An help in the following problem:
Let $f:[-1,1] \longrightarrow \mathbb{R}$ a $C^1$ function, i.e., continuously differentiable. Suppose that we have $$\int_{-1}^{1} f(x)\;dx = \pi \quad\text{and}\quad \int_{-1}^{1} f'(x)\, x^n\;dx = 0,$$ for all integer $n\geq 2014!$. Determine $f(0)$.
I've tried integration by parts directly, but nothing. Thanks for help me!
As Daniel said, $f(x)=\pi/2$ is the solution.
Let $g(x)=f'(x)x^{2014!}$. Then, for any $n=0,1,2,\ldots$ we have $$ \int_{-1}^1g(x)x^ndx=\int_{-1}^1f'(x)x^{n+2014!}\,dx=0. $$ So $g(x)=0$ for all $x\in[-1,1]$ (it is orthogonal to $x^n$ for all $n$), and thus $f'(x)=0$. So $f$ is constant, and now the first equality gives $f(x)=\pi/2$, $x\in[-1,1]$.