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Buna! Bacul bate la usa iar eu am o problema serioasa cu integralele recurente... Ma puteti ajuta va rog? As avea doua exercitii la care nu le gasesc nicicum solutia...
1. Calculati limita:
[tex] \lim_{n \to \infty} n \int\limits^1_0{ x^{n} e^{x} } \, dx [/tex]
2. Se considera functiile
[tex]f(x)= \frac{1}{1-x} [/tex]
[tex]g_{n}(x)= \frac{ x^{n} }{1-x} [/tex]
a) Aratati ca
[tex]0 \leq \int\limits^1_0{g_n(x)} \, dx \leq \frac{1}{ 2^{n} } [/tex]
b) Aratati ca:
[tex] \lim_{n \to \infty} ( \frac{1}{1*2}+ \frac{1}{2* 2^{2} } +...+ \frac{1}{n* 2^{n} } )=ln2[/tex]

Răspuns :

Matepentrutoti
[tex]I_n= \int\limits^1_0 {x^ne^x} \, dx =e-nI_{n-1}\\ I_n+nI_{n-1}=e\\ I_{n+1}-I_{n}= \int\limits^1_0 {x^n(x-1)e^x} \, dx \leq 0=\ \textgreater \ I_n\ descrescator\\ I_n \leq I_{n-1}\\ nI_{n-1} \leq n I_{n-1}\\ Adunam \ inegalitatile\\ e\leq (n-1)I_{n-1}=\ \textgreater \ I_{n-1} \geq \frac{e}{n+1} =\ \textgreater \ I_n \geq \frac{e}{n+2} \\ Analog\\ nI_n \leq nI_{n-1}\\ I_n \leq I_n\\ (n+1)I_n \leq e=\ \textgreater \ I_n \leq \frac{e}{n+1} \\ Concluzie\\ \frac{e}{n+2} \leq I_n \leq \frac{e}{n+1} |\cdot n\\ \frac{en}{n+2} \leq nI_n \leq \frac{en}{n+1}=\ \textgreater \ nI_n-\ \textgreater \ e [/tex]

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