Suma Riemann de integrale improprii
Posted: Sun Jan 06, 2008 2:06 pm
Consideram \( f: (0,1) \to\mathbb{R} \) o functie monotona astfel incat integrala improprie \( \int_0^1f(x)dx \) sa existe. Sa se arate ca
\( \lim_{n\to\infty}\frac{1}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\ldots+f\left(\frac{n-1}{n}\right)\right)=\int_0^1f(x)dx \).
\( \lim_{n\to\infty}\frac{1}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\ldots+f\left(\frac{n-1}{n}\right)\right)=\int_0^1f(x)dx \).