Question

Rezolvati,va rog,.................Multumesc!

Answer 1

Răspuns:

Explicație pas cu pas:

$\displaystyle I_{n+1}-I_n=\int_0^1\dfrac{x^{n+1}}{x^3+1}dx-\int_0^1\dfrac{x^n}{x^3+1}dx=\int_0^1\dfrac{x^{n+1}-x^n}{x^3+1}dx=\\=\int_0^1\dfrac{x^n(x-1)}{x^3+1}dx\leq 0,\texttt{ deoarece }x-1\leq 0~\forall~x\in{[0,1]}\\\texttt{Mai departe avem ca:}\\0\leq x^3\leq 1 |+1\\1\leq x^3+1\leq 2 |()^{-1}\\1\geq \dfrac{1}{x^3+1}\geq\dfrac{1}{2}|\cdot x^n\\x^n\geq\dfrac{x^n}{1+x^3}\geq\dfrac{x^n}{2}$

$\displaystyle\int_0^1x^ndx\geq\int_0^1\dfrac{x^n}{1+x^3}dx\geq\int_0^1\dfrac{x^n}{2}dx\\\dfrac{x^{n+1}}{n+1}|_0^1\geq I_n\geq \dfrac{x^{n+1}}{2(n+1)}|_0^1\\\dfrac{1}{n+1}\geq I_n\geq\dfrac{1}{2(n+1)}\\\text{Cum }\lim_{n\to\infty}\dfrac{1}{n+1}=\lim_{n\to\infty}\dfrac{1}{2(n+1)}=0\texttt{ rezulta din criteriul }\\\texttt{clestelui ca }\lim_{n\to\infty}I_n=0$