How to approach the following problem Let $f:[0,1]\to\mathbb{R}$ be a continuous function then the maximum value of $\int_0^1 f(x)x^2 \,\text{d}x -\int_0^1x(f(x))^2\,\text{d}x$ for all such function(s) is?
I first thought differentiating it but the limits are not variable so no use then I spent a lot of time thinking over it but literally have no idea how to begin.
Please help.
 A: Hint:
$$f(x)x^2 -xf(x)^2 = xf(x) ( x-f(x)) $$
Think of $f(x-f)$ as a quadratic in $f$. Complete the square to get $$\frac{x^2}{4} - \left(f-\frac x2\right)^2 $$
So, your integral is $$\int_0^1 \frac{x^3}{4} - x\left(f(x) -\frac x2\right)^2 dx $$
What function $f$ maximizes this?
A: Seems like functioncal calculus to me:
We want to optmize:
$$ g[f] =  \int_0^1  \left[ f(x)x^2 - x f^2 (x) \right]dx$$
Add a peturbation to $g$:
$$g[f+ \epsilon] = \int_0^1 \left[ (f + \epsilon)x^2 - x (f+\epsilon)^2 \right] dx$$
It is easy to find that terms of order $\epsilon$ are:
$$ \epsilon \int_0^1 \left[x^2 - 2 xf \right] dx$$
So, for the minimum condition, this term in bracket must be zero, which makes our function as:
$$  f(x) = \frac{x}{2}$$
Now you can integrate easily
All of this is based on this article that I read

Meta commentary:
This is completely analogous to how if we are a maxima point in single variable calculus , then first order variation is zero. I.e:
$$ f(x+h) = f(x) + h f'(x) + \text{junk}$$
Now, if you are at maxima then $f'(x)=0$ and hence if you increase $h$ by some teeny amount, the function doesn't change much. Similar notion here though more abstract.
Another notion is that, similar to how we neglect term of $dx^2$, we neglect terms of $\epsilon^2$ saying they are tinnier than tiny petrubations.
