第三章 一元函数积分学(不定积分)?一.?求下列不定积分:? 1. ò - + -? dx?x?x?x? 1?1?ln?1?1?2? 解. = - + - ò? dx?x?x?x? 1?1?ln?1?1?2? c?x?x?x?x?d?x?x + ÷ ? ? ? è ? - + = - + - + ò? 2?1?1?ln?4?1?1?1?ln?1?1?ln?2?1? 2. ò + + × + + +? dx?x?x?x?x?x? cos?1?sin?1?)?cos?1?(? 1?sin?cos? 2? 解.? c?x?x?x?x?d?x?x?dx?x?x?x?x?x + ÷ ? ? ? è ? + + = + + + + = + + × + + + ò ò? 2?2? cos?1?sin?1?2?1?cos?1?sin?1?cos?1?sin?1?cos?1?sin?1?)?cos?1?(? 1?sin?cos? 3. ò +?)?1?(?8?x?x?dx? 解.?方法一:?令? t?x?1 =?,? c?t?t?dt?t?dt?t?t?t?x?x?dx + + - = + - = ÷ ? ? ? è ? + - = + ò òò? )?1?ln(?8?1?1?1?1?1?1?)?1?(? 8?8?7?8?2?8? =? c?x + ÷ ? ? ? è ? + -? 8?1?1?ln?8?1? 方法二: ò ò ò + - - = + = +? dx?x?x?x?x?x?dx?x?x?x?dx? )?1?1?1?(?)?1?(?)?1?(? 8?8?7?8?8?7?8? =? c?x?x?x?x?d?x?dx + + - = + + - ò ò? )?1?ln(?8?1?|?|?ln?1?)?1?(?8?1? 8?8?8? =? c?x + ÷ ? ? ? è ? + -? 8?1?1?ln?8?1? 二.?求下列不定积分:?1. ò + + +? 2?2?)?1?(? 2?2? x?x?x? dx? 解. ò ò + + + + = + + +? 1?)?1(?)?1(? )?1(?2?2?)?1(? 2?2?2?2? x?x? x?d?x?x?x? dx? t?x? tan?1= + 令 ò? t?t?t?dt?sec?tan?cos?2?2? = ò + + + + - = + - =? c?x?x?x?c?t?t?tdt? 1?2?2?sin?1?sin?cos? 2?2? 2. ò +?2?4?1?x?x?dx?解.?令?x?=?tan?t, ò ò ò ò ò + + - = - = = = +? c?t?t?t?t?d?t?t?d?dt?t?t?t?t?t?dt?x?x?dx? sin?1?sin?3?1?sin?sin?sin?sin?sin?cos?sec?tan?cos?1? 3?2?4?4?3?4?2?2?4 =? c?x?x?x?x + + + ÷ ÷ ? ? ? ? è ? + -? 2?3?2? 1?1?3?1? 3. ò + +? 2?2? 1?)?1?2?(? x?x?dx?解.?令? t?x?tan = ò ò ò ò + = + = + = + +? t?t?d?dt?t?t?t?dt?t?t?t?x?x?dx? 2?2?2?2? 2?2?2? sin?1?sin?cos?sin?2? cos?sec?)?1tan?2?(? sec?1?)?12?(? =? c?x?x?c?t + + = +? 2?1?arctan?sin?arctan? 4. ò -?2?2?2? x?a?dx?x? (a?>?0)?解.?令? t?a?x? sin = ò ò ò + - = - = × = -? c?t?a?t?a?dt?t?a?t?a? tdt?a?t?a?x?a?dx?x? 2?sin?4?1?2?1?2?2?cos?1?cos?cos?sin? 2?2?2?2?2?2?2?2? =? c?x?a?a?x?a?x?a + ÷ ? ? ? è ? - -? 2?2?2?2?arcsin?2?5. ò -? dx?x? 3?2?)?1?(?解.?令? t?x?sin = ò ò ò ò + + = + = = -? dt?t?t?dt?t?tdt?dx?x? 4? 2?cos?2?cos?2?1?4?)?2?cos?1?(?cos?)?1?(? 2?2?4?3?2? = ò + + + = + + +? c?t?t?t?dt?t?t?t? 4sin?32?1?2sin?4?1?8?3?)?4cos?1?(?8?1?2sin?4?1?4?1?=? c?t?t?x + + +? )?2cos?4?1?1?(?2sin?4?1?arcsin?8?3? =? c?t?t?t?x + - + +? )?4?sin?2?1?4(?cos?sin?2?4?1?arcsin?8?3? 2?=? c?x?x?x?x + - - +? )?2?5?(?1? 8?1?arcsin?8?3? 2?2?6. ò -?dx?x?x? 4?2? 1?解.?令? t?x?1 = ò ò ò - - = ÷ ? ? ? è ? - - = -? dt?t?t?dt?t?t?t?t?dx?x?x? 2?2?4?2?2?4?2? 1?1?1?1?1? u?t?sin = 令 ò -? udu?u? 2?cos?sin? =? c?x?x?c?u + - = +? 3? 3?2?3? 3?)?1?(?cos?3?1? 三.?求下列不定积分: 1. ò + - +? dx?e?e?e?e?x?x? x?x? 1?2?4?3? 解. ò ò ò + - = + - - = + - + = + - + - - - - -? c?e?e?e?e? e?e?d?dx?e?e?e?e?dx?e?e?e?e? x?x?x?x? x?x?x?x? x?x?x?x? x?x? )?arctan(?1?)?(? )?(?1?1? 2?2?2?2?4?3? 2. ò +?)?4?1?(?2? x?x?dx? 解.?令? x?t?2 =? ,? 2?ln?t?dt?dx =? c?t? t?dt?t?t?t?t?dt?dx?x?x + - - = ÷ ? ? ? è ? + - = + = + ò ò ò? 2?ln?arctan?2?ln?1?1?1?1?2?ln?1?2?ln?)?1?(?)?4?1?(?2? 2?2?2?2?=? c?x?x + + - -? )?2?arctan?2?(2?ln?1? 四.?求下列不定积分:?1. ò -? dx?x?x? 10?5?)?2?(? 解. ò ò ò - - - + - - = - - = -? dx?x?x?x?x?x?d?x?dx?x?x? 9?4?9?5?9?5?10?5? )?2?(?99?5?)?2?(?99?)?2?(?99?1?)?2?(? = ò - - × × + - ′ - - -? dx?x?x?x?x?x?x? 98?3?98?4?9?5? )?2?(?98?99?4?5?)?2?(?98?99?5?)?2?(?99? =? 96?2?97?3?98?4?9?5? )?2?(?96?97?98?99?3?4?5?)?2?(?97?98?99?4?5?)?2?(?98?99?5?)?2?(?99 - × × × × × - - × × × - - × - - -? x?x?x?x?x?x?x?x? c?x?x?x + - × × × × × × × × - - × × × × × × × -? 94?95? )?2?(?95?96?97?98?99? 2?3?4?5?)?2?(?95?96?97?98?99? 2?3?4?5? 2. ò +?4?1?x?x?dx? 解. ò ò ò ò + - = + - = + - = +? 2?2?2?4?4?4?2?4? )?(?1?2?1?1?1?1?1?/?1?1? t?dt?t?tdt?t?t?t?dt?t?t?x?x?x?dx?令? c?x?x?c?u?u?du?u?u?u?t + + - = + + - = - = ò? 2?4?2?2? 1?ln?2?1?|?sec?tan?|?ln?2?1?sec?sec?2?1?tan?令 五.?求下列不定积分:?1. ò? xdx?x? 2?cos? 解. ò ò ò + = + =? x?xd?x?dx?x?x?xdx?x? 2?sin?4?1?4?1?)?2?cos?1?(?2?1?cos? 2?2 ò - + =? xdx?x?x?x? 2?sin?4?1?2?sin?4?1?4?1?2 c?x?x?x?x + + + =? 2?cos?8?1?2?sin?4?1?4?1?2?2. ò? xdx?3?sec? 解. ò ò ò - = =? xdx?x?x?x?x?x?xd?xdx? tan?sec?tan?tan?sec?tan?sec?sec?3?= ò ò - + + = - -? xdx?x?x?x?x?xdx?x?x?x? 3?2? sec?|?tan?sec?|?ln?tan?sec?sec?)?1?(sec?tan?sec? c?x?x?x?x?xdx + + + = ò? |?tan?sec?|?ln?2?1?tan?sec?2?1?sec?3? 3. ò? dx?x?x?2?3?)?(ln? 解. ò ò ò + - = - =? dx?x?x?x?x?x?d?x?dx?x?x? 2?2?3?3?2?3? )?(ln?3?)?(ln?1?1?)?(ln?)?(ln ò + - - =? dx?x?x?x?x?x?x? 2?2?3? ln?6?)?(ln?3?)?(ln ò + - - - =? dx?x?x?x?x?x?x?x? 2?2?3? 6?ln?6?)?(ln?3?)?(ln? c?x?x?x?x?x?x?x + - - - - =? 6?ln?6?)?(ln?3?)?(ln? 2?3?4. ò? dx?x)?cos(ln? 解. ò ò ò - + = + =? dx?x?x?x?x?dx?x?x?x?dx?x? )?cos(ln?)]?sin(ln?)?[cos(ln?)?sin(ln?)?cos(ln?)?cos(ln \? c?x?x?x?dx?x + + = ò? )]?sin(ln?)?[cos(ln?2?)?cos(ln? 六.?求下列不定积分:?1. ò - + +? dx?x?x?x?x? 2?2? 2?)?1?(? )?1?ln(? 解. ò ò - + + = - + +? 2?2?2?2? 2? 1?1?)?1?ln(?2?1?)?1?(? )?1?ln(? x?d?x?x?dx?x?x?x?x? = ò + × - - - + +? dx?x?x?x?x?x? 2?2?2?2? 1?1?1?1?2?1?1?1?)?1?ln(?2?1? t?x?tan = 令? tdt?t?t?x?x?x? 2?2?2? 2? sec?sec?1?tan?1?1?2?1?)?1?(?2 )?1?ln( × × - - - + + ò? =? dt?t?t?x?x?x ò - - - + +? 2?2? 2? sin?2?1?cos?2?1?)?1?(?2 )?1?ln(? = ò - - - + +? t?t?d?x?x?x? 2?2? 2? sin?2?1? sin?2?2?2?1?)?1?(?2 )?1?ln(? =? c?t?t?x?x?x + - + - - + +? sin?2?1? sin?2?1?ln?2?4?1?)?1?(?2 )?1?ln(? 2? 2? =? c?x?x? x?x?x?x?x + - + + + - - + +? 2?1? 2?1?ln?2?4?1?)?1?(?2 )?1?ln(? 2?2?2? 2 2. ò +? dx?x?x?x? 2?1?arctan? 解. ò ò ò + + - + = + = +? dx?x?x?x?x?x?xd?dx?x?x?x? 2?2?2?2?2? 1?1?arctan?1?1?arctan?1?arctan? =? c?x?x?x?x?dx?x?x?x + + + - + = + - + ò? )?1?ln(?arctan?1?1?1?arctan?1? 2?2?2?2? 3. ò? dx?e?e?x?x?2?arctan? 解.? dx?e?e?e?e?e?de?e?dx?e?e? x?x?x?x?x?x?x?x?x ò ò ò + + - = - = - - -? 2?2?2?2?2? 1?2?1?arctan?2?1?arctan?2?1?arctan? dx?e?e?e?e? x?x?x?x ò + + - = - -? 2?2? 1?2?1?arctan?2?1 ò + + - = -? dx?e?e?e?e? x?x?x?x? )?1?(?1?2?1?arctan?2?1? 2?2? c?x?e?e?e?dx?e?e?e?e?e? x?x?x?x?x?x?x?x + + + - = + - + - = - - - ò? )?arctan?arctan?(2?1?)?1?1?(?2?1?arctan?2?1? 2?2?2? 七.?设 ? í ì - + - + = -x?e?x?x?x?x?x?f? )?3?2?(? 3?)?1?ln(?)?(? 2? 2? 0?0 < 3?x?x? ,?求 ò? dx?x?f?)?(? .? 解. ? ? ? í ì - + - + = - ò ò ò? dx?e?x?x? dx?x?x?dx?x?f? x?)?32?(? )?3)?1?ln(?(?)?(? 2? 2 ? ? ? í ì + + + - + - + - - + = -? 1?2? 2?2?2?2? )?1?4?(? 3?)]?1?ln(?[2?1?)?1?ln(?2?1? c?e?x?x? c?x?x?x?x?x? x? 0?0 < 3?x?x?考虑连续性,?所以? c?=－1+?c1,??c1?=?1?+?c ò? dx?x?f?)?( ? ? ? í ì + + + + - + - + - - + = -? c?e?x?x? c?x?x?x?x?x? x? 1?)?1?4?(? 3?)]?1?ln(?[2?1?)?1?ln(?2?1?2? 2?2?2?2? 0?0 < 3?x?x? 八.?设? x?b?x?a?e?f?x? cos?sin?)?(?' + =? ,?(a,?b?为不同时为零的常数),?求?f(x).?解.?令? t?x?e?t?x? ln = = ，? ,? )?cos(ln?)?sin(ln?)?(?'? t?b?t?a?t?f + =? ,?所以 ò + =? dx?x?b?x?a?x?f? )]?cos(ln?)?sin(ln?[?)?(?=? c?x?a?b?x?b?a?x + - + +? )]?cos(ln?)?(?)?sin(ln?)[(?2? 九.?求下列不定积分:?1. ò -? dx?x?x? 2?3?4? 解.?令? t?x? sin?2 = ò ò ò - - = = -? t?td?t?tdt?t?dx?x?x? cos?cos?)?cos?1?(?32?cos?sin?32?4? 2?2?2?3?2?3 =? c?x?x?c?t?t + - - - = + + -? 2?3?2?2?5?2?5?3? )?4?(3?4?)?4?(5?1?cos?5?32?cos?3?32? 2. ò > -? )?0?(?2?2? a?dx?x?a?x?解.?令? t?a?x? sec = ò ò ò + - = = = > -? c?at?t?a?tdt?a?t?t?at?a?t?a?a?dx?x?a?x? tan?tan?tan?sec?sec?tan?)?0?(? 2?2?2?=? c? x?a?a?a?x + - -? arccos?2?2?3.? dx?e?e?e? x?x?x ò - +?2?1?)?1?(? 解. = - + ò? d?e?e?e? x?x?x? 2?1? )?1?( ò -? dx?e?e?x?x?2?1? +? dx?e?e?x?x ò -?2?2?1? = ò -?x?x?e?de?2?1?－? dx?e?e?d?x?x ò - -?2?2?1?)?1?(?2?1? =? c?e?e? x?x + - -? 2?1?arcsin? 4. ò -?dx?x?a?x?x?2? (a>?0)? 解. ò -?dx?x?a?x?x?2? x?u = 令 ò -? du?u?a?u? 2?4?2?2? t?a?u? sin?2 = 令 ò? tdt?a? 4?2?sin?8? = ò ò + - = -? dt?t?t?a?dt?t?a? )?2?cos?2?cos?2?1?(?2?4?)?2?cos?1?(?8? 2?2?2?2? =? c?t?a?t?a?t?a?dt?t?a?t?a?t?a + + - = + + - ò? 4?sin?4?2?sin?2?3?2?4?cos?1?2?2?sin?2?2? 4?2?2?2?2?2?=? c?t?t?t?a?t?t?a?t?a + - + -? )?sin?2?1?(?cos?sin?cos?sin?4?3? 2?2?2?2?=? c?t?t?a?t?t?a?t?a + - -? cos?sin?2?cos?sin?3?3? 3?2?2?2? =? c?a?x?a?a?x?a?x?a?a?x?a?a?x?a?a?x?a + - - - -? 2?2?2?2?2?2?2?2?3?2?arcsin?3? 2?2?2? =? c?x?a?x?x?a?a?x?a + - + -? )?2?(?2?3?2?arcsin?3?2? 十.?求下列不定积分:?1. ò + -? dx?x?x?cos?2?sin?2? 解. ò ò ò + + + + = + -? x?x?d?dx?x?dx?x?x? cos?2? )?cos?2?(?cos?2?1?2?cos?2?sin?2? t?x =?2?tan?令 ò ò + + + = + + + - + +? |?cos?2?|?ln?3?2?2?|?cos?2?|?ln?1?1?2?1?2?2? 2?2?2?2? x?t?dt?x?t?t?t?dt =? c?x?x?c?x?t + + + = + + +? |?cos?2?|?ln?)2?(tan?3?1?arctan?3?4?|?cos?2?|?ln?3?arctan?3?4? 2. ò +? dx?x?x?x?x?cos?sin? cos?sin?解. ò ò + - + = +? dx?x?x?x?x?dx?x?x?x?x? cos?sin? 1?cos?sin?2?1?2?1?cos?sin? cos?sin? = ò ò ò + - + = + - +? dx?x?x?dx?x?x?dx?x?x?x? cos?sin?1?2?1?)?cos?(sin?2?1?cos?sin? 1?cos)?(sin?2?1? 2? = ò + + - -? )4?sin(? )4?(?4?2?)?cos?(sin?2?1 p p?x?x?d?x?x? =? c?x?x?x + + - -? |?)8?2?tan(?|?ln?4?2?)?cos?(sin?2?1 p 第三章 一元函数积分学(定积分)? 一．若?f(x)在[a，b]上连续,?证明:?对于任意选定的连续函数F(x),?均有? 0?)?(?)?( = F ò?b?a dx?x?x?f? ,?则?f(x) o?0.?证明:?假设?f(x)1?0,a?< x<?b,?不妨假设?f(x)?>?0.?因为?f(x)在[a，b]上连续,?所以存在d>?0,?使 得在[x－d, x+ d]上?f(x)?>?0.?令?m?=? )?(?min? x?f?x d x d x + ￡ ￡ -? .?按以下方法定义[a，b]上F(x):?在[x－d, x+ d]上F(x)?=? 2?2? )?( x d - -?x? ,?其它地方F(x)?=?0.?所以?0? 2?)?(?)?(?)?(?)?(? 2 > 3 F = F ò ò + - pd d x d x? m?dx?x?x?f?dx?x?x?f?b?a? .?和? 0?)?(?)?( = F ò?b? a dx?x?x?f? 矛盾.?所以?f(x) o?0.? 二.?设l为任意实数,?证明: ò + =?20? )?(tan?1?1 p l?dx?x?I? =? 4?)?(cot?1?1?20 p p l = + ò? dx?x? .? 证明:?先证:? 4?)?(cos?)?(sin? )?(sin?20 p p = + ò? dx?x?f?x?f? x?f? = ò +?20? )?(cos?)?(sin? )?(cos p? dx?x?f?x?f? x?f?令?t?=? x -? 2 p? ,?所以 = + ò?2 0 )?(cos?)?(sin? )?(sin p? dx?x?f?x?f? x?f ò - +?0?2? )?(?)?(sin?)?(cos? )?(cos p? t?d?t?f?t?f? t?f? = = + ò?20 )?(sin?)?(cos? )?(cos p? dt?t?f?t?f? t?f ò +?20? )?(sin?)?(cos? )?(cos p? dx?x?f?x?f? x?f?于是 = + ò?20 )?(cos?)?(sin? )?(sin?2 p? dx?x?f?x?f? x?f + + ò?20 )?(cos?)?(sin? )?(sin p? dx?x?f?x?f? x?f ò +?20? )?(sin?)?(cos? )?(cos p? dx?x?f?x?f? x?f =? 2?)?(cos?)?(sin? )?(cos?)?(sin? 20?20 p p p = = + + ò ò? dx?dx?x?f?x?f? x?f?x?f? 所以? 4?)?(cos?)?(sin? )?(sin?20 p p = + ò? dx?x?f?x?f? x?f? = ò +?20? )?(cos?)?(sin? )?(cos p? dx?x?f?x?f? x?f? .? 所以 ò + =?20? )?(tan?1?1 p l?dx?x?I? 4?)?(sin?)?(cos? )?(cos?cos?sin?1?1? 20?20 p p l l l p l = + = ÷ ? ? ? è ? + = ò ò? x?x?x?dx?x?x? 同理? 4?)?(cot?1?1?20 p p l = + = ò? dx?x?I? .? 三．已知?f(x)在[0，1]上连续,?对任意?x,?y都有|f(x)－f(y)|?<?M|x－y|,?证明?n?M?n?k?f?n?dx?x?f? n k? 2?1?)?(? 1?1?0 ￡ ÷ ? ? ? è ? - ? ò = 证明: ?ò ò = - =?nk?n?k?n?k? dx?x?f?dx?x?f? 1? 1?1?0? )?(?)?(? , = ? =?nk? n?k?f?n?1? )?(?1? dx?n?k?f?nk?n?k?n?k ?ò = -?1? 1? )?(? n?M?n?M?dx?x?n?k?M? dx?n?k?x?M?dx?n?k?f?x?f? dx?n? k?f?x?f?n?k?f?n?dx?x?f? nk?nk?n?k?n?k? nk?n?k?n?k?nk?n?k?n?k? nk?n?k?n?k?nk? 2?1?2? )?(?)?(?)?(? |?)?(?)?(?|?)?(? 1?)?(? 1? 2?1? 1? 1? 1?1? 1? 1? 1?1?1?0 = = ÷ ? ? ? è ? - = - ￡ - ￡ ú ? ù ê ? é - = - ? ?ò ?ò ?ò ?ò ? ò = = - = - = - = - = 四.?设 ò =?40?tan p? xdx?I? n?n? ,?n为大于?1?的正整数,?证明:? )?1?(?2?1?)?1?(?2?1 - < < +? n?I?n? n? .? 证明:?令?t?=? x?tan?,?则 ò ò + = =? 1?0? 2?40? 1?tan? dt?t?t?xdx?I? n?n?n p 因为? 2?2?2?'?2? )?1?(?1?1? t?t?t?t + - = ÷ ? ? ? è ? +? >?0,?(0?<?t?<?1).?所以? 2?1?1?1?1?1? 2?2 = + < +t?t? 于是 ò ò ò - < + <? 1?0? 1?1?0? 2?1?0? 2?1?1?2?1? dt?t?dt?t?t?dt?t? n?n?n? 立即得到? )?1?(?2?1?2?1?)?1?(?2?1 - < < < +? n?n?I?n? n? .? 五.?设?f(x)在[0,?1]连续,?且单调减少,?f(x)?>?0,?证明:?对于满足?0?< a<b?<?1?的任何 a,b,?有 ò ò > b a a a b? dx?x?f?dx?x?f? )?(?)?(?0?证明:?令 ò ò - =? x? dt?t?f?dt?t?f?x?x?F a a a? )?()?()?(? 0? (x 3 a),? 0?)?(?)?(? 0 > = òa a a? dt?tf?F? . = - = ò? )?(?)?(?)?(?'? 0 x?f?dt?tf?x?F a a ò > - a?0? 0?)]?(?)?(?[? dt?x?f?tf? ,?(这是因为?t ￡ a,?x 3 a,?且?f(x)单减). 所以? 0?)?(?)?( > > a b? F?F? ,?立即得到 ò ò > ba a a b? dx?x?f?dx?x?f? )?(?)?(?0? 六.?设?f(x)在[a,?b]上二阶可导,?且? )?(?'?'?x?f? <?0,?证明: ÷ ? ? ? è ? + - ￡ ò? 2?)?(?)?(? b?a?f?a?b?dx?x?f?b?a证明: "x,?t?[a,?b],? 2?)?(?!?2)?(?'?'?)?)(?(?'?)?(?)?(? t?x?f?t?x?tf?tf?x?f - + - + = x ￡? )?)(?(?'?)?(? t?x?t?f?t?f - + 令? 2?b?a?t + =? ,?所以 ÷ ? ? ? è ? + - ÷ ? ? ? è ? + + ÷ ? ? ? è ? + ￡? 2?2?'?2?)?(? b?a?x?b?a?f?b?a?f?x?f? 二边积分 ò ò ò ÷ ? ? ? è ? + - ÷ ? ? ? è ? + + ÷ ? ? ? è ? + ￡? b?a?b?a?b?a? dx?b?a?x?b?a?f?dx?b?a?f?dx?x?f? 2?2?'?2?)?(? = ÷ ? ? ? è ? + -? 2?)?(? b?a?f?a?b? .? 七.?设?f(x)在[0,?1]上连续,?且单调不增,?证明:?任给a??(0,?1),?有 ò ò 3? 1? 0?0? )?(?)?(? dx?x?f?dx?x?f a a 证明:?方法一:?令 ò ò - =? x?x? dt?t?f?dt?t?f?x?F? 0?0? )?()?(?)?( a a a?(或令 ò ò - =? x? dt?t?f?dt?t?f?x?x?F? 0?0? )?()?()?( a a? )0?)?(?)?(?)?(?' 3 - =? x?f?x?f?x?F a a a? ,?所以?F(x)单增?? 又因为?F(0)?=?0,?所以?F(1) 3?F(0)?=?0.?即?0?)?(?)?(? 1? 0?1?0 3 - ò ò? dt?tf?dt?t?f a a a? ,?即 ò ò 3? 1? 0?0? )?(?)?(? dx?x?f?dx?x?f a a 方法二:?由积分中值定理,?存在x?[0, a],?使? )?(?)?(? 0 x a a? f?dx?x?f = ò? ??由积分中值定理,?存在h?[a,?1],?使? )?1?)(?(?)?(?1 a h a - = ò? f?dx?x?f?因为? )?(?)?(?, x h x h? f?f ￡ 3 所以? .? 所以? )?1?)(?(?)?(?)?(?)?(?)?(? 2?1? 0?1?0 a h a x a a a a a a - + = + = ò ò ò? f?f?dx?x?f?dx?x?f?dx?x?f ò = = - + ￡ a x a a x a x a? 0?2? )?(?)?(?)?1?)(?(?)?(? dx?x?f?f?f?f? 八.设?f(x)在[a,?b]上连续,? )?(?'?x?f?在[a,?b]内存在而且可积,?f(a)?=?f(b)?=?0,?试证: ò ￡? b? a? dx?x?f?x?f? |?)?(?'?|?2?1?|?)?(?|? ,??(a?<?x?<?b)?证明:? |?)?(?'?|?)?(?'?|?)?(?'?|? x?f?x?f?x?f ￡ ￡ -? ,?所以 ò ò ￡ - ￡ -? x?a?x?a? dt?t?f?a?f?x?f?dt?t?f? |?)?('?|?)?(?)?(?|?)?('?|? , 即 ò ò ￡ ￡ -? x?a?x?a? dt?t?f?x?f?dt?t?f? |?)?('?|?)?(?|?)?('?|? ? ò ò ￡ - ￡ -? b? x?b?x? dt?t?f?x?f?b?f?dt?t?f? |?)?('?|?)?(?)(?|?)?('?|?即 ò ò ￡ ￡ -? b? x?b?x? dt?t?f?x?f?dt?t?f? |?)?('?|?)?(?|?)?('?|?所以 ò ò ￡ ￡ -? b? a?b?a? dt?t?f?x?f?dt?t?f? |?)?('?|?)?(?2?|?)?('?|?即 ò ￡? b? a? dx?x?f?x?f? |?)?(?'?|?2?1?|?)?(?|? ,??(a?<?x?<?b)?九.?设?f(x)在[0,?1]上具有二阶连续导数? )?(?'?'?x?f? ,?且? 0?)?(?0?)?1?(?)?0?( 1 = =? x?f?f?f? ，? ,?试证:? 4?)?(?)?(?'?'?1?0 > ò? dx?x?f?x?f?证明:?因为（0，1）上?f(x)1?0,?可设?f(x)>?0? 因为?f(0)?=?f(1)?=?0 $x0 ??(0,1)使?f(x0)?=? 1?0?max￡ ￡x? (f(x))?所以? dx?x?f?x?f ò?1? 0 )?(?)?(?'?'? >? dx?x?f?x?f ò?1?00? )?(?'?'?)?(?1? （1） 在（0，x 0）上用拉格朗日定理? 0?0?(?)?(?'? x?x?f?f? ） = a? )?,?0?(? 0?x ? a 在（x0,?1）上用拉格朗日定理? 0?0?1?)?(?)?(?'? x?x?f?f - - = b? )?1?,?(?0?x ? b 所以? )?(?4?)?1?(? )?(? )?(?'?)?(?'?)?(?'?'?)?(?'?'?)?(?'?'? 0?0?0? 0?1?0? x?f?x?x?x?f? f?f?dx?x?f?dx?x?f?dx?x?f 3 - = - = ￡ 3 ò ò ò ba ba a b （因为? ab?b?a 3 +?2?)?2?(? ） 所以 ò 3? 1?0?0? 4?)?(?'?'?)?(?1? dx?x?f?x?f? 由（1）得 ò >?1? 0? 4?)?(?)?(?'?'? dx?x?f?x?f?十．设?f(x)在[a,?b]上连续,?且?f(x)?>?0,则 ò ò - 3 ú ? ù ê ? é -? b?a?b?a? dx?x?f?a?b?dx?x?f?a?b? )?(?ln?1?)?(?1?ln?证明:?将?lnx?在?x0?用台劳公式展开? )?(?1?ln?)?(?1?)?(?1?ln?ln? 0?0?0?2?0?2?0?0?0?0? x?x?x?x?x?x?x?x?x?x?x - + ￡ - - - + = x （1） 令 ò - =? b?a? dt?t?f?a?b?x? )?(1?0? x?=?f（t） 代入（1）? )?)?(?1?)?(?(? )?(?1?1?)?(?1?ln?)?(?ln ò ò ò - - - + - ￡? b?a?b?a?b?a? dt?t?f?a?b?t?f?dt?t?f?a?b?dt?t?f?a?b?t?f?将上式两边取 ò?b? a，最后一项为?0，得 ò ò - ￡ -? b? a?b?a? dt?t?f?a?b?dt?t?f?a?b? )?(1?ln?)?(ln?1?十一.?设?f(x)在[0,?1]上有一阶连续导数,?且?f(1)－f(0)?=?1,?试证:? 1?)]?(?'?[?1?0? 2 3 ò dx?x?f?证明: = ò?1? 0 2?)]?(?'?[? dx?x?f? 1?)?0?(?)?1(?(?)?1?)?(?'?(?1?)]?(?'?[? 2?2?1?0?1?0?2?1?0? 2 = - = × 3 ò ò ò f?f?dx?x?f?dx?dx?x?f? 十二.?设函数?f(x)在[0,?2]上连续,?且 ò?2?0 )?(?dx?x?f? =?0, ò?2?0 )?(?dx?x?xf? =?a?>?0.?证明: $ x ??[0,?2],?使?|f(x)|3?a.?解.?因为?f(x)在[0,?2]上连续,?所以|f(x)|在[0,?2]上连续,?所以$ x ??[0,?2],?取x使|f(x)|?=?max? |f(x)|?(0 ￡?x ￡?2)使|f(x)|3?|f(x)|.?所以? |?)?(?|?|?1?|?|?)?(?|?|?)?(?||?1?|?|?)?(?)?1(?|? 2? 0?2?0?2?0 x x? f?dx?x?f?dx?x?f?x?dx?x?f?x?a = - ￡ - ￡ - = ò ò ò 第三章 一元函数积分学(广义积分)? 一.?计算下列广义积分:?(1) ò -?2? 0? 3?1?)?1?(? dx?e?e?x?x? (2) ò +￥ + +?0? 2?2? )?4?)(?1?(? 1? dx?x?x? (3) ò ￥ + ￥ - +? 2?3?2?)?1?(? x?dx?(4) ò? 1?0 )?sin(ln?dx?x? (5) ò - - -?1?2? 2? 1?1? dx?x?x? (6)? dx?x?x ò +￥ +?0? 2?3?2?)?1?(?arctan? 解.?(1)? 3?2?2?2? 3?1?0?2?0? 3?1? )?1?(2?3?)?1?(? )?1?(?lim?)?1?( - = - - = - ò ò + ?? e?dx?e?e?d?dx?e?e? x?x?x?x e e?(2)? 12?4?1?1?1?3?1?lim?)?4?)(?1?(? 1? 0? 2?2?0? 2?2 p = ú ? ù ê ? é + - + = + + ò ò +￥ ? ￥ +? b?b? dx?x?x?dx?x?x?(3) ò ￥ + ￥ - +? 2?3?2?)?1?(? x?dx? 因为? 1?)?1?(? 1?lim? 2?3?2?3 = + ￥ ?? x?x?x? ,?所以 ò ￥ + +?0? 2?3?2?)?1?(? x?dx?积分收敛.所以 ò ￥ + ￥ - +? 2?3?2?)?1?(? x?dx? =2? t?x?x?dx? tan?)?1?(?0? 2?3?2 = + ò ￥ + 令? 2?cos?2?sec?sec?2? 20?20? 3?2 = = ò ò p p? tdt?dt?t?t? (4)? 2?1?1?)?cos(ln?)?sin(ln?(2?1?lim?)?sin(ln?lim?)?sin(ln? 0?1?0?1?0 - = - = = + + ? ? ò ò e e e e? x?x?x?x?dx?x?dx?x (5)? 3?tan?sec?tan?sec?1?1? 3?2?1?2? 2 p p p = = - ò ò - -? dt?t?t?t?t?dx?x?x? (6)? 1?2?cos?sec?sec?)?1?(?arctan? 20?20? 3?2?0? 2?3?2 - = = = + ò ò ò ￥ + p p p? tdt?t?dt?t?t?t?dx?x?x