Existence of $c$
- Tolaso J Kos
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Existence of $c$
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 $\displaystyle \int_0^c x f(x) \, {\rm d}x = \frac{c}{2} \int_0^c f(x) \, {\rm d}x$.
\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 $\displaystyle \int_0^c x f(x) \, {\rm d}x = \frac{c}{2} \int_0^c f(x) \, {\rm d}x$.
Imagination is much more important than knowledge.
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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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