JEE Main 2025 — Mathematics Calculus

Let, $\displaystyle \mathrm{I}=\pi^2 \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}$I=π2∫−13/2​∣xsinπx∣dx

$\begin{aligned}
& =\pi^2\left{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right} \
& =\pi^2\left{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right}
\end{aligned}$

Consider

$\begin{aligned}
& \int \mathrm{x} \sin \pi \mathrm{xdx} \
& -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \
& =-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}
\end{aligned}$

$$\begin{aligned} \mathrm{I} & =\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \\ & =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \\ & =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \\ & =3 \pi+1\end{aligned}$$I​=π2{2(−πx​cosπx+π2sinπx​)01​−(−πx​cosπx+π2sinπx​)13/2​}=π2{π2​−(−π21​−π1​)}=π2{π3​+π21​}=3π+1​