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Δημοσιεύτηκε: **Τετ Φεβ 11, 2026 12:50 am**

Έστω $f:\left [ 0,c \right ]\rightarrow \mathbb{R}$ συνεχής συνάρτηση, τέτοια ώστε $\displaystyle \int \limits_{0}^{c}\left ( t-c \right )f\left ( t \right )dt=0.$ Να δείξετε ότι υπάρχουν $\xi _{x},\xi _{3},\xi _{4},\xi ''_{x}\in \left ( 0,c \right ),$ τέτοια ώστε $\displaystyle \left ( \xi _{x}-c \right )\int \limits_{0}^{\xi _{3}}qf\left ( q \right )dq=\xi _{x}\int \limits_{0}^{\xi_{χ} ''}yf\left ( y \right )dy+\frac{\xi _{4}^{2}}{\left ( \int \limits_{0}^{\xi _{4}}yf\left ( y \right )dy \right )}\int \limits_{\xi_{χ} ''}^{\xi _{3}}\left ( \frac{1}{x^{2}}\int \limits_{0}^{x}yf\left ( y \right )dy \right )dx\left ( \xi _{3_{x}} +\xi _{x}''\right )\int\limits_{0}^{\xi _{x}}f\left ( x \right )dx$ και $\xi _{5}\in \left ( 0,2 \right ),$ τέτοιο ώστε $\displaystyle c^2\int \limits_{0}^{\xi _{5}}x^{2}f\left ( x \right )dx=2\xi _{5}^{2}\int \limits_{0}^{c}\left ( y\int \limits_{0}^{y}f\left ( t \right )dt \right )dy.$

Δημοσιεύτηκε: **Σάβ Φεβ 14, 2026 9:25 pm**

Η λύση μου είναι η εξής (έχει ολίγον γράψιμο )
$\displaystyle \int \left ( \int \limits_{0}^{x}tf\left ( t \right )dt \right )dx=x\int \limits_{0}^{x}tf\left ( t \right )dt-\int \limits _{0}^{x}t^{2}f\left ( t \right )dt=\int \limits_{0}^{x}\left ( x-t \right )tf\left ( t \right )dt+c'',\forall x\in \left [ 0,c \right ].$ Θεωρώντας τη συνάρτηση $\displaystyle g\left ( x \right )=\int \limits _{0}^{x}\left ( x-t \right )tf\left ( t \right )dt=\frac{dg\left ( x \right )}{dx}\cdot x-\int \limits_{0}^{x}y^{2}f\left ( y \right )dy,\forall x\in \left [ 0,c \right ]\Rightarrow \frac{d\left ( \frac{1}{x}\int \limits_{0}^{x}\left ( x-t \right )tf\left ( t \right )dt \right )}{dx}$ $\displaystyle =\frac{\int \limits_{0}^{x}y^{2}f\left ( y \right )dy}{x^{2}},\forall x\in \left ( 0,c \right ].$ Έστω $\displaystyle L\left ( x \right )=\left\{\begin{matrix}\frac{1}{x}\int \limits_{0}^{x}\left ( x-t \right )tf\left ( t \right )dt+c'',x\in \left ( 0,c \right ] & \\ & \\c'',x=0 \end{matrix}\right.\Rightarrow \displaystyle \lim_{x \to 0^{+}}L\left ( x \right )=\displaystyle \lim_{x \to 0^{+}}\left(\frac{g\left ( x \right )-g\left ( 0 \right )}{x-0}+c''\right)=$ $=g'\left ( 0 \right )+c''=c''=L\left ( 0 \right ),$ άρα η

είναι συνεχής στο

και αφού $\displaystyle \frac{dL(x)}{dx}=\frac{\int \limits_{0}^{x}y^2f(y)dy}{x^{2}},\forall x\in \left ( 0,c \right ]\overset{\Theta .M.T}\Rightarrow \boxed{\exists \xi _{5_{x}}\in \left ( 0,c \right ):\frac{\int \limits_{0}^{\xi _{5_{x}}}y^{2}f\left ( y \right )dy}{\xi _{5_{x}}^{2}}=\frac{L(c)-L(0)}{c-0}=\frac{1}{c^{2}}\int \limits_{0}^{c}\left ( c-t \right )tf\left ( t \right )dt}\left ( 1 \right ).$ Αφού $\displaystyle \int \limits _{0}^{c}\left ( c-y \right )\left ( \int \limits_{0}^{y} f\left ( t \right )dt\right )'dy=0\Leftrightarrow \left [ \left ( c-y \right )\int \limits_{0}^{y}f\left ( t \right )dt \right ]^{y=c}_{y=0}+\int \limits_{0}^{c}\left ( \int \limits_{0}^{y}f\left ( t \right )dt \right )dy=0$ $\displaystyle \Leftrightarrow \int \limits_{0}^{c}\left ( \int \limits_{0}^{y} f\left ( t \right )dt\right )dy=0\left ( 2 \right ),$ άρα $\displaystyle \int \limits_{0}^{c}\left ( c-y \right )yf\left ( y \right )dy=\int \limits _{0}^{c}\left ( c-y \right )yd\left ( \int \limits _{0}^{y}f\left ( z \right )dz \right )=\int \limits_{0}^{c}\left ( 2y-c \right )\left ( \int \limits_{0}^{y}f\left ( p \right ) dp\right )dy\overset{(2)}=2\int \limits_{0}^{c}y\left ( \int \limits_{0}^{y}f\left ( p \right ) dp\right )dy\overset{(1)}\Rightarrow$ $\displaystyle \boxed{2\xi _{5_{x}}^{2}\int \limits_{0}^{c}y\left ( \int \limits_{0}^{y}f\left ( p \right )dp \right )dy=c^{2}\int \limits_{0}^{\xi _{5_{x}}}q^{2}f\left ( q \right )dq}\left ( 3 \right ).$ Θεωρούμε τη συνάρτηση $\displaystyle q\left ( x \right )=\int \limits_{0}^{x}\left ( t-x \right )f\left ( t \right )dt,\forall x\in \left [ 0,c \right ]\Rightarrow \frac{d\left ( \int \limits_{0}^{x} \left ( t-x \right )f\left ( t \right )dt\right )}{dx}=$ $\displaystyle =\left ( \int\limits_{0}^{x} tf\left ( t \right )dt\right )'-x\left ( \int\limits_{0}^{x}f\left ( t \right ) dt\right )'-\int \limits_{0}^{x}f\left ( y \right )dy=-\int\limits_{0}^{x}f\left ( y \right )dy\Rightarrow$
$\displaystyle q\left ( x \right )=\int\limits_{0}^{x}yf\left ( y \right )dy+\frac{dq\left ( x \right )}{dx}\cdot x\Rightarrow \frac{d\left ( \frac{1}{x}\int\limits_{0}^{x}\left ( t-x \right )f\left ( t \right )dt \right )}{dx}=-\frac{\int\limits_{0}^{x}yf\left ( y \right )dy}{x^{2}},\forall x\in \left ( 0,c \right ].$ Θεωρούμε τη συνάρτηση $\displaystyle G\left ( x \right )=\left\{\begin{matrix} \frac{1}{x}\int\limits_{0}^{x}\left ( t-x \right )f\left ( t \right )dt,x\in \left ( 0,c \right ] & \\ & \\0,x=0 \end{matrix}\right..$ Είναι $\displaystyle \displaystyle \lim_{x \to 0^{+}}\left ( \frac{1}{x}\int\limits_{0}^{x}\left ( t-x \right )f\left ( t \right )dt \right )=\displaystyle \lim_{x \to 0^{+}}\frac{q\left ( x \right )-q\left ( 0 \right )}{x-0}=q'\left ( 0 \right )=0=G\left ( 0 \right ),$ άρα η

είναι συνεχής στο

και ισχύει $\displaystyle \frac{dG\left ( x \right )}{dx}=-\frac{\int\limits_{0}^{x}yf\left ( y \right )dy}{x^{2}},\forall x\in \left ( 0,c \right ],$ και αφού G(c)=0=G(0),

επομένως από Θ.Rolle $\displaystyle \exists \xi _{x}\in \left ( 0,c \right ):G'\left ( \xi _{x} \right )=0\Leftrightarrow -\frac{\int\limits_{0}^{\xi _{x}}yf\left ( y \right )dy}{\xi _{x^{2}}}=0\Leftrightarrow \boxed{\int\limits_{0}^{\xi _{x}}qf\left ( q \right )dq=0}\left ( 4 \right ).$ Από Θ.Μ.Τ για την

στο $\displaystyle \left [ 0,\xi _{x} \right ],\exists \xi _{x}''\in \left ( 0,\xi _{x} \right ):G'\left ( \xi _{x}'' \right )=\frac{G\left ( \xi _{x} \right )-G\left ( 0 \right )}{\xi _{x}-0}=\frac{\int\limits_{0}^{\xi _{x}}\left ( t-\xi _{x} \right )f\left ( t \right )dt}{\xi _{x^{2}}}\overset{(4)}=-\frac{\int\limits_{0}^{\xi _{x}}f\left ( u \right )du}{\xi _{x}}\Leftrightarrow$ $\displaystyle -\frac{\int\limits_{0}^{\xi _{x}''}yf\left ( y \right )dy}{\xi _{x}''^{2}}=-\frac{\int\limits_{0}^{\xi _{x}}f\left ( u \right )du}{\xi _{x}}\Leftrightarrow \boxed{\xi _{x}''^{2}\int\limits_{0}^{\xi _{x}}f\left ( u \right )du=\xi _{x}\int\limits_{0}^{\xi _{x}''}yf\left ( y \right )dy}\left ( 5 \right )$ και από Θ.Μ.Τ για την

στο $\displaystyle \left [ \xi _{x} ,c\right ],\exists \xi _{3_{x}}\in \left ( \xi _{x},c \right ):G'\left ( \xi _{3_{x}} \right )=\frac{G\left ( \xi _{x} \right )-G\left ( c \right )}{\xi _{x}-c}\overset{G(c)=0}=\frac{\frac{1}{\xi _{x}}\int\limits_{0}^{\xi _{x}}\left ( t-\xi _{x} \right )f\left ( t \right )dt}{\xi _{x}-c}\overset{(4)}=-\frac{\int\limits_{0}^{\xi _{x}}f\left ( z \right )dz}{\xi _{x}-c}\Leftrightarrow$
$\displaystyle -\frac{\int\limits_{0}^{\xi _{3_{x}}}yf\left ( y \right )dy}{\xi _{3_{x}}^{2}}=-\frac{\int\limits_{0}^{\xi _{x}}f\left ( z \right )dz}{\xi _{x}-c}\Leftrightarrow \boxed{\xi _{3_{x}}^{2}\int\limits_{0}^{\xi _{x}}f\left ( z \right )dz=\left ( \xi _{x}-c \right )\int\limits_{0}^{\xi _{3_{x}}}yf\left ( y \right )dy}\left ( 6 \right ).$ Είναι $\displaystyle \boxed{\left ( \xi _{3_{x}}^{2}-\xi _{x} ''^{2}\right )\int\limits_{0}^{\xi _{x}}f\left ( p \right )dp\overset{^{\left ( 5 \right )}_{\left ( 6 \right )}}=\left ( \xi _{x}-c \right )\int\limits_{0}^{\xi _{3_{x}}}yf\left ( y \right )dy-\xi _{x}\int\limits_{0}^{\xi _{x}''}yf\left ( y \right )dy}\left ( 7 \right ).$ Από ΘΜΤΟΛ για την $\displaystyle -\frac{\int\limits _{0}^{x}yf\left ( y \right )dy}{x^{2}}$ στο $\displaystyle \left [ \xi _{x}'',\xi _{3_{x}} \right ],\exists \xi _{4_{x}}\in \left ( \xi _{x}'',\xi _{3_{x}} \right ):-\frac{\int\limits _{0}^{\xi _{4_{x}}}yf\left ( y \right )dy}{\xi _{4_{x^{}}}^{2}}=\frac{\int\limits_{\xi _{x}''}^{\xi _{3_{x}}}\left ( -\frac{1}{x^{2}}\int\limits_{0}^{x}wf\left ( w \right )dw \right )dx}{\xi _{3_{x}}-\xi _{x}''}\Leftrightarrow \xi _{3_{x}}-\xi _{x}''=\frac{\xi _{4_{x}}^{2}\int\limits_{\xi _{x}''}^{\xi _{3_{x}}}\left ( \frac{1}{x^{2}}\int\limits_{0}^{x}wf\left ( w \right )dw \right )dx}{\int\limits_{0}^{\xi _{4_{x}}}yf\left ( y \right )dy}\overset{(7)}\Rightarrow$
$\displaystyle \overset{(7)}\Rightarrow \boxed{\left ( \xi _{3_{x}} +\xi _{x}''\right )\frac{\xi _{4_{x}}^2\int\limits_{\xi _{x}''}^{\xi _{3_{x}}}\left ( \frac{1}{x^{2}} \int\limits_{0}^{x}wf\left ( w \right )dw\right )dx}{\left ( \int\limits_{0}^{\xi _{4_{x}}}yf\left ( y \right )dy \right )}\int\limits_{0}^{\xi _{x}}f\left ( p \right )dp+\xi _{x}\int\limits_{0}^{\xi _{x}''}yf\left ( y \right )dy=\left ( \xi _{x}-c \right )\int\limits_{0}^{\xi _{3_{x}}}yf\left ( y \right )dy}\left ( 8 \right ).$

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