Question

Subpunctul c va rog! Realizat corespunzator pentru bac

Answer 1

$\displaystyle I_n = \int_{0}^1\dfrac{x^{n+1}}{x+3}\, dx \\ nI_n = n\int_{0}^1\dfrac{x^{n+1}}{x+3}\, dx = \int_{0}^1\dfrac{nx^{n-1}x^2}{x+3}\, dx =\\ \\=\int_{0}^1(x^{n})'\cdot \dfrac{x^2}{x+3}\, dx =\dfrac{1}{4}-\int_{0}^1x^n\cdot \Big(\dfrac{x^2}{x+3}\Big)'\ dx =\\ \\= \dfrac{1}{4}-\int_{0}^1x^n\cdot \dfrac{x^2+6x}{(x+3)^2}\, dx$

$0\leq x\leq 1\Big|+3 \Rightarrow 3\leq x+3\leq 4\Big|^2 \Rightarrow 9\leq (x+3)^2\leq 16 \Big|^{-1}\Rightarrow \\ \\ \Rightarrow \dfrac{1}{16}\leq \dfrac{1}{(x+3)^2}\leq \dfrac{1}{9}\Big|\cdot(x^2+6x) \Rightarrow\\ \\ \Rightarrow \dfrac{x^2+6x}{16}\leq \dfrac{x^2+6x}{(x+3)^2}\leq \dfrac{x^2+6x}{9}\Big|\cdot x^n \Rightarrow\\ \\ \Rightarrow \dfrac{x^{n+2}+6x^{n+1}}{16}\leq x^n\cdot \dfrac{x^2+6x}{(x+3)^2}\leq \dfrac{x^{n+2}+6x^{n+1}}{9}$

$\displaystyle \lim\limits_{n\to \infty}\int_{0}^1(x^{n+2}+6x^{n+1})\, dx =\lim\limits_{n\to \infty}\Big(\dfrac{1}{n+3}+\dfrac{1}{n+2}\Big) = 0\\ \\ \Rightarrow 0\leq \lim\limits_{n\to \infty }\int_0^1x^n\cdot \dfrac{x^2+6x}{(x+3)^2}\, dx\leq 0 \Rightarrow \lim\limits_{n\to \infty}\int_0^1x^n\cdot \dfrac{x^2+6x}{(x+3)^2}\, dx = 0 \\ \\\\ \Rightarrow \lim\limits_{n\to \infty} nI_n = \dfrac{1}{4} - 0 = \boxed{\dfrac{1}{4}}$

Answer 2

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