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Question Number 125015 by Khalmohmmad last updated on 07/Dec/20

Answered by Dwaipayan Shikari last updated on 07/Dec/20

$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{sinx}{x}dx=Si\left(\frac{\pi }{3}\right)-Si\left(\frac{\pi }{6}\right)=0.4697$$

Answered by mathmax by abdo last updated on 07/Dec/20

$$\mathrm{I}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}\mathrm{let}\mathrm{determine}\mathrm{spproximste}\mathrm{vslue}\mathrm{of}\mathrm{I}$$$$\mathrm{we}\mathrm{hsve}\mathrm{x}-\frac{{\mathrm{x}}^3}{6}⩽\mathrm{sinx}⩽\mathrm{x}\Rightarrow 1-\frac{{\mathrm{x}}^2}{6}⩽\frac{\mathrm{sinx}}{\mathrm{x}}⩽1\Rightarrow$$$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(1-\frac{{\mathrm{x}}^2}{6})\mathrm{dx}⩽{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}⩽\frac{\pi }{3}-\frac{\pi }{6}(=\frac{\pi }{6})\mathrm{but}$$$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(1-\frac{{\mathrm{x}}^2}{6})\mathrm{dx}={[\mathrm{x}-\frac{1}{18}{\mathrm{x}}^3]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{\pi }{3}-\frac{1}{18}\times \frac{{\pi }^3}{27}-\frac{\pi }{6}+\frac{1}{18}\times \frac{{\pi }^3}{6^3}$$$$=\frac{\pi }{6}-\frac{{\pi }^3}{18.27}+\frac{{\pi }^3}{18.216}\Rightarrow \frac{\pi }{6}+(\frac{1}{18.216}-\frac{1}{18.27}){\pi }^3⩽\mathrm{I}⩽\frac{\pi }{6}$$$$$$$$\mathrm{so}{\mathrm{v}}_0=\frac{\pi }{6}+(\frac{1}{36.216}-\frac{1}{36.27}){\pi }^3\mathrm{is}\mathrm{a}\mathrm{approximate}\mathrm{value}\mathrm{of}\mathrm{this}$$$$\mathrm{integral}0$$$$$$

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