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Question Number 205534 by hardmath last updated on 23/Mar/24

$$\mathrm{Find}:$$$$\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }{\int }_0^1\mathrm{n}{\mathrm{x}}^{\mathbf{n}}{\mathrm{e}}^{{\mathbf{x}}^2}\mathrm{dx}=?$$

Answered by Berbere last updated on 24/Mar/24

$$I_0=0$$$$I_1=\frac{1}{2}(e-1)$$$$\frac{I_n}{n}={\int }_0^1{x}^n{e}^{x^2}⩽e{\int }_0^1{x}^n=\frac{e}{n+1}\to 0....\left(E\right)$$$$I_{n+1}\stackrel{IBP}{=}(n+1){\int }_0^1{x}^{n+1}{e}^{x^2}dx=(n+1)\left[{}_0^1\frac{e^{x^2}}{2}{x}^n\right]-\frac{n(n+1)}{2}{\int }_0^1{x}^{n-1}{e}^{x^2}dx$$$$‘‘x^{n+1}{e}^{x^2}={\left(x^n\right)}_u.{\left({xe}^{x^2}\right)}_{v^{\prime }}^{″}$$$$=(n+1)\frac{e}{2}-\frac{(n+1)n}{2(n-1)}{I}_{n-1}$$$$\frac{I_{n+1}}{n+1}=\frac{e}{2}-\frac{n}{2}.\frac{I_{n-1}}{n-1};applie\left(E\right),\frac{I_{n+1}}{n+1}\stackrel{\mathrm{\infty }}{\to }0$$$$\frac{I_{n+1}}{n+1}\to 0\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{I_{n-1}}{n-1}.\frac{n}{e}=1\Rightarrow I_{n-1}\sim \frac{e(n-1)}{n}\stackrel{\mathrm{\infty }}{\to }e$$

Commented by hardmath last updated on 24/Mar/24

$$\mathrm{Yes}\mathrm{answer}=\mathrm{e},\mathrm{cool}\mathrm{dear}\mathrm{professor},\mathrm{thankyou}$$

Commented by Berbere last updated on 24/Mar/24

$$withePleasur$$

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