JEE Main 2026 — Mathematics Calculus

Let $\displaystyle I_r = \int_{0}^{r} x|\sin \pi x| \, dx$Ir​=∫0r​x∣sinπx∣dx .....(1)

Apply King Property

$\displaystyle = \int_{0}^{r} (r - x)|\sin \pi x| \, dx$=∫0r​(r−x)∣sinπx∣dx .....(2)

By (1) + (2)

$$2I_r = \int_{0}^{r} r|\sin \pi x| \, dx \Rightarrow I_r = \dfrac{r}{2} \int_{0}^{r} |\sin \pi x| \, dx$$2Ir​=∫0r​r∣sinπx∣dx⇒Ir​=2r​∫0r​∣sinπx∣dx

$$I_1 = \dfrac{1}{2} \int_{0}^{1} |\sin \pi x| \, dx = \dfrac{1}{2\pi} \int_{0}^{\pi} |\sin t| \, dt = \dfrac{1}{2\pi}(2)$$I1​=21​∫01​∣sinπx∣dx=2π1​∫0π​∣sint∣dt=2π1​(2)

$$I_2 = \dfrac{2}{2} \int_{0}^{2} |\sin \pi x| \, dx = \dfrac{2}{2\pi} \int_{0}^{2\pi} |\sin t| \, dt = \dfrac{2}{2\pi}(4)$$I2​=22​∫02​∣sinπx∣dx=2π2​∫02π​∣sint∣dt=2π2​(4)

$$S = \sqrt{\pi \cdot \dfrac{1}{2\pi} \cdot 2} + \sqrt{\pi \cdot \dfrac{2}{2\pi} \cdot 4} + \sqrt{\pi \cdot \dfrac{3}{2\pi} \cdot 6} + \ldots + \sqrt{\pi \cdot \dfrac{20}{2\pi}(2 \cdot 20)}$$S=π⋅2π1​⋅2​+π⋅2π2​⋅4​+π⋅2π3​⋅6​+…+π⋅2π20​(2⋅20)​

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

$$= \dfrac{20 \times 21}{2} = 210$$=220×21​=210