sin \mathrm{t}| \mathrm{dt}=\frac{2}{2 \pi} \end{aligned}$$

$\mathrm{S}=\sqrt{\pi \cdot \frac{1}{2 \pi} \cdot 2}+\sqrt{\pi \cdot \frac{2}{2 \pi} \cdot 4}+\sqrt{\pi \cdot \frac{3}{2 \pi} \cdot 6}+\ldots+\sqrt{\pi \cdot \frac{20}{2 \pi}(2 \cdot 20)}$

$=1+2+3+\ldots+20$

$=\frac{20 \times 21}{2}=210$