Ακρότατα και ολοκλήρωμα

by mathxl, Dec 29, 2011, 10:23 PM

Έστω η συνάρτηση $f(x)=\displaystyle\int_{0}^{\pi}\sin(\vert x-t\vert+\frac{\pi}{4}) dt$ , $0\le x\le 2\pi$ . Να μελετηθεί ως προς τα ακρότατα.

Η άσκηση είναι από την διεύθυνση http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=454681

This post has been edited 4 times. Last edited by mathxl, Dec 29, 2011, 10:46 PM

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Για $x \in \left[ {0,\pi } \right]$ έχουμε ότι
$f(x) = \int_0^\pi {\sin } (|x - t| + \frac{\pi }{4})dt = \int_0^x {\sin } (|x - t| + \frac{\pi }{4})dt + \int_x^\pi {\sin } (|x - t| + \frac{\pi }{4})dt =$
$= \int_0^x {\sin } (x - t + \frac{\pi }{4})dt + \int_x^\pi {\sin } (t - x + \frac{\pi }{4})dt = \left[ {\cos \left( {x - t + \frac{\pi }{4}} \right)} \right]_0^x - \left[ {\cos \left( {t - x + \frac{\pi }{4}} \right)} \right]_x^\pi =$
$= \sqrt 2 - \cos \left( {x + \frac{\pi }{4}} \right) + \cos \left( {\frac{\pi }{4} - x} \right) = \sqrt 2 + \sqrt 2 \sin x$
Για ${x \in \left( {\pi ,2\pi } \right]}$ έχουμε ότι
$f(x) = \int_0^\pi {\sin } (|x - t| + \frac{\pi }{4})dt = \int_0^\pi {\sin } (x - t + \frac{\pi }{4})dt =$
$= \left[ {\cos \left( {x - t + \frac{\pi }{4}} \right)} \right]_0^\pi = - 2\cos \left( {x + \frac{\pi }{4}} \right)$

Συνεπώς
$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\sqrt 2 + \sqrt 2 \sin x,x \in \left[ {0,\pi } \right]}\\ { - 2\cos \left( {x + \frac{\pi }{4}} \right),x \in \left( {\pi ,2\pi } \right]} \end{array}} \right.$
Για τα ακρότατα έχουμε $- 2 = f\left( {\frac{{7\pi }}{4}} \right) \le f\left( x \right) \le f\left( {\frac{\pi }{2}} \right) = 2\sqrt 2$

by mathxl, Dec 29, 2011, 10:44 PM

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Γεια σας! Βρηκα αυτη την ασκηση απο https://mathematica.gr/forum/viewtopic.php?f=55&t=21122 και με δυσκολευει η ευρεση του τυπου της f. Ειναι ευκολο να δωσετε την απαντηση;
by lina1979, Jun 14, 2020, 3:51 PM
wha?????????????????????????
by dantx5, Nov 10, 2011, 2:09 AM
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by bestmath, Oct 11, 2011, 10:11 PM
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by knittingfrenzy18, Sep 8, 2011, 6:56 PM

4 shouts