prove that $∃ c ∈ (0, 1)$ such that $c^2f(c) = 2\int_0^cxf(x) \,dx$ for a continuous function $f$ Let $f\colon [0, 1] → \mathbb{R}$ be a continuous function. Given, $f(0) = 0$ and $\int_0^1 f(x)\, dx = 0$, prove that $∃ c ∈ (0, 1)$ such that $c^2f(c) = 2\int_0^cxf(x) \,dx.$
I need some help on how to use the fundamental theorem to prove this result.
My attempt: Let $F'(x)=f(x)$, the $\int_0^1 f(x)\, dx = 0$ implies that $F(1)=F(0)$. Now we take $H(x)=xf(x)$ and $\int_0^c xf(x)dx=\frac{c^2f(c)}{2}-\int_0^c\frac{x^2f'(x)}{2}dx$. This is where I'm stuck.
 A: For $x\geq 0$ fixed define $\varphi:[0,x]\longrightarrow \mathbb{R}$ by $\varphi(t)=t$. Since $f,\varphi$ are continous on $[0,x]$ and $\varphi$ is non$-$negative, then by this we can locate $\xi_x\in[0,x]$ such that $$\int_0^xf(t)\varphi(t)dt=f(\xi_x)\int_0^x\varphi(t)dt=\frac{f(\xi_x)x^2}{2}$$ Also, $f$ being continuous on $[0,1]$ with $\int_0^1f(t)dt=0$ implies there exists $a \in (0,1)$ such that $f(a)=0$. (This is easily proved by a straight$-$forward application of Rolle's Theorem to the function $x \longrightarrow \int_0^xf(t)dt$) Define $G:[0,a]\longrightarrow \mathbb{R}$ by $$G(x)=\int_0^xtf(t)dt-\frac{x^2f(x)}{2}$$ From our initial commentary we can express $G$ as $$G(x)=\frac{x^2}{2}\bigg(f(\xi_x)-f(x)\bigg)$$ Now set $m=\min\Big\{f(x):x\in [0,a]\Big\}$ and $M=\max\Big\{f(x):x\in [0,a]\Big\}$. Since $f$ is continuous on the compact set $[0,a]$ we can locate $x_1,x_2\in [0,a]$ such that $f(x_1)=m$ and $f(x_2)=M$. Now $G(x_1)\geq 0$ while $G(x_2)\leq 0$ so by $\text{IVT}$ there exists $c$ between $x_1$ and $x_2$ (inclusive) such that $G(c)=0$. We now need only prove that $c$ can be chosen to belong to the open interval $(0,1)$.
If $mM<0$, then $f(0)=0=f(a)$ implies $x_1,x_2\in (0,a)$; this means $c$ must also belong to $(0,a)\subset (0,1)$ and we're done. On the other hand, if $m=M=0$, then $f\equiv 0$ on $[0,a]$, and taking $c$ to be any real number in $(0,a)$ will work. Now if $m=0$ and $M>0$ we have $x_2 \in (0,a)$ and $G(a)\geq 0$ so by IVT find $c\in [x_2,a] \subset (0,1)$ with $G(c)=0$. A similar argument applies to the case when $m<0$ and $M=0$ completing the proof.
A: First (as here) we prove there exists $\xi \in (0,1)$ such that $\int_0^c tf(t)\,dt = 0$. Define $\phi : [0,1] \to \Bbb{R}$ as $$\phi(x) := \frac1x \int_0^x \int_0^t f(u)\,du\,dt, \quad x \in (0,1]$$
and $\phi(0)=0$. Then $\phi$ is continuous on $[0,1]$ and differentiable on $(0,1)$ with derivative
$$\phi'(x) = \frac{x\int_0^x f(t)\,dt - \int_0^x\int_0^t f(u)\,du\,dt}{x^2} = \frac1{x^2}\int_0^x tf(t)\,dt, \quad x \in (0,1)$$
by integration by parts. Mean value theorem gives a $\theta \in (0,1)$ such that
$$\phi(1) - \phi(0) = \phi'(\theta)$$
so Fubini gives $$\phi'(\theta) = \phi(1) = \int_0^1 \int_0^t f(u)\,du\,dt = \int_0^1 \int_u^1 f(u)\,dt\,du = \int_0^1(1-u)f(u)\,du = -\int_0^1 uf(u)\,du$$ On the other hand $\phi'(1) = \int_0^1 tf(t)\,dt = -\phi(1)$. The derivative $\phi'$ extended by $\phi'(0)=0$ (right derivative) is continuous on $[0,1]$ so from
$$\begin{cases} \phi'(\theta) = \phi(1), \\ \phi'(1)=-\phi(1)\end{cases}$$the intermediate value theorem gives that there exists $\xi \in [\theta,1]$ such that $\phi'(\xi) = 0$.
$\phi'$ is furthermore differentiable on $(0,1)$ with derivative
$$\phi''(x) = \frac{x^3f(x) - 2x\int_0^x tf(t)\,dt}{x^4} = \frac{x^2f(x) - 2\int_0^x tf(t)\,dt}{x^3}, \quad x\in (0,1).$$
Hence the mean value theorem implies there exists $c \in (0,\xi)$ such that
$$\phi'(\xi)-\phi'(0) = \phi''(c)(\xi - 0)$$
or $\phi''(c)=0$. This means
$$c^2f(c) = 2\int_0^c tf(t)\,dt.$$
