Question Number 212899 by vasil92 last updated on 26/Oct/24
Answered by MrGaster last updated on 02/Nov/24
![lim_(n→∞) ∫_0 ^1 (√(1−x^n ))dx=1 =lim_(n→∞) [x−(x^(n−1) /((n+1)∙^2 ))∙ (((n+1)),(1) )∙(1/( (√(1−x^n ))))∣_0 ^1 ] =lim_(n→∞) [1−(1/(n+1))∙(1/( (√(1−1^n ))))−(0−(0^(n+1) /((n+1)∙2))∙ (((n+1)),(1) )∙(1/( (√(1−0^n )))))] =lim_(n→∞) [1−(1/(n+1))∙(1/( (√(1−1))))]= determinant (((lim_(n→∞) [1−(1/(n+1))∙0]=lim_(n→∞) [1−0]=1)))](https://www.tinkutara.com/question/Q213315.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{{n}} }{dx}=\mathrm{1} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[{x}−\frac{{x}^{{n}−\mathrm{1}} }{\left({n}+\mathrm{1}\right)\centerdot^{\mathrm{2}} }\centerdot\begin{pmatrix}{{n}+\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{{n}} }}\mid_{\mathrm{0}} ^{\mathrm{1}} \right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{1}^{{n}} }}−\left(\mathrm{0}−\frac{\mathrm{0}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)\centerdot\mathrm{2}}\centerdot\begin{pmatrix}{{n}+\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{0}^{{n}} }}\right)\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{1}}}\right]=\begin{array}{|c|}{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\centerdot\mathrm{0}\right]=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{1}−\mathrm{0}\right]=\mathrm{1}}\\\hline\end{array} \\ $$