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Question Number 139738 by mathsuji last updated on 30/Apr/21

$$\underset{n\to \mathrm{\infty }}{lim}\underset{0}{\stackrel{1}{\int }}\frac{n\cdot x^n}{2+x^n}dx=?$$

Answered by mindispower last updated on 01/May/21

$${\int }_0^1\frac{{nx}^n}{2+x^n}dx={\int }_0^{1}x.\frac{d(2+x^n)}{2+x^n}$$$$={\left[xln(2+x^n)\right]}_0^1-{\int }_0^{1}ln(2+x^n)dx$$$$=ln\left(3\right)-{\int }_0^{1}ln\left(2(1+\frac{x^n}{2})\right)dx$$$$=ln\left(\frac{3}{2}\right)-{\int }_0^{1}ln(1+\frac{x^n}{2})dx$$$$0⩽ln(1+x)⩽x...\mathrm{\forall }x\in [0,\mathrm{\infty }[$$$$proof0⩽\frac{1}{1+x}⩽1\Rightarrow 0⩽{\int }_0^x\frac{1}{1+t}⩽{\int }_0^{x}1dt$$$$\iff 0⩽ln(1+x)⩽x$$$$proved$$$$\Rightarrow 0⩽ln(1+\frac{x^n}{2})⩽\frac{x^n}{2}\Rightarrow 0⩽An={\int }_0^{1}ln(1+\frac{x^n}{2})dx⩽{\int }_0^1\frac{x^n}{2}$$$$\iff 0⩽A_n⩽\frac{1}{2(n+1)}\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }An=0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1\frac{{nx}^n}{1+x^n}dx=\underset{n\to \mathrm{\infty }}{\lim }ln\left(\frac{3}{2}\right)+{\mathrm{A}}_n=ln\left(\frac{3}{2}\right)$$

Commented by mathsuji last updated on 02/May/21

$$thankyouverymuchsir$$

Answered by mathmax by abdo last updated on 01/May/21

$${\mathrm{U}}_{\mathrm{n}}={\int }_0^1\frac{{\mathrm{nx}}^{\mathrm{n}}}{2+{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}\Rightarrow {\mathrm{U}}_{\mathrm{n}}=\mathrm{n}{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{2+{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}{=}_{{\mathrm{x}}^{\mathrm{n}}=\mathrm{t}}\mathrm{n}{\int }_0^1\frac{\mathrm{t}}{2+\mathrm{t}}\times \frac{1}{\mathrm{n}}{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}\mathrm{dt}$$$$={\int }_0^1\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}}}{2+\mathrm{t}}\mathrm{dt}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{U}}_{\mathrm{n}}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\int }_0^1\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}}}{2+\mathrm{t}}\mathrm{dt}={\int }_0^1\frac{\mathrm{dt}}{2+\mathrm{t}}$$$$={\left[\ln (\mathrm{t}+2)\right]}_0^1=\ln \left(3\right)-\ln \left(2\right)$$

Commented by mathsuji last updated on 02/May/21

$$thankyouverymuchsir$$

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