Question

Am integrat prin parti si am ajuns la x²/2.√1+x² - ax-1. ∫x²/2 . 1/ √1+x²-ax nu stiu daca am ajuns la ce trebuia ca sa continui exercitiul putin ajutor va rog

Answer 1

$f:\mathbb{R}\to \mathbb{R}, ~~f(x) = \sqrt{1+x^2}-ax,\quad a\in \mathbb{R}. \\ \\ \displaystyle \int\limits_{0}^1 x\cdot f(x) \, dx = \int \limits_{0}^1x\cdot \Big(\sqrt{1+x^2}-ax\Big)\, dx = \\ \\ = \int\limits_{0}^1\Big(x\sqrt{1+x^2} - ax^2\Big) \, dx = \int\limits_{0}^1x\sqrt{1+x^2}\, dx -\int\limits_{0}^1ax^2\, dx = \\ \\ = \dfrac{1}{2}\int\limits_{0}^1(1+x^2)'\cdot (1+x^2)^{\dfrac{1}{2}}\, dx - \left a\dfrac{x^3}{3}\right|_{0}^1 =$

$= \dfrac{1}{2}\cdot \left\dfrac{(1+x^2)^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right|_{0}^1 - \dfrac{a}{3} = \left \dfrac{1}{3}(1+x^2)^{\dfrac{3}{2}}\right|_{0}^1-\dfrac{a}{3} = \\ \\ = \dfrac{1}{3}(1+x^2)\sqrt{1+x^2}\Big|_{0}^1-\dfrac{a}{3} = \dfrac{2\sqrt 2}{3}- \dfrac{1}{3} - \dfrac{a}{3} = \\ \\ = \dfrac{1}{3}\Big(2\sqrt 2 -1) - \dfrac{a}{3} \Rightarrow \boxed{c) \text{ r\u{a}spuns corect}}$