JEE Main 2021 — Mathematics Calculus

Let, $\displaystyle I={\pi }^{2}{\int }_{0}^{2}\mathrm{sin}(\frac{\pi x}{2}){(x-[x])}^{[x]}dx$I=π2∫02​sin(2πx​)(x−[x])[x]dx

$$I={\pi }^{2}{\int }_{0}^{1}\mathrm{sin}(\frac{\pi x}{2}){(x-[x])}^{[x]}dx+{\pi }^{2}{\int }_{1}^{2}\mathrm{sin}(\frac{\pi x}{2}){(x-[x])}^{[x]}dx$$I=π2∫01​sin(2πx​)(x−[x])[x]dx+π2∫12​sin(2πx​)(x−[x])[x]dx

$$I={\pi }^{2}{\int }_{0}^{1}\mathrm{sin}(\frac{\pi x}{2}){x}^{0}dx+{\pi }^{2}{\int }_{1}^{2}\mathrm{sin}(\frac{\pi x}{2}){(x-1)}^{1}dx$$I=π2∫01​sin(2πx​)x0dx+π2∫12​sin(2πx​)(x−1)1dx

$$I={\pi }^{2}{\int }_{0}^{1}\mathrm{sin}(\frac{\pi x}{2})dx+{\pi }^{2}{\int }_{1}^{2}\mathrm{sin}(\frac{\pi x}{2})(x-1)dx$$I=π2∫01​sin(2πx​)dx+π2∫12​sin(2πx​)(x−1)dx

$$I={\pi }^{2}{\int }_{0}^{1}\mathrm{sin}(\frac{\pi x}{2})dx+{\pi }^{2}[x{\int }_{1}^{2}\mathrm{sin}(\frac{\pi x}{2})dx-{\int }_{1}^{2}{\frac{d}{dx}x{\int }_{1}^{2}\mathrm{sin}(\frac{\pi x}{2})dx}dx]-{\pi }^{2}{\int }_{1}^{2}\mathrm{sin}(\frac{\pi x}{2})dx$$I=π2∫01​sin(2πx​)dx+π2[x∫12​sin(2πx​)dx−∫12​dxd​x∫12​sin(2πx​)dxdx]−π2∫12​sin(2πx​)dx

$$I={\pi }^{2}{[-\frac{2}{\pi }\mathrm{cos}(\frac{\pi x}{2})]}_{0}^{1}+{\pi }^{2}{[(x)\frac{2}{\pi }(-\mathrm{cos}(\frac{\pi x}{2}))}_{1}^{2}-{\int }_{1}^{2}\frac{2}{\pi }(-\mathrm{cos}\frac{\pi x}{2})dx]+{\pi }^{2}\times \frac{2}{\pi }{[\mathrm{cos}(\frac{\pi x}{2})]}_{1}^{2}$$I=π2[−π2​cos(2πx​)]01​+π2[(x)π2​(−cos(2πx​))12​−∫12​π2​(−cos2πx​)dx]+π2×π2​[cos(2πx​)]12​

$$I={\pi }^{2}{[-\frac{2}{\pi }\mathrm{cos}(\frac{\pi x}{2})]}_{0}^{1}+{\pi }^{2}{[(x)\frac{2}{\pi }(-\mathrm{cos}(\frac{\pi x}{2}))}_{1}^{2}+{(\frac{2}{\pi })}^{2}{[\mathrm{sin}(\frac{\pi x}{2})]}_{1}^{2}]+{\pi }^{2}\times \frac{2}{\pi }{[\mathrm{cos}(\frac{\pi x}{2})]}_{1}^{2}$$I=π2[−π2​cos(2πx​)]01​+π2[(x)π2​(−cos(2πx​))12​+(π2​)2[sin(2πx​)]12​]+π2×π2​[cos(2πx​)]12​

$$I=-{\pi }^{2}[\frac{2}{\pi }\mathrm{cos}(\frac{\pi }{2})-\frac{2}{\pi }\mathrm{cos}(0)]+{\pi }^{2}[2\times \frac{2}{\pi }(-\mathrm{cos}\pi )+1\times \frac{2}{\pi }\times (\mathrm{cos}\frac{\pi }{2})]+4[\mathrm{sin}(\pi )-\mathrm{sin}\frac{\pi }{2}]]+{\pi }^{2}\times \frac{2}{\pi }[\mathrm{cos}(\pi )-\mathrm{cos}\frac{\pi }{2}]$$I=−π2[π2​cos(2π​)−π2​cos(0)]+π2[2×π2​(−cosπ)+1×π2​×(cos2π​)]+4[sin(π)−sin2π​]]+π2×π2​[cos(π)−cos2π​]

$$I=2\pi +4\pi +4(0-1)-2\pi$$I=2π+4π+4(0−1)−2π

$$I=4\pi -4$$I=4π−4

$$I=4(\pi -1)$$I=4(π−1)