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Question Number 212899 by vasil92 last updated on 26/Oct/24

Answered by MrGaster last updated on 02/Nov/24

$$\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1\sqrt{1-x^n}dx=1$$$$=\underset{n\to \mathrm{\infty }}{\lim }[x-\frac{x^{n-1}}{(n+1){\cdot }^2}\cdot \left(\begin{array}{c}n+1\\ 1\end{array}\right)\cdot \frac{1}{\sqrt{1-x^n}}{\mid }_0^1]$$$$=\underset{n\to \mathrm{\infty }}{\lim }[1-\frac{1}{n+1}\cdot \frac{1}{\sqrt{1-1^n}}-(0-\frac{0^{n+1}}{(n+1)\cdot 2}\cdot \left(\begin{array}{c}n+1\\ 1\end{array}\right)\cdot \frac{1}{\sqrt{1-0^n}})]$$$$=\underset{n\to \mathrm{\infty }}{\lim }[1-\frac{1}{n+1}\cdot \frac{1}{\sqrt{1-1}}]=\overline{)\begin{array}{c}\underset{n\to \mathrm{\infty }}{\lim }[1-\frac{1}{n+1}\cdot 0]=\underset{n\to \mathrm{\infty }}{\lim }[1-0]=1\end{array}}$$

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