The integral ${\int }_{-1}^{\frac{3}{2}}\left(\left|{\pi }^{2}x\sin (\pi x)\right|\right)dx$ is equal to : [JEE Main 2025, 8 Apr (Shift 1)]

Question

YESPYQ · Accepted Answer

Correct answer: c) $1+3\pi$. $I={\pi }^{2}\int _{−1}^{3/2}|x\sin \pi x|dx$ $={\pi }^{2}\left{\int _{−1}^{1}x\sin \pi xdx−\int _{1}^{3/2}x\sin \pi xdx\right}$ $={\pi }^{2}\left{2\int _{0}^{1}x\sin \pi xdx−\int _{−1}^{3/2}x\sin \pi xdx\right}$ $\int x\sin \pi xdx−x⋅\frac{1}{\pi }\cos \pi x+$$\int 1⋅\frac{1}{\pi }\cos \pi xdx$ $={\pi }^{2}\left{2\left(−\frac{x}{\pi }\cos \pi x+{\left.\frac{\sin \pi x}{{\pi }^{2}}\right)}_{0}^{1}\right.\right.$ $−\left(−\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^{2}}\right)$ $={\pi }^{2}\left{\frac{2}{\pi }−\left(−\frac{1}{{\pi }^{2}}−\frac{1}{\pi }\right)\right}$ $={\pi }^{2}\left{{\left(−\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^{2}}\right)}_{0}^{1}\right.$ $−\left(−\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^{2}}\right)$ $={\pi }^{2}\left{\frac{2}{\pi }−\left(−\frac{1}{{\pi }^{2}}−\frac{1}{\pi }\right)\right}$ $={\pi }^{2}\left{\frac{3}{\pi }+\frac{1}{{\pi }^{2}}\right}$ $=3\pi +1$

The integral \({\int }_{-1}^{\frac{3}{2}}\left(\left|{\pi }^{2}x\sin (\pi x)\right|\right)dx\) is equal to : [JEE Main 2…

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