Q32 / 75 · JEE Main 2025 (8 Apr Shift 2) · Mathematics
The integral $\int_{-1}^{\frac{3}{2}}\left(\left|\pi^2 x \sin (\pi x)\right|\right) d x$ is equal to:
- A$3+2 \pi$
- B$4+\pi$
- Correct answerC$1+3 \pi$
- D$2+3 \pi$
Solution
Let, $\mathrm{I}=\pi^2 \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}$
$\begin{aligned}<br/>& =\pi^2\left\{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \\ & =\pi^2\left\{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\}<br/>\end{aligned}$
Consider
$\begin{aligned}<br/>& \int \mathrm{x} \sin \pi \mathrm{xdx} \\ & -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \\ & =-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}<br/>\end{aligned}$
$\begin{aligned} \mathrm{I} & =\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \\ & =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \\ & =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \\ & =3 \pi+1\end{aligned}$
$\begin{aligned}<br/>& =\pi^2\left\{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \\ & =\pi^2\left\{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\}<br/>\end{aligned}$
Consider
$\begin{aligned}<br/>& \int \mathrm{x} \sin \pi \mathrm{xdx} \\ & -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \\ & =-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}<br/>\end{aligned}$
$\begin{aligned} \mathrm{I} & =\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \\ & =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \\ & =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \\ & =3 \pi+1\end{aligned}$
Hence, the correct answer is option: 'C'.