Limita integrala

Felixx · Mesaj de **Felixx** » 15 Iun 2019, 00:08

Din $0\leq x\leq 1\Rightarrow 0\leq x^{2}\leq 1\Rightarrow \frac{1}{2}\leq \frac{1}{x^{2}+1}\leq 1\Rightarrow \frac{1}{2}x^{n}\leq \frac{x^{n}}{x^{2}+1}\leq x^{n}\Rightarrow \frac{1}{2}\int_{0}^{1}x^{n}dx\leq \int_{0}^{1}\frac{x^{n}}{x^{2}+1}dx\leq \int_{0}^{1}x^{n}dx$
unde:
$\int_{0}^{1}x^{n}dx=\frac{1}{n+1}\underset{n\rightarrow \infty }{\rightarrow}0$
si
$\frac{1}{2}\int_{0}^{1}x^{n}dx=\frac{1}{2}\frac{1}{n+1}\underset{n\rightarrow \infty }{\rightarrow}0$
Conform criteriul clestelui limita este 0.

Kierkegaard · Mesaj de **Kierkegaard** » 15 Iun 2019, 00:36

Multumesc frumos dl @Felixx, apreciez enorm.

Forum AniDeȘcoală.ro

Limita integrala

Limita integrala

Re: Limita integrala

Re: Limita integrala