Functii

Moderators: Bogdan Posa, Beniamin Bogosel, Marius Dragoi

Post Reply
User avatar
Beniamin Bogosel
Co-admin
Posts: 710
Joined: Fri Mar 07, 2008 12:01 am
Location: Timisoara sau Sofronea (Arad)
Contact:

Functii

Post by Beniamin Bogosel »

Fie \( f:[0,1]\to \mathbb{R} \) cu proprietatea ca \( \int_0^1f(x)dx=\frac{1}{2} \) si \( \int_0^1f^2(x)dx=\frac{1}{3} \).
Demonstrati ca \( \int_0^1xf(x)=\frac{1}{3} \).
Marius Mainea
Gauss
Posts: 1077
Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Functia \( f(x)=1-x \) nu verifica concluzia.
User avatar
Beniamin Bogosel
Co-admin
Posts: 710
Joined: Fri Mar 07, 2008 12:01 am
Location: Timisoara sau Sofronea (Arad)
Contact:

Post by Beniamin Bogosel »

Era dintr-o carte mai veche si n-am reusit s-o rezolv. M-am gandit ca era gresita, dar n-am gasit contraexemplu... :)
User avatar
Beniamin Bogosel
Co-admin
Posts: 710
Joined: Fri Mar 07, 2008 12:01 am
Location: Timisoara sau Sofronea (Arad)
Contact:

Post by Beniamin Bogosel »

Daca in plus \( f \) este crescatoare, atunci este rezolvabila problema?
User avatar
Vlad Matei
Pitagora
Posts: 58
Joined: Wed Sep 26, 2007 6:44 pm
Location: Bucuresti

Post by Vlad Matei »

Ma rog, problema nu poate fi dreasa. Tot ce se poate preciza este \( \displaystyle \int_{0}^{1}xf(x)dx\leq\frac{1}{3} \).
Un exemplu de functie strict crescatoare este \( \displaystyle \frac{\sqrt{15}}{4}x^{2}+\frac{1}{2}-\frac{\sqrt{15}}{12} \). :D
Post Reply

Return to “Analiza matematica”