Question

nu stiu sa rezolv in continuare, va rog sa ma ajutati ​

Answer 1

Răspuns:

Explicație pas cu pas:

$I_{n+1} - I_{n}=\int\limits^1_0 {\frac{x^{n+1} }{x^{2}+3x+2 }} \, dx -\int\limits^1_0 {\frac{x^{n} }{x^{2}+3x+2 }} \, dx=\int\limits^1_0 {\frac{x^{n}(x-1) }{x^{2}+3x+2 }} \, dx$

Cum $x\leq 1$ ⇒$I_{n+1}-I_{n}\leq 0$⇒$I_{n+1}\leq I_{n}$

Deci sirul In este descrescator

$I_{n+2}+3I_{n+1}+2I_{n}= \int\limits^1_0 {\frac{x^{n+2} }{x^{2}+3x+2 }} \, dx+ 3\int\limits^1_0 {\frac{x^{n+1} }{x^{2}+3x+2 }} \, dx+2\int\limits^1_0 {\frac{x^{n} }{x^{2}+3x+2 }} \, dx=\int\limits^1_0 {\frac{x^{n}(x^{2} +3x+2_ }{x^{2}+3x+2 }} \, dx=\int\limits^1_0 {x^{n} } \, dx=\frac{x^{n+1} }{n+1}|^{1} _{0} =\frac{1}{n+1}$

$I_{n+2}+3I_{n+1}+2I_{n}=\frac{1}{n+1}$, pentru orice n natural       (*)

$I_{n+2}\leq I_{n}\\ I_{n+1}\leq I_{n}\\ I_{n+2}+3I_{n+1}+2I_{n}\leq I_{n}+3I_{n}+2I_{n}$⇒

⇒$\frac{1}{n+1}\leq 6I_{n}$⇒$\frac{1}{6(n+1)}\leq I_{n}$      (1)

In relatia  (*) inlocuim n cu n-2

$I_{n}+3I_{n-1}+2I_{n-2}=\frac{1}{n-1}\\ I_{n-2}\geq I_{n}\\ I_{n-1}\geq I_{n}\\ I_{n-2}+3I_{n-1}+2I_{n}\geq I_{n}+3I_{n}+2I_{n}$⇒

⇒$\frac{1}{n-1}\geq 6I_{n}$⇒$\frac{1}{6(n-1)}\geq I_{n}$      (2)

din (1) si (2) ⇒ $\frac{1}{6(n+1)}\leq I_{n}\leq \frac{1}{6(n-1)}$⇒

⇒$\frac{n}{6(n+1)}\leq nI_{n}\leq \frac{n}{6(n-1)}$⇒

⇒$\lim_{n \to \infty} nI_n =\frac{1}{6}$

Answer 2

$\displaystyle n\int_0^1\dfrac{x^n}{x^2+3x+2}\, dx = n\int_{0}^1\dfrac{x^n}{(x+1)(x+2)}\, dx =\\ \\ = \int_{0}^1\Big(\dfrac{2}{x+2}-\dfrac{1}{x+1}\Big)\cdot nx^{n-1}\, dx = \\ \\ = \int_{0}^1\Big(\dfrac{2}{x+2}-\dfrac{1}{x+1}\Big)\cdot (x^{n})'\, dx= \\ \\ = \dfrac{1}{6}-\int_{0}^1 \Bigg(-\dfrac{2}{(x+2)^2}+\dfrac{1}{(x+1)^2}\Bigg)x^{n}\, dx =\\ \\ = \dfrac{1}{6}-\int_{0}^1 \Bigg(\dfrac{x^n}{(x+1)^2}-\dfrac{2x^n}{(x+2)^2}\Bigg)\, dx$

$\displaystyle \begin{align*} \displaystyle \Big|\int_0^1 \Bigg(\frac{x^n}{(x+1)^2} - \frac{2x^n}{(x+2)^2}\Bigg) \, dx\Big| &\leq \int_0^1 \Bigg|\frac{x^n}{(x+1)^2} - \frac{2x^n}{(x+2)^2}\Bigg| \, dx\\ &\leq \displaystyle \int_0^1 \Bigg(\frac{x^n}{(x+1)^2} + \frac{2x^n}{(x+2)^2}\Bigg) \, dx\\ &\leq \displaystyle \int_0^1 \Bigg(\frac{x^n}{(0+1)^2} + \frac{2x^n}{(0+2)^2}\Bigg) \, dx\\ &= \displaystyle \frac{3}{2(n+1)} \end{align*}\tag*{}$

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