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Question Number 81853 by jagoll last updated on 16/Feb/20

$$\underset{x\to \mathrm{\infty }}{\lim }\left\{n\underset{0}{\stackrel{1}{\int }}\frac{x^n}{x^3+1}dx\right\}=?$$

Commented by john santu last updated on 16/Feb/20

$$\frac{1}{2}istheanswer$$

Commented by mathmax by abdo last updated on 16/Feb/20

$$let{U}_n=n{\int }_0^1\frac{x^n}{1+x^3}dxchangement{x}^n=tgivex=t^{\frac{1}{n}}\Rightarrow$$$$U_n=n{\int }_0^1\frac{t}{1+t^{\frac{3}{n}}}\times \frac{1}{n}{t}^{\frac{1}{n}-1}dt={\int }_0^1\frac{t^{\frac{1}{n}}}{1+t^{\frac{3}{n}}}dt={\int }_0^1{f}_n\left(t\right)dt$$$$f_{n}c.stof\left(x\right)=\frac{1}{2}and0⩽f_n⩽1\Rightarrow 0⩽{\int }_0^1{f}_n(t⩽1theoremof$$$$convergencedomineegive{lim}_{n\to +\mathrm{\infty }}{U}_n={\int }_0^1{lim}_{n\to +\mathrm{\infty }}{f}_n\left(t\right)dt$$$$=\frac{1}{2}$$

Answered by mind is power last updated on 16/Feb/20

$$u=x^n$$$$\Rightarrow dx=\frac{u^{\frac{1}{n}-1}du}{n}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\left\{{\int }_0^1\frac{u^{\frac{1}{n}}du}{u^{\frac{3}{n}}+1}\right\}$$$${\int }_0^1\frac{u^{\frac{1}{n}}}{u^{\frac{3}{n}}+1}⩽{\int }_0^1\frac{1}{1}du=1\Rightarrow absolutecv$$$$wecanswitchLimand\int$$$$={\int }_0^1\underset{n\to \mathrm{\infty }}{\lim }\frac{u^{\frac{1}{n}}}{u^{\frac{3}{n}}+1}du$$$$\lim \frac{u^{\frac{1}{n}}}{u^{\frac{3}{n}}+1}=1\mu ppu\in [0,1[lesbegueintegralandnotation$$$$$$$$n\to \mathrm{\infty }$$$$weget={\int }_0^1\frac{1}{2}du=\frac{1}{2}$$$$$$

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