Let $f:[0, 1] \rightarrow \mathbb{R}$ be a continuous function such that $f(0)=0$ and

\begin{equation} \int_0^1 x f(x) \, {\rm d}x = \int_0^1 f(x) \, {\rm d}x \end{equation}

Prove that there exists a $c \in (0, 1)$ such that $\displaystyle \int_0^c x f(x) \, {\rm d}x = \frac{c}{2} \int_0^c f(x) \, {\rm d}x$.

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## Existence of $c$

- Tolaso J Kos
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### Existence of $c$

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### Re: Existence of $c$

Let $I(x) = \bigintsss_{0}^{x} f(t) \, {\rm d}t$ and $G(x) = \bigintsss_{0}^{x} I(t) \, {\rm d}t$. Integrating by parts $(1)$ reveals that

$$ \int_{0}^{1} I(t) \, {\rm d}t =0 \implies G(1) =0$$

Now let us consider the function $\displaystyle K(x) = \frac{G(x)}{x^2}$ . It holds that $K(1)=0$. As we can also see using two consecutive DeL’ Hospital’s Rules , it also holds that $\lim \limits_{x \rightarrow 0} K(x)=0$. So, by Rolle’s theorem there exists a $c \in (0, 1)$ such that

$$G\left ( c \right ) = \frac{c}{2} I(c)$$

However integration by parts reveals that

$$ G(c) = c I(c) - \int_{0}^{c} x f(x) \, {\rm d}x$$

and thus $\displaystyle \int_{0}^{c} x f(x) \, {\rm d}x = \frac{c}{2} \int_{0}^{c} f(x) \, {\rm d}x$ which is the desired output.

$$ \int_{0}^{1} I(t) \, {\rm d}t =0 \implies G(1) =0$$

Now let us consider the function $\displaystyle K(x) = \frac{G(x)}{x^2}$ . It holds that $K(1)=0$. As we can also see using two consecutive DeL’ Hospital’s Rules , it also holds that $\lim \limits_{x \rightarrow 0} K(x)=0$. So, by Rolle’s theorem there exists a $c \in (0, 1)$ such that

$$G\left ( c \right ) = \frac{c}{2} I(c)$$

However integration by parts reveals that

$$ G(c) = c I(c) - \int_{0}^{c} x f(x) \, {\rm d}x$$

and thus $\displaystyle \int_{0}^{c} x f(x) \, {\rm d}x = \frac{c}{2} \int_{0}^{c} f(x) \, {\rm d}x$ which is the desired output.

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$