# Thread: find the max value of integral

1. ## find 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 $\displaystyle \int^1_0 x f(x) dx$

You are so nice^^

2. Originally Posted by GTK X Hunter
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 $\displaystyle \int^1_0 x f(x) dx$

You are so nice^^
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 $\displaystyle \max_f \int_0^1 xf(x) \ dx = 1.$

3. Originally Posted by GTK X Hunter
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... explain to me..
Hint: Did you ever come across the Cauchy–Schwarz inequality?

4. Originally Posted by NonCommAlg
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 $\displaystyle \max_f \int_0^1 xf(x) \ dx = 1.$
How can you conclude that $\displaystyle \sqrt{\int_0^1 x^2 dx \cdot \int_0^1 (f(x))^2 dx} = 1.$

5. first assumption was $\displaystyle \int^1_0\left(f(x)\right)^2 dx = 3,$ and $\displaystyle \int_0^1x^2\,dx=\frac13.$