ÈÇÓÊÚãÇá ÇáãæÞÚ
https://www.codecogs.com/latex/eqneditor.php
ÊãÊ ßÊÇÈÉ ÇáÓÄÇá æßãÇ íáí: æÇáãæÞÚ ãæÌæÏ Ýí ÇáãäÊÏì ãäÐ 4 ÇÚæÇã
\[Find \int {\color{Red} e^x sin x}\: dx\]
æÊßÈíÑ ÍÌã ÇáÍÑæÝ Ýí ÇáãäÊÏì
áÇÍÙ Ðáß
\[Find \int {\color{Red} e^x sin x}\: dx\]
ÈÇÓÊÚãÇá ÇáãæÞÚ
https://www.codecogs.com/latex/eqneditor.php
ÊãÊ ßÊÇÈÉ ÇáÓÄÇá æßãÇ íáí: æÇáãæÞÚ ãæÌæÏ Ýí ÇáãäÊÏì ãäÐ 4 ÇÚæÇã
\[Find \int {\color{Red} e^x sin x}\: dx\]
æÊßÈíÑ ÍÌã ÇáÍÑæÝ Ýí ÇáãäÊÏì
áÇÍÙ Ðáß
\[Find \int {\color{Red} e^x sin x}\: dx\]
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ßÇãá ãæÓì ÇáäÇÕÑí ; 05-28-2016 ÇáÓÇÚÉ 11:16 AM
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå
$\int {\color{Red} e^x sin x}\: dx$
$ Let\\ e^x =dv , \mapsto \space v= e^x ,u=sinx \mapsto du= cosx$
$\int {\color{Red} e^x sin x}\: dx= sinx \space e^x - \int {\color{Red} e^x cosx}\: dx$
$ Let\\\ e^x =dv , \mapsto \space v= e^x ,u=cosx \mapsto du= -sinx$
$\int {\color{Red} e^x sin x}\: dx= e^x sinx –(e^xcosx+ \int {\color{Red} e^x sinx \: dx}) $
$\int {\color{Red} e^x sin x}\: dx=\frac{1}{2}( e^x sinx –e^xcosx)$
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ÇÍãÏ ÇáÌÇÈÑí ; 05-28-2016 ÇáÓÇÚÉ 10:31 PM
ÞÇá ÇÍÏ ÇáÍßãÇÁ
(( ÇáÑíÇÖíÇÊ ØÑíÞí áãÚÑÝÉ Çááå æÃãÇä ãä ÇáÔÑß Èå))
ÔßÑÇ ÇÓÊÇÐ ÍÇÌã
ÇÑÌæ ØÑÍ ØÑíÞÉ ÇÎÑì áÍá ÇáÓÄÇá
ÔßÑÇ ÇÓÊÇÐ ÍÇÌã
ÇÑÌæ ØÑÍ ØÑíÞÉ ÇÎÑì áÍá ÇáÓÄÇá
$\int {\color{Red} e}^{x}\sin xdx$
$\int e^{x}.\frac{e^{ix}-e^{-ix}}{2i}dx=\frac{1}{2i}\int [e^{(1+i)x}-e^{(1-i)x}]dx$
$=\frac{1}{2i}[\frac{e^{(1+i)x}}{1+i}-\frac{e^{(1-i)x}}{1-i}]+C$
$=\frac{e^{x}}{2i}[\frac{e^{ix}-ie^{ix}-e^{-ix}-ie^{-ix}}{2}]+C$
$=\frac{e^{x}}{2}[\frac{e^{ix}-e^{-ix}}{2i}-\frac{e^{ix}+e^{-ix}}{2}]+C$
$= \frac{e^{x}}{2}(\sin x-\cos x)+C$
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ÇÍãÏ ÇáÌÇÈÑí ; 05-29-2016 ÇáÓÇÚÉ 05:33 PM
[URL="http://alnasiry.net/forums"][IMG]http://alnasiry.net/forums/uploaded/2_iraqiflag.gif[/IMG][/URL]
$\int e^x sinx dx$
$=\int( \frac{1}{2}e^x cosx-\frac{1}{2}e^x cosx+\frac{1}{2}e^x sinx+\frac{1}{2}e^x sinx)dx$
$=\int( \frac{1}{2}e^x (cosx+sinx)+\frac{1}{2}e^x (sinx-cosx))dx$
$\int d(\frac{1}{2}e^x (sinx-cosx))=\frac{1}{2}e^x (sinx-cosx)+c$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
\[\LARGE {\color{Magenta} Evaluate}:\int_{0}^{2} x \sqrt{4-2x} dx\]
$\int_{0}^{2} x\sqrt{4-2x}dx $
$=1/2\int_{0}^{2}\sqrt{4-2x}((4-(4-2x))dx$
$=\int_{0}^{2}(2\sqrt{4-2x}-1/2(4-2x)^{3/2})dx$
$=-2/3(4-2x)^{3/2}+1/10(4-2x)^{5/2}\left. \right \}_{0}^{2}$
$=32/15$
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ßÇãá ãæÓì ÇáäÇÕÑí ; 05-31-2016 ÇáÓÇÚÉ 12:57 AM
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
\[\LARGE Q3.\int_{1}^{2} x \sqrt{3x-2} \: dx\]
$\int_{1}^{2}x\sqrt{3x-2}dx $
$=1/3\int_{1}^{2}\sqrt{3x-2}(2+3x-2)dx$
$=2/3\int_{1}^{2}\sqrt{3x-2}dx+1/3\int_{1}^{2}(3x-2)^{3/2}dx$
$=4/27(3x-2)^{3/2}+2/45(3x-2)^{5/2}]_{1}^{2}$
$=326/135$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
\[\LARGE Q4.\int_{1}^{5} \frac{x}{\sqrt{2x-1}}\: dx\]
ÇáÐíä íÔÇåÏæä ÇáãæÖæÚ ÇáÂä: 1 (0 ãä ÇáÃÚÖÇÁ æ 1 ÒÇÆÑ)
Powered by vBulletin® Version 4.2.3
Copyright © 2026 vBulletin Solutions, Inc. All rights reserved.
Translate By
Almuhajir
Developed By Marco Mamdouh
Style
ÊØæíÑ æÏÚã ÔÑßÉ
Zavord