find the max value of integral

**GTK X Hunter** · October 13th 2009, 06:32 AM

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

Please explain to me...

You are so nice^^

**NonCommAlg** · October 13th 2009, 07:36 AM

Originally Posted by GTK X Hunter

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

Please explain to me...

You are so nice^^

by Holder's inequality: $\int_0^1 xf(x) \ dx \leq \int_0^1 x|f(x| \ dx \leq \sqrt{\int_0^1 x^2 dx \cdot \int_0^1 (f(x))^2 dx} = 1.$ also if $f(x) = 3x,$ then $\int_0^1 (f(x))^2 dx = 3$ and $\int_0^1 xf(x) dx = 1.$ thus $\max_f \int_0^1 xf(x) \ dx = 1.$

**Opalg** · October 13th 2009, 10:08 AM

Originally Posted by GTK X Hunter

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

Please explain to me... explain to me..

Hint: Did you ever come across the Cauchy–Schwarz inequality?

**GTK X Hunter** · October 13th 2009, 05:54 PM

Originally Posted by NonCommAlg

by Holder's inequality: $\int_0^1 xf(x) \ dx \leq \int_0^1 x|f(x| \ dx \leq \sqrt{\int_0^1 x^2 dx \cdot \int_0^1 (f(x))^2 dx} = 1.$ also if $f(x) = 3x,$ then $\int_0^1 (f(x))^2 dx = 3$ and $\int_0^1 xf(x) dx = 1.$ thus $\max_f \int_0^1 xf(x) \ dx = 1.$

How can you conclude that $\sqrt{\int_0^1 x^2 dx \cdot \int_0^1 (f(x))^2 dx} = 1.$