If $\displaystyle f: \mathbb{R}\mapsto\mathbb{R}$ is continuous, and $\displaystyle \int^1_0\left(f(x)\right)^2 dx = 3$.

Find the maximum value of $\displaystyle \int^1_0 x f(x) dx$

Please explain to me... (Bow)

You are so nice^^ (Rofl)

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- Oct 13th 2009, 06:32 AMGTK X Hunterfind the max value of integral
If $\displaystyle f: \mathbb{R}\mapsto\mathbb{R}$ is continuous, and $\displaystyle \int^1_0\left(f(x)\right)^2 dx = 3$.

Find the maximum value of $\displaystyle \int^1_0 x f(x) dx$

Please explain to me... (Bow)

You are so nice^^ (Rofl) - Oct 13th 2009, 07:36 AMNonCommAlg
by Holder's inequality: $\displaystyle \int_0^1 xf(x) \ dx \leq \int_0^1 x|f(x| \ dx \leq \sqrt{\int_0^1 x^2 dx \cdot \int_0^1 (f(x))^2 dx} = 1.$ also if $\displaystyle f(x) = 3x,$ then $\displaystyle \int_0^1 (f(x))^2 dx = 3$ and $\displaystyle \int_0^1 xf(x) dx = 1.$ thus $\displaystyle \max_f \int_0^1 xf(x) \ dx = 1.$

- Oct 13th 2009, 10:08 AMOpalg
- Oct 13th 2009, 05:54 PMGTK X Hunter
- Oct 14th 2009, 05:19 PMKrizalid
first assumption was $\displaystyle \int^1_0\left(f(x)\right)^2 dx = 3,$ and $\displaystyle \int_0^1x^2\,dx=\frac13.$