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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.

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  • $\begingroup$ Is the $[\ldots ]$ just a parenthesis, or does it have any other meaning? $\endgroup$ – John Alexiou Apr 11 '18 at 20:26
  • $\begingroup$ Any function of the form $f(x) = (x-1) g(x)$ can be made to fit the constraints. $\endgroup$ – John Alexiou Apr 11 '18 at 20:33
  • $\begingroup$ @ja72 [...] are just parentheses. I think it's not that straightforward, because we are dealing with real functions only. $\endgroup$ – Vincent J. Ruan Apr 11 '18 at 21:26
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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.

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You need to have more than just $f(x)$ being differentiable etc. For example it needs to be polynomial or in form of $e^{g(x)}$.

The problem is that you can create whatever function $f(x)=f(a,b,c,x)$ where $a, b, c$ are parameters and then your conditions would require solving $a,b,c$.

Take out of nowhere $f(x)=ax^5+bx^2+c$, it might create a needed system. Since there are more solutions, the only way that this would work is to prove that the final integral is for some reason unique.

To refute that it is sufficient to find two examples that give different integral and I am sure that you will find it as soon as you start looking into all possible forms the function can have.

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  • $\begingroup$ I've also thought like you. The thing is, when I try to solve for $(a,b,c)$, they turn out to be complex numbers. Then, obviously the final integral is not the same as of $\frac{7}{4}(1-x^4)$, but the problem still stands. $\endgroup$ – Vincent J. Ruan Apr 11 '18 at 20:02

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