# $f(1)=\frac{1}{2}\int_{0}^{\frac{1}{2}}e^{1-x^2}f(x)dx$, prove that there exists $\xi\in(0,\,1)$ such that $f'(\xi)=2\xi f(\xi)$.

$$f(x)$$ is continuous on $$[0,\,1]$$, differentiable on $$(0,\,1)$$, and satisfies $$f(1)=\frac{1}{2}\int_{0}^{\frac{1}{2}}e^{1-x^2}f(x)dx.$$ Prove that there exists $$\xi\in(0,\,1)$$ such that $$f'(\xi)=2\xi f(\xi)$$.

Let $$F(x)=e^{1-x^2}f(x)$$, then it's equivalent to prove that $$F(\xi)'=0$$. Also, we have $$F(1)=\frac{1}{2}\int_{0}^{\frac{1}{2}}F(x)\,dx$$. But I don't know how to continue.

I think there is a mistype. The correct equation is $$f(1)=2\int_0^{1/2} e^{1-x^{2}}f(x)\, dx$$. I will answer the question with this correction. I would like to thank Martin R for suggesting this correction.
Suppose $$F'$$ does not vanish at any point. Then $$F'(x) >0$$ for all $$x$$ or $$F'(x) <0$$ for all $$x$$. This is because any derivative has IVP. In the first case $$F$$ is strictly increasing so $$2 \int_0^{\frac 1 2} F(x)\, dx < 2 \int_0^{\frac 1 2} F(1)\, dx =F(1)$$, a contradiction. Similarly in the second case we get $$2 \int_0^{\frac 1 2} F(x)\, dx > 2 \int_0^{\frac 1 2} F(1)\, dx =F(1)$$.
The statement is wrong, a counterexample is $$F(x) = 1-\frac 67 x$$. $$F(1)=\frac{1}{2}\int_{0}^{\frac{1}{2}}F(x)\,dx = \frac 17 \, ,$$ but $$F'$$ is nowhere zero.
If you actually meant the equation $$f(1)= 2\int_{0}^{\frac{1}{2}}e^{1-x^2}f(x)dx$$ then the conclusion follows from the mean-value theorem for integrals, as demonstrated in Rolle's Theorem related exercise.