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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Definite Integration

The integral ∫_ -1 ^ (3)/(2) (|π^2 x (π x)|) d x is equal to:

The integral ∫−132(∣π2xsin⁡(πx)∣)dx\int_{-1}^{\frac{3}{2}}\left(\left|\pi^2 x \sin (\pi x)\right|\right) d x is equal to:

Options

  1. A3+2π3+2 \pi
  2. B4+π4+\pi
  3. C1+3π1+3 \pi
  4. D2+3π2+3 \pi

Correct answer

C. 1+3π1+3 \pi

Step-by-step solution

Let, I=π2∫−13/2∣xsin⁡πx∣dx\mathrm{I}=\pi^2 \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}
<br/>=π2{∫−11xsin⁡πxdx−∫13/2xsin⁡πxdx}=π2{2∫01xsin⁡πxdx−∫−13/2xsin⁡πxdx}<br/>\begin{aligned}<br/>& =\pi^2\left\{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \\ & =\pi^2\left\{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\}<br/>\end{aligned}
Consider
<br/>∫xsin⁡πxdx−x⋅1πcos⁡πx+∫1⋅1πcos⁡πxdx=−xπcos⁡πx+sin⁡πxπ2<br/>\begin{aligned}<br/>& \int \mathrm{x} \sin \pi \mathrm{xdx} \\ & -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \\ & =-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}<br/>\end{aligned}
I=π2{2(−xπcos⁡πx+sin⁡πxπ2)01−(−xπcos⁡πx+sin⁡πxπ2)13/2}=π2{2π−(−1π2−1π)}=π2{3π+1π2}=3π+1\begin{aligned} \mathrm{I} & =\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \\ & =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \\ & =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \\ & =3 \pi+1\end{aligned}
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