Question Number 212600 by MrGaster last updated on 18/Oct/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{e}^{{x}^{\mathrm{3}} } {dx}=\frac{\pi}{\mathrm{2}}. \\ $$
Answered by Berbere last updated on 21/Oct/24
![=lim_(n→∞) ∫^1 _0 ((d(nx))/(1+(nx)^2 ))e^x^3 A=lim_(n→∞) {[^1 _0 tan^(−1) (nx)e^x^3 ]−∫_0 ^1 3x^2 e^x^3 tan^(−1) (nx)dx} =(π/2)e−lim_(n→∞) ∫_0 ^1 3x^2 e^x^3 tan^(−1) (nx)dx (π/2)−(1/x)≤tan^(−1) (x)=(π/2)−tan^(−1) ((1/x))≤(π/2) ((π/2)−(1/(nx)))3x^2 e^x^3 ≤3x^2 e^x^3 tan^(−1) (nx)≤(π/2).3x^2 e^x^3 lim_(n→∞) (π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^3 ≤u=lim_(n→∞) ∫_0 ^1 3x^2 e^x^3 tan^(−1) (nx)≤(π/2)e−(π/2) (π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^2 dx≤(π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^3 ≤u≤(π/2)e−(π/2) lim_(n→∞) (π/2)e−(π/2)−(3/(2n))(e−1)≤u≤(π/2)e−(π/2) u=(π/2)e−(π/2);A=(π/2)e−u=(π/2)](https://www.tinkutara.com/question/Q212705.png)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{{d}\left({nx}\right)}{\mathrm{1}+\left({nx}\right)^{\mathrm{2}} }{e}^{{x}^{\mathrm{3}} } {A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\underset{\mathrm{0}} {\left[}^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}} \left({nx}\right){e}^{{x}^{\mathrm{3}} } \right]−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{x}^{\mathrm{2}} {e}^{{x}^{\mathrm{3}} } \mathrm{tan}^{−\mathrm{1}} \left({nx}\right){dx}\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}{e}−\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{x}^{\mathrm{2}} {e}^{{x}^{\mathrm{3}} } \mathrm{tan}^{−\mathrm{1}} \left({nx}\right){dx} \\ $$$$\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{x}}\leqslant\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{nx}}\right)\mathrm{3}{x}^{\mathrm{2}} {e}^{{x}^{\mathrm{3}} } \leqslant\mathrm{3}{x}^{\mathrm{2}} {e}^{{x}^{\mathrm{3}} } \mathrm{tan}^{−\mathrm{1}} \left({nx}\right)\leqslant\frac{\pi}{\mathrm{2}}.\mathrm{3}{x}^{\mathrm{2}} {e}^{{x}^{\mathrm{3}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{xe}^{{x}^{\mathrm{3}} } \leqslant{u}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{x}^{\mathrm{2}} {e}^{{x}^{\mathrm{3}} } \mathrm{tan}^{−\mathrm{1}} \left({nx}\right)\leqslant\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{xe}^{{x}^{\mathrm{2}} } {dx}\leqslant\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{xe}^{{x}^{\mathrm{3}} } \leqslant{u}\leqslant\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}{n}}\left({e}−\mathrm{1}\right)\leqslant{u}\leqslant\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}} \\ $$$${u}=\frac{\pi}{\mathrm{2}}{e}−\frac{\pi}{\mathrm{2}};{A}=\frac{\pi}{\mathrm{2}}{e}−{u}=\frac{\pi}{\mathrm{2}} \\ $$