$Q1)\space prove\ that\\ \int_{0}^{\infty }\frac{1-e^{-\sqrt{2}x}}{xe^x}dx=sinh^{-1}1$
$Q1)\space prove\ that\\ \int_{0}^{\infty }\frac{1-e^{-\sqrt{2}x}}{xe^x}dx=sinh^{-1}1$
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ÌÈÇÑ ÇáÍÓíäí ; 06-11-2016 ÇáÓÇÚÉ 01:02 AM
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
ÇáÓáÇã Úáíßã
ÇÚÇäí ãÔßá ÈÑÝÚ ÇáãáÝÇÊ
æÍÇæáÊ ÇáßÊÇÈÉ ÈØÑíÞÉ latex æáßä ÎÑÌ ÇáÍá ÛíÑ ãÑÊÈ
ÇÖØÑÑÊ Çáì ÇÊÈÇÚ åÐå ÇáØÑíÞÉ Ýí ÊÍãíá ÇáãÔÇÑßÇÊ ÈÓÈÈ ÇáãÔßáÉ ÇáÊí áÏí
ÝÛÇáÈÇ ãÇíÙåÑ áí ÚäÏãÇ ÇÍÇæá ÑÝÚ ãáÝ ( ÝÔá ÇáÊÍãíá ÇáÕÝÍÉ áÇÊÚãá ) [IMG] Image 67.jpg (90.9 ßíáæÈÇíÊ)[/IMG]
$Q2 \int_{0}^{\infty }\frac{e^{-3x^2}-e^{-4x^2}}{x}dx$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
$\int_{0}^{\infty }\frac{e^{-3x^{2}}-e^{-4x^{2}}}{x}dx$
$=\int_{0}^{\infty }\frac{1}{x}(e^{-ax^{2}})dx$
$=\int_{0}^{\infty }(\int_{4}^{3}\frac{x^{2}e^{-ax^{2}}}{x}da)dx$
$=\int_{4}^{3}(\int_{0}^{\infty }-xe^{-ax^{2}}dx)da$
$=\int_{4}^{3}[\frac{e^{-ax^{2}}}{2a}]_{0}^{\infty }da$
$=\int_{4}^{3}\frac{-1}{2a}da=\frac{-1}{2}lna=\frac{-1}{2}ln\frac{3}{4}$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
$Q3\int_{0}^{\infty }\frac{\arctan(2x)-\arctan (3x) }{x}dx$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
ááÊÐßíÑ ¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿ ¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
$Q4)\int_{0.2}^{3.5}\left \lfloor x \right \rfloor dx$
$Q5)\int_{0.2}^{3.5}\left \lceil x \right \rceil dx$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
ááÊÐßíÑ ¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿ ¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿¿
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
ÇáÐíä íÔÇåÏæä ÇáãæÖæÚ ÇáÂä: 2 (0 ãä ÇáÃÚÖÇÁ æ 2 ÒÇÆÑ)
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