Question

10 daca se poate multumesc mult! banuiesc ca nu putem aplica l hopital pt ca e sir. am incercat sa integrez insa dupa ce fac limita imi da 0. Rasp. corect e pi/2 A​

Answer 1

$\displaystyle a_n = \dfrac{1}{n^2}\int_{-n}^n(x\arctan x)\, dx \quad n\geq 1,\quad n\in \mathbb{N}\\ \\ \\ I = \int_{-n}^n(x\arctan x)\, dx = 2\int_{0}^n(x\arctan x)\, dx\quad \text{(functia este para)}\\ \\ I =2\int_{0}^n\Big(\dfrac{x^2}{2}\Big)'\arctan x \, dx =2\cdot \dfrac{x^2\arctan x}{2}\Big|_{0}^n - \int_{0}^n\dfrac{x^2+1-1}{1+x^2}\, dx \\ \\ I = n^2\arctan n - x\Big|_{0}^n+\int_{0}^n\dfrac{1}{x^2+1}\, dx$

$I =n^2\arctan n-n+\arctan n\\ \\ l = \lim\limits_{n\to \infty}\dfrac{n^2\arctan n-n+\arctan n}{n^2} = \\ \\ =\lim\limits_{n\to \infty}\dfrac{n^2\Big(\arctan n-\dfrac{1}{n}+\dfrac{\arctan n}{n^2}\Big)}{n^2} = \\ \\ = \lim\limits_{n\to \infty }\Big(\arctan n-\dfrac{1}{n}+\dfrac{\arctan n}{n^2}\Big) = \\ \\ = \dfrac{\pi}{2}-\dfrac{1}{\infty}+\dfrac{\dfrac{\pi}{2}}{\infty} = \dfrac{\pi}{2}-0+0 = \boxed{\dfrac{\pi}{2}}$