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Inegalitate integrala
Posted: Sun Sep 20, 2009 10:10 am
by Marius Mainea
Fie \( a,b\in\mathbb{R}, a<b \) si \( f,g:[a,b]\rightarrow\mathbb{R} \) functii continue. Aratati ca:
\( \(\int_a^bf(x)dx\)^2+\(\int_a^bg(x)dx\)^2\le \(\int_a^b\sqrt{f^2(x)+g^2(x)}dx\)^2 \)
C. Buse, ,,Traian Lalescu'' 2009
Posted: Sun Sep 20, 2009 1:53 pm
by opincariumihai
Fie \( z=\int_a^bf+i\int_a^bg=r( \cos t+i \sin t) \)
Atunci
\( r=\sqrt{\(\int_a^bf(x)dx\)^2+\(\int_a^bg(x)dx\)^2}=(\int_a^bf+i\int_a^bg)(\cos t-i\sin t)=\int_a^b(f\cos t+g\sin t)\le \(\int_a^b\sqrt{f^2(x)+g^2(x)}dx\) \)
Posted: Sun Sep 20, 2009 3:49 pm
by Marius Mainea
Folosind CBS
\( \(\int_a^b\sqrt{f^2(x)+g^2(x)}ds\)^2-\(\int_a^bf(x)dx\)^2=\(\int_a^b\sqrt{f^2(x)+g^2(x)}-f(x)dx\)\(\int_a^b\sqrt{f^2(x)+g^2(x)}+f(x)dx\)\ge\(\int_a^b\sqrt{\(\sqrt{f^2(x)+g^2(x)}-f(x)dx\)\(\sqrt{f^2(x)+g^2(x)}+f(x)dx\)}dx\)^2=\(\int_a^b|g(x)|dx\)^2\ge\(\int_a^bg(x)dx\)^2 \)