Question

putin ajutor la 2c va rog, stiu ca se face cu substitutie insa nu mi da ca in barem ​

Answer 1

$\displaystyle \int_{0}^1\Big(1-\dfrac{x}{n}\Big)^n\, dx = -n\int_{0}^1\Big(1-\dfrac{x}{n}\Big)'\cdot \Big(1-\dfrac{x}{n}\Big)^n \, dx = \\ \\ = -n\cdot \dfrac{\Big(1-\dfrac{x}{n}\Big)^{n+1}}{n+1}\Bigg|_{0}^1 \\ \\ \Rightarrow \lim\limits_{n\to \infty}\Bigg( -\dfrac{n}{n+1}\cdot \Big(1-\dfrac{x}{n}\Big)^{n+1}\Bigg|_{0}^1\Bigg) = \\ \\ = -1\cdot \lim\limits_{n\to \infty}\Bigg[\Big(1-\dfrac{1}{n}\Big)^{n+1}-(1-0)^{n+1}\Bigg] =$

$= -\lim\limits_{n\to \infty}\Big(1-\dfrac{1}{n}\Big)^{n+1}+1=-\lim\limits_{n\to \infty}\Big(1-\dfrac{1}{n}\Big)^{n(n+1)\cdot \frac{-1}{n}} +1 = \\ \\ = -e^{\lim\limits_{n\to \infty}{(-\frac{n+1}{n}})}+1 = -e^{-1}+1 = 1-\dfrac{1}{e} = \boxed{\dfrac{e-1}{e}}$