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Question Number 147688 by mathmax by abdo last updated on 22/Jul/21
$$\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \int_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\sqrt{\mathrm{n}}} \:\:\:\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{arctan}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$
Answered by ArielVyny last updated on 24/Jul/21
![according that lim ∫f(x)dx=∫limf(x)dx we have lim_(n→+∞) ∫_(1/n) ^(√n) xe^(−x^2 ) arctan(nx)dx =(π/2)∫_0 ^(+∞) xe^(−x^2 ) dx x^2 =t→2xdx=dt ∫_0 ^(+∞) e^(−t) (1/2)dt=(1/2)[−e^(−t) ]_0 ^(+∞) =(1/2)[0+1]=(1/2) lim_(n→+∞) ∫_(1/n) ^(√n) xe^(−x^2 ) arctg(nx)dx=(π/4)](Q147874.png)
$${according}\:{that}\:{lim}\:\int{f}\left({x}\right){dx}=\int{limf}\left({x}\right){dx} \\ $$$${we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \int_{\frac{\mathrm{1}}{{n}}} ^{\sqrt{{n}}} {xe}^{−{x}^{\mathrm{2}} } {arctan}\left({nx}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{+\infty} {xe}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${x}^{\mathrm{2}} ={t}\rightarrow\mathrm{2}{xdx}={dt} \\ $$$$\int_{\mathrm{0}} ^{+\infty} {e}^{−{t}} \frac{\mathrm{1}}{\mathrm{2}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left[−{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{0}+\mathrm{1}\right]=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{n}\rightarrow+\infty} \int_{\frac{\mathrm{1}}{{n}}} ^{\sqrt{{n}}} {xe}^{−{x}^{\mathrm{2}} } {arctg}\left({nx}\right){dx}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$