I've encountered this problem:
Let $f(x)$ be a continuously differentiable (real) function on $[0,1]$ satisfying these equations: $$f(1)=0$$ $$\int_0^1 [f'(x)]^2 dx = 7$$ $$\int_0^1 x^2f(x) dx = \frac{1}{3}$$. Compute $\int_0^1f(x) dx$.
I've managed to find a $f(x) = \frac{7}{4}(1-x^4)$ in a few trials. However, I cannot find any other solution (or at least any other elementary solution), which seems weird to me because these equations are not enough to uniquely define a function. Moreover, assume that there are some other solutions, how can the problem be so sure that $\int_0^1f(x) dx$ are all the same among those solutions? Is there any neat way to solve the problem without finding a solution?
I highly doubt these two questions. I think the problem is wrong. But I'm not sure, so I post it here to discuss.
Thanks in advance.
Integrating by parts gives $1=\displaystyle\int_0^1 3x^2f(x)\,\mathrm dx=\left[x^3f(x)\right]_0^1-\int_0^1 x^3f'(x)\,\mathrm dx$, so that $\displaystyle\int_0^1 x^3f'(x)\,\mathrm dx=-1$.
Therefore $\displaystyle\int_0^1(f'(x)+7x^3)^2\mathrm dx=\int_0^1[f'(x)]^2\mathrm dx+14\int_0^1 x^3f'(x)\,\mathrm dx+\int_0^1 49x^6\mathrm dx=7-14+7=0$.
Thus $f'(x)+7x^3=0$ for all $x\in[0,1]$, so $f(x)=\frac74(1-x^4)$ is the unique continuously differentiable solution.