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

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 $f''(x)$? – Mann May 7 '15 at 14:06
• No it is the square of $f$ – usermath May 7 '15 at 14:13
• Uh, thanks I guess this problem beyond high school anyway :), well tried anyway. Found hints from other answer however , seems interesting. – Mann May 7 '15 at 14:15

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?
• FYI, I get $f(x) = \frac4{\pi}\arctan(x)$ if that helps – Alexandre Halm May 7 '15 at 14:17