TITLE: Proving $\int^c_0 xf(x)\,dx=(c-1)\int^c_0 f(x)\,dx$
QUESTION [5 upvotes]: Let $f\colon [0,1]\to\mathbb{R}$ be a continuous function such that $$\int^1_0 xf(x)\,dx = \int^1_0 f(x)\,dx.$$
Prove that exists a $c\in(0,1)$ such that
$$\int^c_0 xf(x)\,dx=(c-1)\int^c_0 f(x)\,dx.$$

I tried to apply the Mean Value Theorem but I did no get to any point.

REPLY [7 votes]: Let $$F : x \mapsto \int_0^x f(t)dt$$
By hypothesis, one has
$$F(1)=\int_0^1 f(x) \,dx= \int_0^1 xf(x) \,dx= \int_0^1 \int_0^x f(x) \,dt \,dx = \int_0^1 \int_t^1 f(x) \,dx \,dt$$ $$= \int_0^1 F(1)-F(t) \,dt = F(1)-\int_0^1F(t)\,dt$$
(you can also make this calculation with an integration by parts on $\int_0^1 xf(x) \,dx$).
So you deduce that
$$\int_0^1 F(t) \,dt = 0$$
Now let $$G : x \mapsto \int_0^x F(t) \,dt, \quad \quad \text{and }\quad h : x \mapsto e^{-x}G(x)$$
Because $G(0)=G(1)=0$, then $h(0)=h(1)=0$, so by Rolle's theorem, there exists $c \in (0,1)$ such that $h'(c)=0$. And one has
$$h'(x)=e^{-x}\left(G'(x)-G(x) \right)$$
so you get $c \in (0,1)$ such that $G'(c)=G(c)$, i.e. such that
$$F(c)=\int_0^c F(t) \,dt$$
You deduce that
$$\int_0^c xf(x) \,dx = \int_0^c \int_0^x f(x) \,dt \,dx = \int_0^c \int_t^c f(x) \,dx \,dt = \int_0^c F(c)-F(t) \,dt =cF(c)-F(c)$$
i.e. that
$$\int_0^c xf(x) \,dx = (c-1)\int_0^c f(x)\,dx$$