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Existence of constant (3)

Let f:[0, 1] \rightarrow \mathbb{R} be a differentiable function with continuous derivative such that f(0)=f'(0)=0. If

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

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

    \[\int_0^c x f(x) \; \mathrm{d}x = \frac{2c}{3} \int_0^c f(x) \; \mathrm{d}x\]

Solution

Consider the function

    \[F(x)= \left\{\begin{matrix} \displaystyle \frac{1}{x^3} \int_{0}^{x} t f(t) \, \mathrm{d}t - \frac{1}{x^2} \int_{0}^{x} f(t) \, \mathrm{d}t & , & x \in (0, 1] \\ 0 & , & x =0 \end{matrix}\right.\]

which is clearly continuous on [0, 1], since:

    \begin{align*} \lim_{x\rightarrow 0^+} F(x)&= \lim_{x\rightarrow 0^+} \left ( \frac{1}{x^3} \int_{0}^{x} t f(t) \, \mathrm{d}t - \frac{1}{x^2} \int_{0}^{x} f(t) \, \mathrm{d}t \right ) \\ &=\lim_{x\rightarrow 0^+} \left ( \frac{xf(x)}{3x^2} - \frac{f(x)}{2x}\right ) \\ &= \lim_{x\rightarrow 0^+} \left ( \frac{f(x)}{3x} - \frac{f(x)}{2x} \right )\\ &= \lim_{x\rightarrow 0^+} \left ( \frac{f'(x)}{3} - \frac{f'(x)}{2} \right )\\ &=0\\ &=F(0) \end{align*}

Furthermore, F is differentiable in (0, 1). The derivative is given by

    \begin{align*} F'(x) &= \frac{x^4 f(x)-3x^2 \int_{0}^{x} tf(t) \; \mathrm{d}t}{x^6} - \frac{x^2 f(x)-2x\int_{0}^{x}f(t) \, \mathrm{d}t}{x^4} \\ &=\frac{x^4 f(x)-3x^2 \int_{0}^{x} tf(t) \; \mathrm{d}t}{x^6} -\frac{x^2 \left ( x^2 f(x)-2x\int_{0}^{x}f(t) \, \mathrm{d}t \right )}{x^6} \\ &= \frac{x^4 f(x)-3x^2 \int_0^x t f(t) \, \mathrm{d}t-x^4f(x)+2x^3 \int_0^x f(t) \, \mathrm{d}t}{x^6} \\ &=\frac{2x^3 \int_0^x f(t) \, \mathrm{d}t - 3x^2 \int_0^x t f(t) \, \mathrm{d}t}{x^6} \end{align*}

Since F(1)=F(0) it follows from Rolle’s theorem that there exists a c \in (0, 1) such that F'(c)=0. The result now follows.

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