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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Definite Integration

The value of ∑_ r=1 ^ 20 ( | π(∫_ 0 ^ r x| π x| d x) |) is \_\_\_\_

The value of ∑r=120\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r} x|\sin \pi x| d x\right)}\right|\right) is ____\_\_\_\_

Correct answer

0

Step-by-step solution

Let Ir=∫0rx∣sin⁡πx∣ dxI_r = \int_{0}^{r} x|\sin \pi x| \, dx .....(1)
Apply King Property
=∫0r(r−x)∣sin⁡πx∣ dx= \int_{0}^{r} (r - x)|\sin \pi x| \, dx .....(2)
By (1) + (2)
2Ir=∫0rr∣sin⁡πx∣ dx⇒Ir=r2∫0r∣sin⁡πx∣ dx2I_r = \int_{0}^{r} r|\sin \pi x| \, dx \Rightarrow I_r = \dfrac{r}{2} \int_{0}^{r} |\sin \pi x| \, dx
I1=12∫01∣sin⁡πx∣ dx=12π∫0π∣sin⁡t∣ dt=12π(2)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)
I2=22∫02∣sin⁡πx∣ dx=22π∫02π∣sin⁡t∣ dt=22π(4)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=π⋅12π⋅2+π⋅22π⋅4+π⋅32π⋅6+…+π⋅202π(2⋅20)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+…+20= 1 + 2 + 3 + \ldots + 20
=20×212=210= \dfrac{20 \times 21}{2} = 210
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