Find the minimum value of $ \int_0^1 (1+x^2)f^2(x)dx$

Question

I was trying to solve a question of an entrance exam. I have taken help for a similar problem from MSE but again I am stuck in the problem. Please help me.

Let $ D= \{f \in [0,1] : f \text{continuous and} \displaystyle \int_0^1f(x)dx = 1\}$. Find the value of $$\text{min}_{f\in D} \displaystyle \int_0^1 (1+x^2)f^2(x)dx$$

I have tried the same eay as described in this problem but I failed to apply Euler-Lagrange equation.I can not find any other way to proceed. Please help me. Thnx in advance.

Just a question does, $f^2(x)$ means second derivative $f''(x)$? — Someone, May 07 '15 at 14:06
Uh, thanks I guess this problem beyond high school anyway :), well tried anyway. Found hints from other answer however , seems interesting. — Someone, May 07 '15 at 14:15

score 3 · Accepted Answer · answered May 07 '15 at 14:03

3

Here's a way:

Write $F(x) = \int_0^x f(t)dt$.

$D$ now corresponds to those $F$ for which $F \in \mathcal C^1$, $F(0)=0$ and $F(1)=1$. And you're looking for $\min_F \int_0^1(1+x^2)(F')^2(x)dx$.

Your Lagrangian can be $\mathcal L = (1+x^2)(y')^2$. Can you take it from there?

answered May 07 '15 at 14:03

Alexandre Halm

3,792

Thank you very much sir, I will try now. I think now I can do it. – usermath May 07 '15 at 14:04
FYI, I get $f(x) = \frac4{\pi}\arctan(x)$ if that helps – Alexandre Halm May 07 '15 at 14:17
I think it's $F(x)=\frac4\pi \arctan(x)$, so $f(x)=\frac{4}{\pi(1+x^2)}$? – shwsq Apr 08 '25 at 13:39

Find the minimum value of $ \int_0^1 (1+x^2)f^2(x)dx$

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