Home » Uncategorized » Existence of constant (4)

Existence of constant (4)

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

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

Prove that there exists a c \in (0, 1) such that

    \[c^2 f(c) = \int_0^c x f(x) \, \mathrm{d}x\]

Solution

For starters let us consider the function G(x) = \int_0^x t f(t) \, \mathrm{d}t and F(x) = \int_0^x f(t)\, \mathrm{d}t. Trivially , it is F(0)=G(0)=0 and we note that:

    \[\int_{0}^{1} f(x) \, \mathrm{d}t = \int_{0}^{1} t f(t) \, \mathrm{d}t \Rightarrow F(1) = G(1)\]

Consider now the function

    \[H(x) = \left\{\begin{matrix} \dfrac{G(x)}{x} - F(x) & , & x \in (0, 1] \\\\ 0 & , & x =0 \end{matrix}\right.\]

Clearly H is continuous on [0, 1] and differentiable on (0, 1) with derivative

    \[H'(x) = \frac{x^2 f(x)-G(x)}{x^2} - f(x)\]

Furthermore , H(1)=G(1)-F(1)=0=H(0). Hence , by Rolle’s theorem there exists a x_0 \in (0, 1) such that H'(x_0)=0; that is G(x_0)=0. Finally, let us consider the function

    \[\Phi(x) = \left\{\begin{matrix} \dfrac{G(x)}{x} & , & x \in (0, x_0] \\\\ 0& , & x=0 \end{matrix}\right.\]

which satisfies Rolle’s conditions on [0, x_0]. Hence, there exists a c \in (0, x_0) \subseteq (0, 1) such that \Phi'(c)=0. However,

    \begin{align*} \Phi'(c) =0 &\Leftrightarrow \frac{c G'(c) - G(c)}{c^2} =0 \\ &\Leftrightarrow \frac{c \cdot c f(c) - G(c)}{c^2} =0 \\ &\Leftrightarrow c^2 f(c) = G(c) \\ &\Leftrightarrow c^2 f(c) = \int_0^c x f(x) \, \mathrm{d}x \end{align*}

Read more

Leave a comment

Donate to Tolaso Network