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Question Number 147688 by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{find}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\int }_{\frac{1}{\mathrm{n}}}^{\sqrt{\mathrm{n}}}{\mathrm{xe}}^{-{\mathrm{x}}^2}\arctan \left(\mathrm{nx}\right)\mathrm{dx}$$

Answered by ArielVyny last updated on 24/Jul/21

$$accordingthatlim\int f\left(x\right)dx=\int limf\left(x\right)dx$$$$wehave{lim}_{n\to +\mathrm{\infty }}{\int }_{\frac{1}{n}}^{\sqrt{n}}{xe}^{-x^2}arctan\left(nx\right)dx$$$$=\frac{\pi }{2}{\int }_0^{+\mathrm{\infty }}{xe}^{-x^2}dx$$$$x^2=t\to 2xdx=dt$$$${\int }_0^{+\mathrm{\infty }}{e}^{-t}\frac{1}{2}dt=\frac{1}{2}{[-e^{-t}]}_0^{+\mathrm{\infty }}=\frac{1}{2}[0+1]=\frac{1}{2}$$$${lim}_{n\to +\mathrm{\infty }}{\int }_{\frac{1}{n}}^{\sqrt{n}}{xe}^{-x^2}arctg\left(nx\right)dx=\frac{\pi }{4}$$$$$$

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