Υπολογισμός ορισμένου ολοκληρώματος

by mathxl, Mar 15, 2012, 9:12 PM

Ένα θέμα ΄προτεινόμενο από τον Kunny εδώ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=469349 2012 Osaka Prefecture University /Science, 2nd exam
Εκφώνηση
Να υπολογίσετε το ολοκλήρωμα $\int_{0}^{\frac{2n}{t}\pi}|x\sin\ tx|\ dx.$ όπου $n$ θετικός ακέραιος και $t$ θετικός πραγματικός αριθμός.

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Είναι ${I_t}\left( n \right) = \int_0^{\frac{{2n}}{t}\pi } | x\sin \;tx|\;dx\mathop = \limits_{dx = \frac{{du}}{t}}^{tx = u} \frac{1}{{{t^2}}}\int_0^{2n\pi } | u\sin \;u|\;du$
και ${I_t}\left( {n + 1} \right) = \frac{1}{{{t^2}}}\int_0^{2n\pi + 2\pi } | u\sin \;u|\;du$
Έτσι έχουμε
${I_t}\left( {n + 1} \right) - {I_t}\left( n \right) = \frac{1}{{{t^2}}}\left( {\int_0^{2n\pi + 2\pi } | u\sin \;u|\;du - \int_0^{2n\pi } | u\sin \;u|\;du} \right) = \frac{1}{{{t^2}}}\int_{2n\pi }^{2n\pi + 2\pi } | u\sin \;u|\;du =$
$\frac{1}{{{t^2}}}\int_{2n\pi }^{2n\pi + \pi } {u\sin u} \;du - \frac{1}{{{t^2}}}\int_{2n\pi + \pi }^{2n\pi + 2\pi } {u\sin u} \;du\mathop = \limits_{du = dk}^{u = \pi + k} \frac{1}{{{t^2}}}\int_{2n\pi }^{2n\pi + \pi } {u\sin u} \;du - \frac{1}{{{t^2}}}\int_{2n\pi }^{2n\pi + \pi } {\left( {\pi + k} \right)\sin \left( {\pi + k} \right)} \;dk =$
$\frac{1}{{{t^2}}}\int_{2n\pi }^{2n\pi + \pi } {u\sin u} \;du - \frac{1}{{{t^2}}}\int_{2n\pi }^{2n\pi + \pi } {\left( {\pi + k} \right)\sin \left( {\pi + k} \right)} \;dk = \frac{1}{{{t^2}}}\int_{2n\pi }^{2n\pi + \pi } {\left( {\pi + 2u} \right)\sin u} \;dk =$
$= \frac{\pi }{{{t^2}}}\left[ { - \cos u} \right]_{2n\pi }^{2n\pi + \pi } + \frac{2}{{{t^2}}}\left[ {\sin u - u\cos u} \right]_{2n\pi }^{2n\pi + \pi } = \frac{{2\pi }}{{{t^2}}} + \frac{{2\pi + 8n\pi }}{{{t^2}}} = \frac{{4\pi \left( {2n + 1} \right)}}{{{t^2}}}$
Επίσης είναι
${t^2}{I_t}\left( n \right) - {t^2}{I_t}\left( {n - 1} \right) = - 4\pi + 8n\pi = {a_1} + 8n\pi$
${t^2}{I_t}\left( {n - 1} \right) - {t^2}{I_t}\left( {n - 2} \right) = - 12\pi + 8n\pi = {a_2} + 8n\pi$
$........................................................................$
${t^2}{I_t}\left( 2 \right) - {t^2}{I_t}\left( 1 \right) = {a_{n - 1}} + 8n\pi = 12\pi - 8n\pi + 8n\pi$
${t^2}{I_t}\left( n \right) - {t^2}{I_t}\left( 1 \right) = {S_{n - 1}} + 8n\left( {n - 1} \right)\pi = - 4\pi {\left( {n - 1} \right)^2} + 8n\left( {n - 1} \right)\pi = 4\pi \left( {{n^2} - 1} \right) \Rightarrow$
${I_t}\left( n \right) = {I_t}\left( 1 \right) + \frac{{4\pi \left( {{n^2} - 1} \right)}}{{{t^2}}} = \frac{1}{{{t^2}}}\int_0^{2\pi } | u\sin \;u|\;du + \frac{{4\pi \left( {{n^2} - 1} \right)}}{{{t^2}}} = \frac{1}{{{t^2}}}\int_0^\pi {u\sin u} \;du - \frac{1}{{{t^2}}}\int_\pi ^{2\pi } {u\sin u} \;du + \frac{{4\pi \left( {{n^2} - 1} \right)}}{{{t^2}}} = \frac{{4\pi {n^2}}}{{{t^2}}}$

by mathxl, Mar 15, 2012, 9:14 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
?????? ???????? ???? ???????? ????????????, ???????? ?????????? ??????????????
by bestmath, Oct 11, 2011, 10:11 PM
?????? ???????????? ???? ?????????????????? ??????! ?????????? ?????? ????????????????! ?????????? ??????????????.
by knittingfrenzy18, Sep 8, 2011, 6:56 PM

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